# Another brief word on the significance of Topology

Topology is the area of mathematics which studies topological spaces, that is, spaces where continuity can be defined. But topological spaces can be very weird, and at some point, one may wonder whether spaces with strange topologies serve any purpose, apart from being helpful to understand the theory. Natural spaces are those attached to geometrical problems (that is, about figures that can be drawn and manipulated our space) or physical problems (that is, particles and systems that moving according to some physical laws). In these spaces, we can naturally fix coordinates to locate the figures or the particles (at least in a local region of the space) as happens for the surface of the Earth. This leads to the notion of the manifold as the most important object to study in topology.

Cheers,

Nalin Pithwa.

# Compact sets, perfect sets, connected sets

Reference: Principles of Mathematical Analysis by Walter Rudin

2.31 Definition: By an open cover of a set E in a metric space X we mean a collection $\{ G_{\alpha} \}$ of open subsets of X such that $E \subset \bigcup_{\alpha}G_{\alpha}$.

2.32 Definition A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover.

More explicitly, the requirement is that if $\{ G_{\alpha}\}$ is an open cover of K, then there are finitely many indices $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ such that

$K \subset G_{\alpha_{1}} \bigcup G_{\alpha_{2}} \ldots \bigcup G_{\alpha_{n}}$

The notion of compactness is of great importance in analysis, especially in connection with continuity.

It is clear that every finite set is compact. The existence of a large class of infinite compact sets in $R^{k}$ will follow with Theorem 2.41

We observed earlier (section 2.29) that if $E \subset Y \subset X$ then E may be open relative to Y without being open relative to Y. The property of being open thus depends on the space in which it is embedded. Note that the same is true of the property of being closed.

Compactness, however, behaves better as we shall see now. To formulate the next theorem, let us say, temporarily that K is compact relative to X if the requirements of Definition 2.32 are met.

2.33 Theorem: Suppose $K \subset Y \subset X$. Then K is compact relative to X if and only if K is compact relative to Y.

By the virtue of the theorem we are able, in many situations, to regard compact sets as metric spaces in their own right, without any paying any attention to any embedding space. In particular, although it makes little sense to talk of open spaces or of closed spaces (every metric space X is an open subset of itself, and is a closed subset of itself), it does make sense to talk of compact metric spaces.

Proof:

Suppose K is compact relative to X, and let $\{ V_{\alpha}\}$ be a collection of sets, open relative to Y, such that $K \subset \bigcup_{\alpha}V_{\alpha}$. By theorem 2.30 there are sets $G_{\alpha}$ open relative to X, such that $V_{\alpha} = Y \bigcap G_{\alpha}$ for all x, and since K is compact relative to X, we have

(22)…..$K \subset \subset G_{\alpha_{1}} \bigcup \ldots \bigcup G_{\alpha_{n}}$

for some choice of finitely many indices $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$. Since $K \subset Y$, the above equation 22 implies that

(23) $K \subset V_{\alpha_{1}} \bigcup \ldots \bigcup V_{\alpha_{n}}$

This proves that K is compact relative to Y.

Conversely, suppose K is compact relative to Y and let $\{ G_{\alpha}\}$ be a collection of open subsets of X which covers K, and put $V_{\alpha} = Y \bigcap G_{\alpha}$. Then (23) will hold for some choice of $\alpha_{1}$ of $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$; and since $V_{\alpha} \subset G_{\alpha}$ (23) implies (22).

This completes the proof. QED.

2.34 Theorem Compact subsets of metric spaces are closed.

Proof:

Let K be a compact subset of a metric space X. We shall prove that the complement of K is an open subset of qX.

Suppose $p \in X$, and $p \notin K$. If $q \in K$, let $V_{q}$ and $W_{q}$ be neighbourhoods of p and q, respectively, of radius less than $\frac{1}{2}d(p,q)$. Since K is compact, there are finitely many points $q_{1}, \ldots, q_{n}$ in K such that

$K \subset W_{q_{1}} \bigcup \ldots \bigcup W_{q_{n}}$

If $V = V_{q_{1}} \bigcap \ldots V_{q_{n}}$, then V is a neighbourhood of p which does not intersect W. Hence, $V \subset K^{c}$ so that p is an interior point of $K^{c}$. Thus, we have proved the theorem. QED.

2.35 Theorem Closed subsets of compact sets are compact.

Proof: Suppose $F \subset K \subset X$, F is closed (relative to X), and K is compact. Let $\{ V_{\alpha}\}$ be an open cover of F. If $F^{c}$ is adoined to $\{ V_{\alpha}\}$, we obtain an open cover $\Omega$ of K. Since K is compact, there is a finite subcollection $\Phi$ of $\Omega$ which covers K, and hence, F. If $F^{c}$ is a member of $\Phi$, we may remove it from $\Phi$ and still retain an open cover of F. We have thus proved that a finite subcollection of $\{ V_{\alpha}$ covers F.

PS: Remark: Note the technique of proof here.

Corollary: If F is closed and K is compact, then $F \bigcap K$ is compact.

Proof: Theorem 2.24b and 2.34 show that $F \bigcap K$ is closed since $F \bigcap K \subset K$. Theorem 2.35 shows that $F \bigcap K$ is compact. Note: Theorem 2.24b is as follows: For any collection $\{ F_{\alpha}\}$ of closed sets, $\bigcap_{\alpha}F_{\alpha}$ is closed. Theorem 2.34 just proved above says that compact subsets of metric spaces are closed.

2.36 Theorem If $\{ K_{\alpha}\}$ is a collection of compact subsets of a metric space X such that the intersection of every finite subcollection of $\{ K_{\alpha}\}$ is nonempty, then $\bigcap K_{\alpha}$ is nonempty.

Proof:

PS: This proof of Rudin is not v clear to me. If one of my readers can help, I would be obliged.

2.37 Theorem: If E is an infinite subset of a compact set K, then E has a limit point in K.

2.37 Proof: If no point of K were a limit point of E, (proof by contradiction), then each $q \in K$ would have a neighbourhood which contains at most one point of E (namely, q, if $q \in E$). (This follows just from the definition of limit point). It is clear that no finite subcollection of $V_{q}$ can cover E and the same is true of K, since $E \subset K$. This contradicts the compactness of K. (recall definition of a compact set).

2.38 Theorem If $\{ I_{n} \}$ is a sequence of intervals in $R^{1}$, such that $I_{n} \supset I_{n+1}$ where $n \in \mathcal{N}$, then $\bigcap_{1}^{\infty} I_{n}$ is not empty.

2.38 Proof: If $I_{n} = [a_{n}, b_{n}]$ let E be the set of all $a_{n}$. Then E is nonempty and bounded above by $b_{1}$. Let x be the sup of E. If m and n are positive integers, then

$a_{n} \leq a_{m+n} \leq b_{m+n} \leq b_{m}$

so that $x \leq b_{m}$ for each m. Since it is obvious that $a_{m} \leq x$, we see that $x \in I_{m}$ for m=1,2,3….

QED.

2.39 Theorem: Let k be a positive integer. If $\{ I_{n}\}$ is a sequence of k-cells such that $I_{n} \supset I_{n+1}$ (n=1,2,3,…) then $\bigcap_{1}^{n} I_{n}$ is not empty.

Proof: Let $I_{n}$ consist of all points $x = (x_{1}, x_{2}, \ldots, x_{n})$ such that

$a_{n,j} \leq x_{j} \leq b_{n,j}$ ($1 \leq j \leq k$) (where $n = 1,2,3,\ldots$)

and put $I_{n,j} = [a_{n,j}. b_{n,j}]$

For each j, the sequence $\{ I_{n,j}\}$ satisfies the hypotheses of previous theorem 2.38. Hence there are real numbers $x_{j}^{*}$ with $1 \leq j \leq k$ such that

$a_{i,j} \leq x_{j}^{*} \leq b_{n,j}$ with $1 \leq j \leq k$ and $n=1,2,3,\ldots$

Setting $x^{*} = (x_{1}^{*}, \ldots, x_{k}^{*})$ we see that $x^{*} \in I_{n}$ for $n=1,2,3,\ldots$. Thus the theorem is proved. QED.

2.40 Theorem Every k-cell is compact.

Proof:

Let I be a k-cell, consisting of all points $x = (x_{1}, \ldots, x_{k})$ such that $a_{j} \leq x_{j} \leq b_{j}$ with $1 \leq j \leq k$.

Put $\delta = (\Sigma_{1}^{k}(b_{j}-a_{j})^{2})^{1/2}$

Then $|x-y| \leq \delta$ if $x \in I, y \in I$.

Suppose, to get a contradiction, that there exists an open cover $\{ G_{\alpha} \}$ of I which contains no finite subcover of I. Put $c_{j} = \frac{(a_{j}+b_{j})}{2}$. The intervals $[a_{j}, c_{j}]$ and $[c_{j}, b_{j}]$ then determine $2^{k}$ k-cells $Q_{i}$ whose union is I. At least one of these sets $Q_{i}$, call it $I_{1}$, cannot be covered by any finite subcollection of $\{ G_{\alpha}\}$ (otherwise I could be so covered). We next subdivide $I_{1}$ and continue the process. We obtain a sequence $I_{n}$ with the following properties:

(a) $I \supset I_{1} \supset I_{2} \supset I_{3} \supset \ldots$

(b) $I_{n}$ is not covered by any finite subcollection of $\{ G_{\alpha}\}$

(c) if $x \in I_{n}$ and $y \in I_{n}$, then $|x-y| \leq 2^{-n} \delta$

By (a) and Theorem 2.39 there is a point $x^{*}$ which lies in every $I_{n}$. For some x, $x^{*} \in G_{\alpha}$. Since $G_{\alpha}$ is open, there exists $r >0$ such that $|y-x^{*}| implies that $y \in G_{\alpha}$. If n is so large that $2^{-n} \delta (there is such an n, for otherwise $2^{n} \leq \frac{\delta}{r}$ for all positive integers n, which is absurd since R is archimedean), then $(c)$ implies that $I_{n} \subset G_{\alpha}$ which contradicts (b).

This completes the proof.

QED.

The equivalence of (a) and (b) in the next theorem is known as the Heine Borel Theorem.

2.41 Theorem: If a set E in $R^{k}$ has one of the following three properties then it has the other two:

(a) E is closed and bounded;

(b) E is compact.

(c) Every infinite subset of E has a limit point in E.

Proof:

Let us assume (a) is true.

Then $E \subset I$ for some k-cell I, and (b) follows from Theorems 2.40 and 2.35. Theorem 2.37 shows that (b) implies (c). It remains to be shown that (c) implies (a).

If E is not bounded, then E contains points $x_{n}$ with

$|x_{n}| >n$ where $n=1,2,3,\ldots$

The set S consisting of these points $x_{n}$ is infinite and clearly has no limit point in $R^{k}$ hence has none in E. Thus (c) implies that E is bounded.

If E is not closed, then there is a point $x_{0} \in R^{k}$ which is a limit point of E but not a point of E. For $n=1,2,3,\ldots$ there are points $x_{n} \in E$ of E such that $|x_{n}-x_{0}|<\frac{1}{n}$. Let S be the set of these points $x_{n}$. Then S is infinite (otherwise $|x_{n}-x_{0}|$ would have a constant positive value for infinitely many n). S has $|x_{0}|$ as a limit point, and S has no other limit point in $R^{k}$. For if $y \in R^{k}$, $y \neq 0$, then

$|x_{n}-y| \geq |x_{n}-y|-|x_{n}-x_{0}| \geq |x_{n}-y| - \frac{1}{n} \geq \frac{1}{2}|x_{n}-y|$

for all but finitely many n, this shows that y is not a limit point of S (use the following Theorem 2.20: if p is a limit point of E, then every neighbourhood of p contains infinitely many points of E).

Thus, S has no limit points in E, hence E must be closed if (c) holds.

QED.

Remarks: (b) and (c) are equivalent in any metric space (prove this as an exercise ) but that (a) does not, in general imply, (b) and (c) both. Examples are furnished in Exercise 16 of this chapter and by the space $L^{2}$ which is discussed in Chapter 11.

2.42 Theorem (Weierstrass) : Every bounded infinite subset of $R^{k}$ has a limit point in $R^{k}$.

Proof: Being bounded, the set E in question is a subset of a k-cell $I \subset R^{k}$. By Theorem 2.40, I is compact, and so E has a limit point in I, by Theorem 2.37 (If E is an infinite subset of a compact set K, then E has a limit point in K).

Perfect Sets:

2.41 Theorem: Let P be a nonempty perfect set in $R^{k}$. Then P is uncountable.

Proof:

((Note: Definition of a perfect set: E is perfect if E is closed and if every point of E is a limit point of E. ))

Since P has limit points, P must be infinite. Suppose P is countable and denote the points of P by $x_{1}, x_{2}, x_{3}, \ldots$ We shall construct a sequence $\{ V_{n}\}$ of neighbourhoods, as follows:

Let $V_{1}$ be any neighbourhood of $x_{1}$. If $V_{1}$ consists of all $y \in R^{k}$ such that $|y-x_{1}|, the closure $\overline{V_{1}}$ of $V_{1}$ is the set of all $y \in R^{k}$ such that $|y-x| \leq r$. (note this) (this is true or makes sense : we have used the fact that P is also closed, being perfect so the limit point of P can belong to E itself which is the case when $y=x_{1}$).

Suppose $V_{n}$ has been constructed so that $V_{n} \bigcap P$ is not empty. Since every point of P is a limit point of P (note that here we have used the other part of the definition a perfect set), there is a neighbourhood $V_{n+1}$ such that (i) $\overline{V_{n+1}} \subset V_{n}$ (ii) $x_{n} \notin \overline{V_{n+1}}$ (iii) $V_{n+1} \bigcap P$ is not empty. By (iii) $V_{n+1}$ satisfies our induction hypothesis, and the construction can proceed.

Put $K_{n}=\overline{V_{n}} \bigcap P$. Since $\overline{V_{n}}$ is closed and bounded. $\overline{V_{n}}$ is compact. And, by definition of $K_{n}$, it is an infinite set. Since $V_{n}$ is closed and bounded, $\overline{V_{n}}$ is compact. Since $x_{n} \notin K_{n+1}$ no point of P lies in $\bigcap_{1}^{\infty}$. But each $K_{n}$ is nonempty, by (iii) and $K{n} \supset K_{n+1}$ by (i), this contradicts the Corollary to theorem 2.36 (If $\{ K_{n}\}$ is a sequence of nonempty compact sets such that $K_{n} \supset K_{n+1}$ (where $n \in \mathcal{N}$) then $\bigcap_{1}^{\infty}K_{n}$ is not empty. )

Corollary : Every interval $[a,b]$ where $a < b$ is uncountable. In particular, the set of all real numbers is uncountable.

2.44 The Cantor Set: The set which we are now going to construct shows that there exist perfect sets in $R^{1}$ which contain no segment.

Let $E_{0}$ be the interval $[0,1]$.. Remove the segment $(\frac{1}{3}, \frac{2}{3})$ and and let $E_{1}$ be the union of the intervals

$[0,\frac{1}{3}]$ and $[\frac{2}{3}, \frac{1}{1}]$

Remove the middle thirds of these intervals, and let $E_{2}$ be the union of the intervals

$[0,\frac{1}{9}]$ and $[\frac{2}{9}, \frac{3}{9}]$ and $[\frac{6}{9}, \frac{7}{9}]$ and $[\frac{8}{9}, 1]$

Continuing in this way, we obtain a sequence of compact sets $E_{n}$ such that

(a) $E_{1} \supset E_{2} \supset E_{3} \ldots$

(b) $E_{n}$ is the union of $2^{n}$ intervals each of length $3^{-n}$

The set

$P = \bigcap_{n=1}^{\infty}E_{n}$

is called the Cantor set. P is clearly compact, and Theorem 2.36 shows that P is not emtpy. Also, we have shown that P is closed being compact. Now, we have to prove that every point of P is a limit point of P: we can also show that P contains no isolated points: let us therefore look at the construction of P:

No segment of the form

(24) $(\frac{3k+1}{3^{n}}, \frac{3l+2}{3^{n}})$

where k and m are positive integers, has a point in common with P. Since every segment $(\alpha, \beta)$ contains a segment of the form (24) if $3^{-m} < \frac{\beta - \alpha}{6}$, P contains no segment.

To show that P is perfect, it is enough to show that P contains no isolated point. Let $x \in P$ and let S be any segment containing x. Let $I_{n}$ be that interval of $E_{n}$ which contains x. Choose n large enough so that $I_{n} \subset S$. Let $x_{n}$ be an endpoint of $I_{n}$ such that $x_{n} \neq x$.

It follows from the construction of P that $x_{n} \in P$. Hence, x is a limit point of P and P is perfect.

QED.

One of the most interesting properties of the Cantor set is that it provides us with an example of an uncountable set of measure zero. (the concept of measure will be discussed in Chapter 11).

CONNECTED SETS:

2.45 Definition: Two subsets A and B of a metric space X are said to be separated if both $A \bigcap \overline{B}$ and $\overline{A} and \bigcap B$ are both empty; that is, if no point of A lies in the closure of B and no point of B lies in the closure of A.

A set $E \subset X$ is said to be connected if E is not a union of two nonempty separated sets.

2.46 Remark: Separated sets are of course disjoint, but disjoint sets need not be separated. For example, the interval $[0,1]$ and the segment $(1,2)$ are not separated since 1 is a limit point of $(1,2)$. However, the segments $(0,1)$ and $(1,2)$ are separated.

The connected subsets of the line have a particularly simpe structure.

2.47 Theorem: A subset E of the real line $R^{1}$ is connected if and only if it has the following property: if $x \in E$, $y \in E$ and $x, then $z \in E$.

Proof:

If there exist $x \in E$, $y \in E$, and soe $z \in (x,y)$ such that $z \notin E$, then $E = A_{1} \bigcup B$, where

$A_{z} = E \bigcap (-\infty, z)$ and $B_{z} = E \bigcap (z,\infty)$

Since $A_{z} \subset (-\infty, z)$ and $B_{z} \subset (z, \infty)$, they are separated. Hence E is not connected.

To prove the converse (note we prove the contrapositive of the converse) suppose E is not connected. So is separated. Then there are nonempty separated sets A and B such that $A \bigcup B = E$. Choose $x \in A$ and $y \in B$ and assume WLOG that $x < y$. Define

$z = \sup {(A \bigcap [x,y])}$

By theorem 2.28 (Let E be a nonempty set of real numbers which is bounded above Let $y=\sup {E}$. Then $y \in E$ if E is closed), $z \in \overline{A}$ hence $z \notin B$. In particular $x \leq z .

If $z \notin A$, it follows that $x < z and $z \notin E$.

If $z \notin A$, then $z \notin \overline{B}$, hence there exists $z_{1}$ such that $z < z_{1} and $z_{1} \notin B$. Then $x and $z_{1} \notin E$.

QED.

Cheers,

Nalin Pithwa

# Topology (Hocking and Young) two problems and their solutions

Reference: Chapter 1, Topology by Hocking and Young. Dover Publications, Inc. Available in Amazon India also.

Question 1:

Show that the mapping f used in the proof of the Theorem 1-18 is continuous.

Proof 1:

The mapping referred to f is the following:

Consider $f: E^{n+1} \rightarrow E^{n+1}$ carrying each point $(x_{1}, x_{2}, \ldots, x_{n+1}) \in E^{n+1}-0$ to $(\frac{x_{1}}{|x|^{2}}, \frac{x_{2}}{|x|^{2}}, \ldots, \frac{x_{n+1}}{|x|^{2}})$

We have to prove that this map f is continuous. Clearly, the $\epsilon - \delta$ definition of continuity will work here.

Consider any point $p \in E^{n+1}-0$ where $p = (\frac{x_{1}}{|x|^{2}}, \frac{x_{2}}{|x|^{2}}, \ldots, \frac{x_{n+1}}{|x|^{2}})$ where clearly $p \neq 0$

Let the n-tuple $x \in E^{n+1}$ and $x \neq 0$

Consider an $\epsilon$ ball, viz, $d(p,x) < \epsilon$ That is, we are considering a neighbourhood about the point x. Let the n-tuple x be given by $(y_{1}, y_{2}, \ldots, y_{n})$

Then, by definition of usual metric, $\sqrt{\frac{x_{1}-y_{1}}{|x|^{2}} + \frac{x_{2}-y_{2}}{|x|^{2}} + \ldots +\frac{x_{n+1}-y_{n+1}}{|x|^{2}}} < \epsilon$ where $|x|$ is the usual norm in $E^{n+1}$

So, by usual algebra, we get

$\sqrt{(x_{1}-y_{1})+(x_{2}-y_{2}) + \ldots + (x_{n+1}-y_{n+1})} < |x| \epsilon$. But this is an open ball of radius $|x|\epsilon$. So the given mapping f maps an open ball to open ball in given domain. Hence, f is open.

Question 2:

Given any closed interval $[a,b]$ in $E^{1}$ find a continuous mapping of $E^{1}$ into $[a,b]$ thereby proving that $[a,b]$ is connected.

Solution/Proof:

We prove the classic case $[a,b]=[0,1]$. Consider the set of all points x in the open unit interval $(0,1)$ is equivalent to the set of all points y on the whole real line. For example, the formula:

$y = \frac{1}{x}\arctan{x} + \frac{1}{2}$
establishes a 1-1 correspondence between these two sets. As this map is continuous, and it maps the real line, which is connected to the above interval, the given interval is also connected.

Cheers,

Nalin Pithwa

# Topological Spaces and Groups: part 3: Fast Review

Reference: Topological Transformation Groups by Deane Montgomery and Leo Zippin, Dover Publications, available Amazon India.

1.21 COSET SPACE of LOCAL GROUPS

DEFINITION. A closed subset H of a local group G is called a subgroup if a neighbourhood V of e exists such that $V^{2}$ is defined, V open and symmetric, and:

i) $x \in H \bigcap V$ implies $x^{-1} \in H \bigcap V$

ii) $x, y \in H \bigcap V$ implies $xy \in H$.

The subgroup is called invariant if a V exists satisfying (i) and (ii) and also

(iii) $y \in V, h \in H \bigcap V$ implies $y^{-1}ky \in H$

LEMMA.

Let G be a local group, H a subgroup, V a neighbourhood as described above, and let W be a symmetric open neighbourhood of a e with $W^{16} \subset V$.. Then for $x, y \in W$ and $x^{-1}y \in H$ is an equivalence relation and $x^{-1}y \in H$ if and only if $xH \bigcap W = yH \bigcap W$.

Proof:

The relation $x^{-1}y \in H$ is reflexive since $x^{-1}x = e \in H$. It is symmetric since if $x, y \in W$ then $x^{-1}y \in W^{2}$ hence $y^{-1}x$ is defined and belongs to $W^{2}$. Now $x^{-1}y \in H$ implies $y^{-1}x \in H$. The relation is transitive because $x, y, z \in W$ and $x^{-1}y , y^{-1}z \in H$ implies $x^{-1}z \in H$ by condition (2).

It can be seen that the sets $xH \bigcap W$ where $x \in W$ are the equivalence classes of this equivalence relation. They are called the local cosets and are the points of a coset space $W/(H \bigcap W)$. We shall now define the topology for this space.

Let T denote the natural map of W onto $W/(H \bigcap W)$ defined by

$T : x \rightarrow (xH \bigcap W)$

Note that T (for local groups) is defined only on W. Let the open sets of the coset space be those which have an open set in W as inverse under T. Then T is open and continuous.

COROLLARY.

With the same assumptioins and notation as in the preceding Lemma, let H be invariant. Then $W/(H \bigcap W)$ is a local group when the product is defined in the natural way.

The product function is defined for $x, y \in W$ by:

$(xH \bigcap W) (yH \bigcap W) = (xyH \bigcap W)$

To prove that this is a single valued function (or a well-defined function) on cosets it needs to be shown that for $h, h^{'} \in H$ and $xh, yh^{'} \in W$ the product $(xy)^{-1}(xhyh^{'})$ exists and is in H. The product is in $W^{4}$ and defines an element of V, This element can be seen to be in $H \bigcap V$ by the use of the associative law and condition (3) above; it must be verified that all indicated products are well-defined. Finally let $U_{0}$ be an open neighbourhood of e in G, with $U_{0}^{2} \subset W$. It can be seen that the product of each pair of elements in the open set $T(U_{0}H \bigcap W)$ is defined as an element of $W/(H \bigcap W)$. The remaining details can be verified by the reader. ( I have yet to ! :-))

THEOREM.

A local group G possesses such a family of neighbourhoods as is defined in Section 1.17 (previous blog). For every subgroup H of G and neighbourhood W of e (this neighbourhood chosen as above) the coset space $W/(W \bigcap H)$ is a Hausdorff space.

The proof is the same as in 1.16 and 1.17. (previous blog)

Remarks on the side: Some of the principal results of the succeeding chapters are valid for local groups as well as global groups. However, the consideration of local groups in each preliminary Lemma and Theorem is not feasible in a work of this kind. In the sequel we shall only occasionally need to make explicit mention of the local groups.

1.22 INVARIANT METRIC

The construction of metrics and invariant metrics in groups was carried out by Garrett Birkhoff and Kakutani independently.

A local group is called metric if some neighbourhood of the identity is metric.

THEOREM.

Let G be a topological group whose open sets at e have a countable basis. Then G is metrizable (Section 1.9) and moreover, there exists a metric which is right invariant.

(From Section 1.9 we reproduce the definition of metrizable: A space is called metrizable if a metric can be defined for it which induces in it the original topology.)

Let $U_{i}$ for $i=1,2, \ldots$ be a countable basis for open sets at e and for each positive integer n let $O_{n}=\bigcap_{1}^{n}U_{i}$. The sets $O_{n}$ are monotonic decreasing and form a basis at e. Let $V_{r}$ for each dyadic rational $r=k/2^{n}$, where $1 \leq k \leq 2^{n}$, where $n \in \mathcal{N}$ be a family of neighbourhoods of e in G such as is constructed in Section 1.17 (previous blog).

Define a function

$f(x,y)$ on G as follows:

(i) $f(x,y) =0$ if and only if $y \in V_{r}V_{r}^{-1}x$ for every r.

(ii) otherwise $f(x,y) =lub(r)$ where $y \in V_{r}V_{r}^{-1}x$

For any set U and element a in G, $y \in Ux$ if and only if $ya \in Uxa$. From this it follows that $f(x,y)$ is right invariant:

(iii) $f(x,y) = f(xa,ya)$

Next from the fact that $V_{r}V_{r}^{-1}$ is symmetric it follows that $xy^{-1} \in V_{r}V_{r}^{-1}$ if and only if $yx^{-1} \in V_{r}V_{r}^{-1}$ for the same r. It follows that f is symmetric

$f(x,y) = f(y,x)$

The sets $V_{1/2^{n}}$ are symmetric so that $V_{1/2^{n}}V_{1/2^{n}}^{-1} = V_{1/2^{n}}^{2} \subset V_{1/2^{n-1}}$ by (5) of section 1.17. Also $V_{1/2^{n-1}} \subset O_{n-1}$ and $\bigcap O_{n-1}=e$. It follows now that

$f(x,y)=0$ if and only if $x=y$

We next define the distance function:

*) $d(x,y) = lub_{u}|f(x,u)-f(y,u)|$

The definition shows that $d(x,y) = d(y,x)$

$d(x,y) \geq f(x,y) \geq 0$ and $d(x,x)=0$

If $d(x,y)=0$ then $f(x,y)=0$ and $x=y$. Finally the triangle inequality:

$d(x,z) =lub_{u}|f(x,u) -f (y,u) + f(y,u)-f(z,u) | \leq d(x,y) + d(y,z)$

The right invariance of the metric is shown as follows:

$d(xa, ya ) = lub_{u}|f(xa,u) - f(ya,u)| = lub_{va}|f(xa,va) - f(ya,va)| = lub_{v}|f(x,v)-f(y,v)|=d(x,y)$

Finally we must show the equivalence of the original neighbourhoods of G and the sphere neighbourhoods of the metric. It is sufficient to verify this at a e because of the invariance of the metric on the one hand and the translation properties of G (Section 1.13 previous blog) on the other. Let $S_{1/2^{n}}$ denote the set of elements whose distance from a is less than $1/2^{n}$. The fact that

$V_{1/2^{n+1}} \subset S_{1/2^{n}}$

shows that the metric spheres are neighbourhoods of e in the original topology. It remains to show that they form a basis at e. Let a neighbourhood O of e be given. Since the sets $O_{n}$ are a basis at e there is an integer k such that $O_{k} \subset O$. Then if $x \in S_{1/2^{k+1}}$, and $d(e,x) < 1/2^{k+1}$ and $f(e,x) < 1/2^{k+1}$, and finally

$x \in V_{1/2^{k+1}}^{2} \subset V_{1/2^{k}}\subset O_{k}$ hence

$S_{1/2^{k+1}}\subset O_{k}$

This completes the proof.

1.23 METRIC HOMOGENEOUS SPACE

THEOREM.

Let H be a closed subgroup of a metric group G. Then G/H is metrizable.

Proof.

Let d(x,y) be a right invariant metric in G. Define $D(xH, yH)$ as follows:

$D(xH, yH) = glb \hspace{0.1in} d(xa, yb)$ where $a, b \in H$

then for every $\epsilon >0$, H contains elements a, b, c, d such that

$D(xH, yH) + D(yH, zH) +2\epsilon \geq d(xa,yb) + d(yc, zd) = d(xa,yb) + d(yb,zdc^{-1}b) \geq d(xa,zdc^{-1}b) \geq D(xH,zH)$

This shows that D satisfies the triangle inequality. It can be verified that D is single-valued in G/H x G/H, that is , D is symmetric and that $D(xH,yH) =0$ if and only if $xH=yH$. Then D is a metric.

The set yH belongs to the set $S_{1/2^{n}}(e).xH$ if $D(xH, yH) < 1/2^{n}$. Therefore any neighbourhood of xH in the original topology includes a metric neighbourhood. The converse can also be shown. Therefore the metric D induces the topology of G/H.

1.24 CONNECTEDNESS

A subset M of a topological space S is called connected if whenever $M = A \bigcup B$ where A and B are open relative to M and are not empty, it follows that $A \bigcap B$ is not empty.

LEMMA.

If M is connected and A is a subset which is both open and closed relative to M, then $A = M$, or A is the empty set.

This follows from the definition.

LEMMA.

If G is a topological group and H is a subgroup which is open, then H is also closed.

Since H is open each coset of H is open. The compliment of H is the union of cosets of H. Hence, the complement of H is open so H must be closed.

THEOREM.

If G is a connected group and W is an open neighbourhood of e, then $G = \bigcup_{n}W^{n}$.

The set $H = \bigcup_{n}(W \bigcap W^{-1})^{n}$ is an open subgroup of G. Therefore H is a closed subgroup, and then it follows from the connectedness of G that $H=G$.

A union of an arbitrary number of connected subsets is connected, provided every two of the sets have a point in common. It follows that to each point $x \in S$ there is uniquely associated a maximal connected subset of M containing x; these maximal connected subsets are called components of S. The closure of a component is connected — therefore the component is closed. A space is called totally-disconnected when each point if a component.

LEMMA.

If M is connected and f is a continuous map of M into a space N then $f(M)$ is connected.

If $f(M) = A \bigcup B$ and $A \bigcap B=\Phi$

Then $M = f^{-1}(A) \bigcup f^{-1}(B)$ and $f^{-1}(A) \bigcap f^{-1}(B) = \Phi$

COROLLARY.

If G is a connected group and H is a closed subgroup, then G/H is a connected space.

DEFINITION.

A space is called normal if any two disjoint closed sets are contained in disjoined open sets; two sets are disjoined (disjunct, mutually separated) if they have no common points.

LEMMA.

A compact Hausdorff space is normal.

The proof is straight forward and is omitted.

LEMMA.

Let S be a compact Hausdorff space and x a point in S. Let Q be a set of indices $\{ q\}$. If $\{ A_{q}\}$ is the set of all compact open subsets containing x, then $C = \bigcap_{a}A_{a}$ is the x-component of S.

Let

$\bigcap_{a}A_{a} = X \bigcup Y$ where $X \bigcap Y = \Phi$

and X and Y are closed. There exist open sets $V \supset X$ and $W \supset Y$ such that $V \bigcap W = \Phi$. Hence,

*) $(S - (V \bigcup W)) \bigcap (\bigcap _{q}A_{q}) = \Phi$

and there is a finite set of indices $Q^{'} \subset Q$ such that *) continues to hold if the intersection $\bigcap_{a}$ os restricted to $q \in Q^{'}$. Let $A = \bigcap_{q^{'}}A_{q^{'}}$ where $q^{'} \in Q^{'}$. Then, A is compact and open and $x \in A$.

Finally, since

$A = (A \bigcap V) \bigcup (A \bigcap W)$

it follows that $A \bigcap V$ and $A \bigcap W$ are compact open sets only one of which can contain x. Since the system $\{ A_{q}\}$ where $q \in Q$ is maximal, one of $A \bigcap V$ and $A \bigcap W$ is in $\{ A_{q}\}$, say $A \bigcap V \in \{ A_{q}\}$ and therefore $\bigcap_{a}A_{a} \subset A \bigcap V \subset V$; hence, Y is empty.

This shows that $C = \bigcap_{q}A_{q}$ is connected. It is now clear that Cis the component containing x.

COROLLARY.

Let S be compact Hausdorff space, C be a component of S, let F be a closed set and suppose that $F \bigcap C = \Phi$. Then there is a compact open set $A^{'}$, $C \subset A^{'}$, $F \bigcap A^{'} = \Phi$.

It follows from the hypothesis that there is a finite set $Q^{''} \subset Q$ such that

$F \bigcap (\bigcap_{q^{''}}A_{q^{''}}) = \Phi$ where $q^{''} \in Q^{''}$

Let

$A^{'} = \bigcap_{a^{''}}A_{q^{''}}$.

This is the set required by the corollary.

1.25 THE IDENTITY COMPONENT $G_{0}$

THEOREM.

Let g be a topological group and $G_{0}$ the component of G containing the identity. Then $G_{0}$ is closed invariant subgroup and the factor group $G/G_{0}$ is totally disconnected.

If M is a connected subset of G, $M^{-1}$ is connected, Mx and xM where $x \in G$ are connected (section 1.13). It follows that the identity component $G_{0}$ is a group and that the cosets $xG_{0}$ are also components. We have already remarked that components are closed sests. Then $G_{0}$ is a closed subgroup. Since $x^{-1}G_{0}x$ is connected and contains the identity e for every $x \in G$, it is clear that $G_{0}$ is invariant.

Now let M denote a connected subset of $G/G_{0}$. Let T be the natural map of G onto $G/G_{0}$. We shall show that $T^{-1}(M)$ is connected, thus proving that it is in a coset of $G/G_{0}$. Suppose that

$T^{-1}(M) = A \bigcup B$

where A and B are relatively open (and closed ) in $T^{-1}(M)$ and $A \bigcap B$ is empty. Then $T(T^{-1}(M)) = M = T(A) \bigcup T(B)$. Since there exists an open set, say U, in G, such that $A = U \bigcap T^{-1}(M)$ and since T is an open set, say U, in G, such that $A = U \bigcap T^{-1}(M)$ and since T is an open map, the set $T(A)$ is open relatively to M. This is true also of $T(B)$. Since M is connected it follows that $T(A) \bigcap T(B)$ is not empty. Let $xG_{0} \in T(A) \bigcap T(B)$. Since $xG_{0}$ is a connected subset of G,

$xG_{0} = (xG_{0} \bigcap A) \bigcup (xG_{0} \bigcap B$

leads to a contradiction.

A subgroup of a group G is called central if each of its elements communtes with every element in the whole group. Subgroups H and $g^{-1}Hg$ are called conjugate.

THEOREM.

If G is a connected group and H is an invariant totally disconnected subgroup then H is central.

Let x be an element of H and consider the map of G into H depending on x:

$g \rightarrow gxg^{-1}$ where $g \in G$.

Since H contains no connected set with more than one point, the image of G which contains x, must coincide with x. This completes the proof.

1.26 TRANSFORMATION GROUPS

We shall soon confine our attention to groups which are locally compact, and we shall be particularly interested in the transformation groups of locally-euclidean spaces. For the time being, we continue to study the more general phenomenon.

DEFINITION I.

Let M denote a Hausdorff space and G a topological group each element of which is a homeomorphism of M onto itself:

(1) $f(g; x) = g(x) = x^{'} \in M$; $g \in G$, and $x \in M$

The pair $(G,M)$, or sometimes G itself will be called a topological transformation group if for every pair of elements of G, and every x of M,

(2) $g_{1}(g_{2}(x)) = (g_{1}g_{2})(x)$ ;

and if $x^{'} = g(x) = f(g; x)$ is continuous simultaneously in $x \in M$ and $g \in G$.

From (2) and from the fact that each g is one-one on M, it follows that for every $x \in M$

(3) $e(x)=x$; e the identity in G.

If e is the only element in G which leaves all of M fixed, that is if e is the only element satisfying (3) for all x, then G is called effective.

DEFINITION I’

A pair (G,M) will be called a local transformation group if all conditions of the preceding Definition are fulfilled excepting only that G is assumed to be a local group, and condition (2) holds whenever $g_{1}g_{2}$ is defined.

Some of the remarks below apply both to local and global case for the most part, except when noted.

Let G be a transformation group, $g \in G$, $x, y \in M$ and suppose that

$g(x) = y \neq x$

Since M is a Hausdorff space, there is a neighbourhood Y of y not containing x. By the definition of the transformation group, there is a neighbourhood U of g such that for $g^{'} \in U$, $g^{'}(x) \in Y$. Therefore, $g^{'}(x) \neq x$, and it follows that the set of elements of G which leave x fixed is closed. Therefore this set is a closed subgroup of G. We shall denote it by $G_{x}$. Similarly, the set of elements of G leaving fixed every point of M is a closed subgroup of $G_{M}$.

THEOREM

Let $(G,M)$ be a transformation group, and let K be the closed subgroup of G leaving all of M fixed. Then K is invariant and G/K is an effective transformation group of M under the action:

$T^{*}: (gK)(x) = g(x)$ where $g \in G$.

For $x \in M$, $h \in K$, $g \in G$ we must always have: $(g^{-1}hg)(x)=x$. This shows the invariance of K. If $(gK)x=y$ and a neighbourhood Y of y is given, there is a neighbourhood U of g and X of x such that $g^{'} \in X$, $x^{'} \in X$ imply $g^{'}(x^{'}) \in Y$. But then $(g^{'}(K)(x^{'}) \in Y$. From this it is clear that $T^{*}$ is continuous simultaneously in gK and x. It can be seen that

$g_{1}K(g_{2}K(x)) = g_{1}g_{2}K(x)$

It follows also that $gK(x)=x$ for every x implies $g \in K$, equivalently $gK=K$.

1.28.1

If G is a transformation group of M and h is a homeomorphism of M onto itself, then the homeomorphisms:

$\{ hgh^{-1}\}$

of M onto itself form a transformation group which is said to be topologically equivalent to G.

1.26.2

A topological group G can be regarded as a transformation group on itself as space in several ways, in particular by associating with $a \in G$.

1. $a(g) = ag$ (left translation)
2. $a(g) = ga^{-1}$ (right translation)
3. $a(g) = aga^{-1}$ (conjugation, taking of transforms.
Also G is a transformation group of a left coset space G/H where H is a closed subgroup, by
4. $a(gH) = agH$ In cases (1) and (2) G is effective. In case (4) if $agH = gH$ for every $g \in G$ then $a \in gHg^{-1}$ for every $g \in G$. Then $K = \bigcap gHg^{-1}$ is an invariant subgroup of G, and G/K is effective on G/H. Of course, K is a subgroup of H which depends on H as well as G and which may be trivial.

1.26.3

Further examples of transformation groups are given below, proofs are omitted. (PS: I will supply the proofs in a later blog, most probably after this blog):

1. Let $G = GL(n,R)$ be the real n x n matrices where $a = (a_{ij})$ with $|a_{ij}| neq 0$. Let $E_{n}$ be the space of n real variables $u_{1}, u_{2}, \ldots, u_{n}$. Then G is a transformation group of $E_{n}$ whose elements are $T_{a}^{*}: a(u) = u^{'}$ where $u_{i}^{'} = \Sigma a_{ij}u_{j}$
2. With G as above, let $S^{n-1}$ in $E_{n}$ be the (n-1)-sphere defined by $\Sigma u _{i}^{2}=1$. Let G act on $S^{n-1}$ as follows: $T_{a}^{''}: a(u)=u^{''}$ where $u_{i}^{''}=u_{i}^{'}/(\Sigma u_{i}^{2})^{1/2}$ and $u_{i}^{'}$ is as above. Here the effective group is G/Kn where Kn consists of scalar matrices: $(h\delta_{ij})$ where $\delta_{ij}$ is the Kronecker delta and h is positive.
3. Let G be the group of two by two real matrices with determinant one and let it act on $E_{2}$ as in (1).
4. Let G be the group of two by two real matrices of determinant one and let G act on itself by inner automorphisms. As a space G is the product of a circle and a plane. One parameter groups fill a neighbourhood of the identity and in the large are closed sets which are either circles or lines. No two of them cross and they are permuted by G.
5. For fixed integers m, n let G be the circle group acting on $E_{4}$ as follows: $x_{1}^{1}=x_{1}\cos{2\pi m t} + x_{2}\sin{2\pi mt}$; $x_{2}^{1} = -x_{1}\sin{2\pi mt} + x_{2}\cos{2\pi mt}$ ; $x_{3}^{1}=x_{3}\cos{2 \pi n t} + x_{4}\sin{2 \pi n t}$; $x_{4}^{1}=-x_{3}\sin{2\pi n t}+x_{4}\cos{2 \pi nt}$. This can also be viewed as a transformation group on the unit sphere $S^{3}$ in $E^{4}$ since it leaves $S^{3}$ invariant. It is known that the simple closed curves swept out by points of $S^{3}$ are linked.
6. A quasi-relation in $E_{3}$ in a fixed cylindrical coordinate system $(z,r, \theta)$ is a group of homeomorphisms depending on a positive continuous function $F(r,z)$, where $0 < r, < \infty$, $-\infty < z < \infty$, F bounded on compact sets in $E_{3}$ and given by (*) $h_{t}: (z,r,\theta) \rightarrow (z,r, \theta + 2\pi F (r,z)t)$ for all real t.

Each point which is not a fixed point moves in a circle about the z-axis. The period of a moving point, that is the least positive t for which it is left fixed by (*) above varies continuously. If h is an arbitrary homeomorphism of $E_{3}$ upon itself, then $\{ h^{-1}h_{1}h\}$ defines a topological quasi-rotation group. As we shall mention later, these groups can be characterized abstractly.

1.26.4

The transformation group G is called transitive on M if for every $x, y \in M$ there is at least one $g \in G$ such that $g(x)=y$. As remarked earlier every topological group G is transitive on G/H. where H is a closed subgroup.

THEOREM.

Let $(G,M)$ be a topological transformation group which is transitive on M. Then the groups of stability $G_{x}$ where $x \in M$ are conjugate and for any one of them $G/G_{x}$ is mapped in a continuous one-one way onto M by the map:

$T_{1}: gG_{x} \rightarrow g(x)$

Let $x, y \in M$ be given with $g(x)=y$. If $g^{'}(x)=x$, $gg^{'}g^{-1}(y)=y$. It follows that $G_{x}$ and $G_{y}$ are conjugate.

Now let x be fixed. If $g^{'}(x)=x$, $gg^{'}(x)=g(x)$ for every $g \in G$ and this shows that the map

$T_{1}: G/G_{x} \rightarrow M$ defined in the theorem is one-one. It maps $G/G_{x}$ onto M because G is transitive.

Let T be the natural map $G \rightarrow G/G_{x}$. Then $T_{1}T$ maps G onto M $T_{1}Tg=g(x)$.

This map is continuous in G by the definition of a transformation group. Let U be open in M. Then $(T_{1}T)^{-1}U$ is open in G and $T(T_{1}T)^{-1}U$ is open in $G/G_{x}$, since F is an open map. This shows that $T_{1}$ is continuous as well as one-one. In the cases of most interest it will turn out that $T_{1}$ is a homeomorphism.

1.27 LOCALLY EUCLIDEAN SPACES

We have used $E_{n}$, where $n \in \mathcal{N}$ to denote euclidean n-space, with real coordinates $x_{1}, x_{2}, \ldots, x_{n}$. It is the topological class of spaces homeomorphic to $E_{n}$ which we have in mind, rather than the space endowed with a standard euclidean metric. By the Brouwer theorem, an open subset of $E_{m}$ and an open subset of $E_{n}$ cannot be homeomorphic if $m \neq n$. This shows that the possibility of the one-one bicontinuous coordinatization $(x_{1}, x_{2}, \ldots, x_{n})$ of $E_{n}$ is a topological property. This number n is a topological invariant and is called the dimension.

The term locally euclidean is used to describe a topological space E of fixed dimension n each point of which has a neighbourhood that is homeomorphic to an open set in $E_{n}$. The simplest examples of such spaces are the open subsets of $E_{n}$. If a locally euclidean space is connected, it is called a MANIFOLD. For example, the spheres of all dimensions, the ordinary torus, the cylinder, etc. are manifolds.

A locally euclidean space can be covered by a certain number (not necessarily finite) of open sets each homeomorphic to an open set in $E_{n}$; let us call such sets each with a fixed homeomorphism, a coordinate neighbourhood. The circle $C_{1}$ can be covered by two (or more) coordinate neighbourhoods, the two dimensional sphere $S^{2}$ by two or more. However, to describe classical euclidean space one uses the entire family of those coordinate systems which are related to each other by orthogonal transformations. Similarly to define affine n-dimensional space, one uses a larger family, namely, all coordinate systems which are affinely related to each other.

We describe a topological manifold one could use in it the family of all coordinate neighbourhoods. However, if for some purpose a restricted class of coordinate neighbourhoods covering the manifold is specified, then one can speak of admissible coordinate systems. In general, where two coordinate neighbourhoods overlap, the coordinate systems will not be found to be in any simple relation. It may happen that the admissible coordinate have been so selected that in every region of overlap of two such systems the two sets of coordinates are related by functions which are differentiable or analytic.

A manifold is said to be a differentiable manifold and to have a differentiable structure of class $C^{r}$ where r is greater than or equal to 1, if there is given a covering family of coordinate neighbourhoods in such a way that where any two of the neighbourhoods overlap the coordinate transformation in both directions is given by n functions with continuous, partial derivatives of order r. A manifold may have essentially different differentiable structures (Milnor). A manifold need not possess a differentiable structure (Kervaire, Smale). The n-sphere (with the possibe exception of n=3) has only a finite number of such structures (see Kervaire-Milnor).

In the same way a manifold is said to be a real analytic manifold and to have a real analytic structure if there is given a covering family of coordinate neighbourhoods in such a way that where any two overlap the coordinate transformation in both directions is given by n functions which are real analytic, that is in some neighbourhood of each point of the overlap they can be expanded in power series.

The definition of a complex analytic manifold and structure is similar to the above. Such a manifold of course has an even number of real dimensions; it is automatically real analytic. However, there are many real analytic manifolds of even dimension which cannot be given a complex analytic structure; thus the existence of a complex analytic structure is a much stronger property than the existence of a differentiable or even real analytic structure.

1.27.1

Suppose that M is an n-dimensional manifold and that x and y are points of M belonging to a set U which is homeomorphic to an open n-sphere. Then it is not difficult to describe a homeomorphism of M onto itself which keeps fixed all points of M not inside of U, and which maps x onto y. Using this and using the connectedness of M, and being given an arbitrary pair of points x and y of M one can find a homeomorphism of M onto itself mapping x on y. The details are not presented here.

1.27.2

Suppose that M and N are manifolds and that there is given a continuous map f of M onto $M^{'}$. The map f is called a covering map and M is said to cover $M^{'}$ if the following conditions are satisfied:

a) for each y in $M^{'}$ there is an open neighbourhood V of y such that $f^{-1}(V)$ is the union of disjoined open sets $U_{x}$ where there is a $U_{x}$ for each $x \in f^{-1}(y)$ and $x \in U_{x}$

b) f is a homeomorphism of $U_{x}$ onto V for each x in $f^{-1}(y)$. For each $y \in M^{'}$, each $x \in f^{-1}(y)$ is called a covering point. It is clear that M and $M^{'}$ are of the same dimension, and it is clear that each point of $f^{-1}(y)$ is an isolated point of $f^{-1}(y)$ (each point is a relatively open subset) so that $f^{-1}(y)$ is a discrete set.

By way of example, let $M^{'}$ be the ordinary torus with momentarily convenient coordinates u, v: $0 \leq u, v <1$ and let M be the ordinary plane with real coordinates x and y. Define the map $f(M) = M^{'}$ by

$(x,y) \rightarrow (u,v)$ if and only if $x \equiv u$ and $y \equiv v {\pmod 1}$ This pair of manifolds can also be regarded as an example of a group M covering a factor-group $M^{'}$. Thus let M now denote the two-dimensional vector space $V_{2}$ and let $x_{1}$ and $y_{1}$ denote two independent vectors in $V_{2}$. Let D be the countable, discrete subgroup of $V_{2}$ consisting of the linear combinations of $x_{1}$ and $y_{1}$ with integral coefficients. Finally, let $M^{'}$ now denote the toral group $V_{2}/D$.

A more general example is the following:

If G is any connected locally euclidean group and H is a discrete subgroup of G then G covers the coset-space G/H under the natural map $G \rightarrow G/H$.

A manifold $M^{'}$ is called simply connected if whenever it is covered by a manifold M, the covering map $f: f(M) = M^{'}$ is a homeomorphism (in that case $f^{-1}(y)$ is single-valued). Euclidean spaces of all dimensions and the sphere-spaces of dimension greater than one are simply connected manifolds; the one-dimensional sphere (circumference of a circle) and more generally the toral spaces of all dimensions are not simply connected.

Cheers,

Nalin Pithwa

# Topological Spaces and Groups: Part 2: Fast Review

Reference: Chapter 1, Topological Transformation Groups by Deane Montgomerry and Leo Zippin

1.10 Sequential Convergence:

The proof of the principal theorem of this section illustrates a standard technique. It will involve choosing an infinite sequence, then an infinite subsequence, then again an infinite subsequence and so on repeating this construction a countably infinite number of times. A special form of this method is called the Cantor diagonalization procedure. To facilitate the working of this technique we shall sometimes use the following notation:

1.10.1

The letter I will denote the sequence of natural numbers 1, 2, 3, …When subsequences of I need to be chosen, they will be labelled in some systematic way: $I_{1}, I_{2}, \ldots$ or $I^{'}, I^{''}, \ldots$ or $I^{*}, I^{**}, \ldots$ and so on. Then given a sequence of elements $x_{n}$, $n \in I$, we can refer to a subsequence as : $x_{n}$, $n \in I_{1}$, or : $x_{n}, n \in I^{*}$ and so on.

DEFINITION:

Let S denote a metric space and let $K_{n}, n \in I$ be a sequence of subsets of S. The sequence $K_{n}$ is said to converge to a set K if for every $\epsilon >0$

1.10.3 $K_{n} \subset S_{\epsilon}(K)$ and $K \subset S_{\epsilon}(K_{n})$

for n sufficiently large (depending only on $\epsilon$)

If $K_{n}$ is given and if a K exists satisfying relation 1.10.3 then $\overline{K}$ also satisfies 1.10.3. If two closed sets $K^{'}$ and $K^{''}$ satisfy 1.10.3 for the same sequence $K_{n}$ then $K^{'}=K^{''}$. In the special case that the sets $K_{n}$ are single points, the set K if it exists is a point and is unique.

1.10.4 THEOREM

Every sequence of non-empty subsets of a compact metric space S has a convergent subsequence.

Proof:

Let $K_{n}, n \in I$ be an arbitrary sequence of subsets of S and let $W_{n}, n\in I$ be a basis for open sets in S. (Theorem 1.9)

Let I be called $I_{0}$ and suppose a sequence $I_{m-1}$ has been defined. Consider $W_{m} \bigcap K_{n}$, $n \in I_{m-1}$ m fixed. Then either $W_{m} \bigcap K_{n}$ is not empty for an infinite subsequence of integers $n \in I_{m-1}$ or on the contrary $W_{m} \bigcap K_{n}$ is empty for almost all $n \in I_{n-1}$. In the first of these cases define $I_{m}$ as the set of indices n in $I_{m-1}$. In the first of these cases define $I_{m}$ as the set of indices n in $I_{m-1}$ such that $W_{m} \bigcap K_{n}$ is not empty for $n \in I_{m}$. Then in all possible cases $I_{m} \subset I_{m-1}$ is uniquely defined. We now consider $I_{m}$ to be defined by induction for all $m \in I$.

We can now specify what subsequence of $K_{n}$ we may take as convergent. Let $I^{*} \subset I$ denote the diagonal sequence of the sequences $I_{m}$, that is $I^{*}$ contains the m-th element of $I_{m}$ for each m. We shall show that $K_{n}, n \in I^{*}$ is convergent. It follows from the definition of I^{*} that for each $m \in I$, if we except at most the first m integers in I^{*},

$W_{m} \bigcap K_{n}, n \in I^{*}$,

is always empty or is never empty depending on m.

We next define the set K to which the sets $K_{n}, n \in I^{*}$ will be shown to converge. Let W denote the union of those sets $W_{m}, m \in I$ for which $W_{m} \bigcap K_{n}, n \in I^{*}$ is almost empty. Let $K = S - W$ Since each $W_{m}$ forming W meets at most a finite number of the sets $K_{n}$ no finite number of these $W_{m}$ can cover S. Therefore W cannot cover S, since S is compact, and hence K is not empty. The set K is closed and therefore compact.

Let $\epsilon >0$ be given. There is a covering of K by sets $W_{k_{1}}, W_{k_{2}}, \ldots, W_{k_{s}}, k_{i} \in I$

each meeting K and each of diameter less than $\epsilon$. None of the sets $W_{k_{i}}$ can belong to W and each must intersect almost all the $K_{n}, n \in I^{*}$. Therefore for sufficiently large n, $n \in I^{*}$

$W_{k_{i}} \bigcap K_{n} \neq \Phi$

It follows that $K \subset S_{\epsilon}(K_{n})$, $n \in I^{*}$ for all n sufficiently large. Finally it can be seen that the closed set $S - S_{\epsilon}(K) \subset W$. It follows that the complement of $S_{\epsilon}(K)$ is covered by sets $W_{j_{1}}, W_{j_{2}}, \ldots, W_{j_{t}}$ each of which is an element in the union defining W. Therefore $W_{j_{k}} \bigcap K_{n}$, $n \in I^{*}$ $k=1,2, \ldots t$

is almost always empty. Therefore for sufficiently large $n \in I^{*}$

$K_{n} \subset S_{\epsilon}(K)$.

This completes the proof of the theorem. QED.

1.10.5

Let X and Y be closed subsets of a compact metric space S. Define Hausdorff metric: $d(X,Y)$ as the greatest lower bound of all $\epsilon$ such that symmetrically $X \subset S_{\epsilon}(Y), Y \subset S_{\epsilon(Y)}$

This is a metric for the collection of closed subsets of S.

THEOREM:

The set F of all closed subsets of a compact metric space S is a compact metric space in the metric defined above.

Proof.

The set F is a metric space in the metric defined above. It $A_{n}$ is in F then $A_{n}$ has a subsequence converging to a set A in the sense of convergence defined above and the set A may be assumed closed. It follows that the subsequence also converges to A in the sense of the metric of F. Hence, every sequence in F has a convergent subsequence.

Let $W_{n}$ where $n \in \mathcal{N}$ be a basis for open sets in S as in the preceding Theorem. For each n and each choice of integers $k_{1}, k_{2}, \ldots, k_{n}$ let $W(k_{1}, k_{2}, \ldots, k_{n})$ denote the union of the sets $W_{k_{1}}, W_{k_{2}}, \ldots, W_{k_{n}}$. Now in F let $W^{*}(k_{1}, \ldots, W_{k_{n}})$ consist of all the compact sets in S which belong to $W(k_{1}, \ldots, k_{n})$ and meet each $W_{k_{i}}$. This gives a countable collection of subsets of F. The proof of the preceding theorem shows that this collection is a basis for open sets in F.

Finally, let $\{ O_{n}\}$ where n=1, 2, …be a countable collection of open sets of F which cover F. We have to find an integer m such that $\bigcup_{1}^{m}O_{i}$ covers F. If no such integer existed we could find a sequence of points $x_{n}$ where $x_{n} \in F - \bigcup_{1}^{n}U_{i}$. This sequence would have to have a susbsequence converging to some point x. Since $x \in O_{m}$ for some m, it follows that infinitely many of the $x_{n}$ belong to $O_{m}$; this contradiction proves the theorem.

QED.

1.10.6

Note that separability implies that every collection of covering sets has a countable covering subcollection. It also implies that there exists a countable set of points which is everywhere dense in the space.

THEOREM.

A metric space S is compact if and only if every infinite sequence of points has a convergent subsequence.

Proof:

Suppose that every infinite subsequence of points of S has a convergent subsequence. We shall prove that S is separable. The last paragraph of the preceding section then shows that S is compact. The converse is shown in 1.10.4

For each positive integer n construct a set $P_{n}$ such that (1) every point of $P_{n}$ is at a distance at least $\frac{1}{n}$ from every other point of $P_{n}$ (2) every point of S not in $P_{n}$ is at a distance less than 1/n from some point of $P_{n}$. It is easy to see that no sequence of points in any one $P_{n}$ can be convergent and it follows that $P_{n}$ is a finite point set. Let $P= \bigcup P_{n}$. Then P is countable and every point of S is a limit point of P. For each rational $r >0$ and each point of P construct the “sphere” with that point as centre and radius r. The set of these spheres is countable and is a basis for open sets. This concludes the proof.

QED.

EXAMPLE.

Let S be a compact metric space let H be the space of real continuous functions defined on S with values in $R_{1}$. Each continuous function $f(x)$ determines a closed subset of $S \times R_{1}$ namely the graph consisting of the pairs $(x, f(x))$, $x \in S$. Hence H is a subset of a compact metric space (see examples 2 and 3 in Sec 1.9).

Example 2 of Section 1.9 reproduced below:

The set F of continuous functions defined on a compact space S with values in a metric space M becomes a metric space by defining for f, g in F $d(f,g) = lub (x \in S) [d_{M}(f(x), g(x))]$ where $d_{M}$ is the metric in M. Recall also the following here in this example: Theorem: Let S be a compact space and $\{ D_{a}\}$ a collection of closed subsets such that $\bigcap_{a}D_{a}$ is empty. Then there is some finite set $D_{a_{1}}, \ldots, D_{a_{n}}$ such that $\bigcap_{i}D_{a_{i}}$ is empty. From this theorem it follows that: A lower semi-continuous (upper semi-continuous) real valued function on a compact space has finite g.l.b. (and L.u.b) and always attains these bounds at some points of space.

Example 3 of Section 1.9 reproduced below:

If S is a compact metric space and $E_{1}$ denotes the real line, then $S \times E_{1}$ is a metrizable locally compact space with a countable basis for open sets.

Remarks: Notice that the metric defined for H in example 2 of 1.9 and the metric which it gets from $S \times H_{1}$ are topologically equivalent. This proves that H itself is a separable metric space.

QED.

1.11 TOPOLOGICAL GROUPS

Topological groups were first considered by Lie, who was concerned with groups defined by analytic relations (to be discussed later). Around 1900-1910 Hilbert and others were interested in more general topological groups. Brouwer showed that the Cantor middle third set can be made into an abelian topological group. Later Schreier and Leja gave a definition in terms of topological spaces whose theory had been developed in the intervening time.

A topological group is a topological space whose points are elements of an abstract group, the operations of the group being continuous in the topology of the space. A detailed definition containing some redundancies is as follows:

DEFINITION:

A topological group G is a space in which for $x, y \in G$ there is a unique product $xy \in G$, and

i) there is a unique identity element e in G such that $xe = ex =x$ for all $x \in G$

ii) to each $x \in G$ there is an inverse $x^{-1} \in G$ such that $xx^{-1} = x^{-1}x=e$

iii) $x(yz) = (xy)z$ for $x, y, z \in G$

iv) the function $x^{-1}$ is continuous on G and $xy$ is continuous on $G \times G$

Familiar examples are the real or complex numbers under addition with the usual topologies for $E_{1}$ and $E_{2}$ respectively, and the complex numbers of absolute value one under multiplication with their usual topology as a subset of $E_{2}$. The space of this last group is homeomorphic to the circumference of a circle.

1.11.1

Of course, properties 1, 2 and 3 define a group in the customary sense and a topological group may be thought of as a set of elements which is both an abstract group and a space, the two concepts being united through 4. Wnen a subset H of G is itself a group we shall call H a subgroup of G, but we shall understand that H is to be given the relative topology. It is easy to see that then H becomes a topological group.

If H is a subgroup of G and if x and y are points of G belonging to the closure of H, then every neighbourhood of the product xy contains points of H. For, let U be a neighbourhood of xy. Then by property 4, there exists neighbourhoods V of x and W of y such that every product of an element of V and an element of W is contained in U. We see from this that xy belongs to the closure of H. Similarly, $x^{-1}$ belongs to the closure. Thus, $\overline{H}$ is a group, and we shall call it a closed subgroup.

1.11.2

If $G_{a}$ where $a \in \{ a\}$ is a collection of topological groups and if G denotes the product space defined in 1.6, (previous blog), then G can be regarded as a group (the product of two elements of G being defined by the product of their components in each factor $G_{a}$). Because each neighbourhood of the PROD $G_{a}$ depends on only a finite number of the factors it is easy to see that G becomes a topological group. We shall call it the topological group of the factors $G_{a}$.

By way of examples, note that the product of an arbitrary number of groups each isomorphic to $C_{1}$: the group of reals modulo one (isomorphic to the complex numbers of modulus one under multiplication) is a compact topological group. The product of an arbitrary number of factors each isomorphic to the group of reals is a topological group which is locally compact if the number of factors is finite.

A finite group with the discrete topology is compact and the topological product of any collection of finite groups is therefore compact.

1.11.3

If x and y are in topological group, $x \neq y$, then as will be seen (section 1.16) we may choose a neighbourhood W of e such that

$y \notin WW^{-1}$

Hence, yW and xW are disjoined, and thus two distinct points of a topological group are in disjoined open sets. This is called the Hausdorff property (see section 1.10); it implies that a point is a closed set.

It is easy to see that if H is an abelian subgroup of a topological group G then $\overline{H}$ is also abelian. Thus, if x, y $\in \overline{H}$ and $xy \neq yx$ there are neighbourhoods $U_{1}$ of xy and $U_{2}$ of yx with $U_{1} \bigcap U_{2} = \Phi$. There exist neighbourhoods V of x and W of y such that for every $v \in V$ and $w \in W$ $vw \in U_{1}$ and $wv \in U_{2}$. However, if v, w $\in H$ then $wv = vw$ and we are led to a contradiction proving that the closure of H is abelian.

QED.

1.11.4

Important examples of topological groups are given below:

EXAMPLE 1.

The sets $M_{n}(R)$ and $M_{n}(C)$ of all $n \times n$ matrices of real and complex elements under addition with the distance of $A = (a_{ij})$, $B = (b_{ij})$ defined by

$d(A, B) = max_{i,j}|a_{ij} - b_{ij}|$

The spaces of these two groups are homeomorphic to $E_{n^{2}}$ and $E_{2n^{2}}$. They are in fact the sets of real or complex vectors with $n^{2}$ co-ordinates, and hence are vector spaces as well as groups. Another example is the set H of continuous real valued functions on a metric space under addition.

EXAMPLE 2.

The sets of non-singular real or complex $n \times n$ matrices $GL(n,R), GL(n,C)$ under multiplication; these are subsets of $M_{n}(R)$ and $M_{n}(C)$ respectively and are given the induced topology. They are open subsets and are therefore locally compact and locally euclidean. (see 1.27 later)

EXAMPLE 3.

Let S be a compact metric space and let G be the group of all homeomorphisms of S onto itself topologized as a subspace of the space of continuous maps of S into itself. (Section 1.9 previous blog, example 2)

1.12 ISOMORPHISM of TOPOLOGICAL GROUPS

The spaces associated with two topological groups may be homeomorphic but the groups essentially different; for example, one abelian and the other not.

EXAMPLE 1.

The matrices $\left |\begin{array}{cc}a & 0 \\ 0 & b \end{array}\right |$ where a and b are real numbers under addition. This is an abelian group with $E_{2}$ as space.

EXAMPLE 2.

The matrices $\left| \begin{array}{cc} e^{a} & b \\ 0 & e^{-a} \end{array} \right |$ where a and b are real under multiplication. This is a non-abelian group with $E_{2}$ as space.

If we give the space in this example (or example 1) the discrete topology we obtain a new topological group with the same algebraic structure.

EXAMPLE 3.

In the additive group of integers, for each pair of integers h and k, $k \neq 0$ let the set $\{ h \pm nk\}$ where n=0, 1,2, …, be called an open set and let the collection of all these sets be taken as a basis for open sets.

EXAMPLE 4.

Introduce a metric into the additive gorup of integers, depending on the prime number p, defined thus:

$d(a,b) = \frac{1}{p^{n}}$

if $a \neq b$ and $p^{n}$ is the highest power of p which is a factor of $a-b$.

EXAMPLE 5.

Let G be the integers under addition with any set called open if it is the complement of a finite set (or is the whole space or the null set). Algebraically G is a group and it is also a space. However, it is not a topological group because addition is now not simultaneously continuous. It is true however that addition is continuous in each variable separately.

For some types of group spaces separate continuity implies simultaneous continuity. It is not known whether this is true for a compact Hausdorff group space.

DEFINITION.

Two topological groups will be called isomorphic if there is a one-one correspondence between their elements which is a group isomorphism (preserves products and inverses) and a space homeomorphism (preserves open sets).

An isomorphic map of G onto G is called an automorphism.

In examples (3) and (4) the abstract group structure of the additve group of integers is embodied in infinitely many non-isomorphic topological groups.

1.13 SET PRODUCTS

If G is a group $A \subset G$ let $A^{-1}$ denote the inverse set $\{ a^{-1}\}$ where $a \in A$. Clearly, $(A^{-1})^{-1}=A$. If $B \subset G$ let AB denote the set $\{ ab\}$ where $a \in A, b \in B$. It is understood that the product set is empty if either factor is empty. We shall write $AA = A^{2}$ and so on. It can be seen that $(AB)C = A (BC)$ and $(AB)^{-1} = B^{-1}A^{-1}$. The set $AA^{-1}$ satisfies

$(AA^{-1})^{-1}=AA^{-1}$

that is, it is symmetric. Similarly, the set intersection $A \bigcap A^{-1}$ is symmetric. The intersection of symmetric sets is symmetric.

A set H in G is called invariant if $gH = Hg$ for every $g \in G$ equivalently if $gHg^{-1}=H$.

THEOREM 1.

Let G be a topological group and let $A \subset G$ be an open set. Then $A^{-1}$ is open.

Proof:

Let $a^{-1}$ be in A. By the continuity of the inverse there exists an open set B containing a such that $b \in B$ implies $b^{-1} \in A$. This means that $B^{-1} \subset A$ and therefore $B \subset A^{-1}$. Thus $A^{-1}$ is a union of open sets and is open.

COROLLARY:

The map $x \rightarrow x^{-1}$ is a homeomorphism.

LEMMA.

Let G be a topological group, A an open subset, b an element. Then Ab and bA are open.

Proof:

Let $a \in A$ and let $c=ab$. Then $a = cb^{-1}$. Because “a” regarded as a product is continuous in c there must exist an open set C containing c such that if $c^{'} \in C$ then $c^{'}b^{-1} \in A$. But then $c^{'} \in Ab$ and it follows that $C \subset Ab$. Therefore Ab is a union of open sets and is open. The proof that bA is open is similar.

QED.

COROLLARY.

For each $a \in G$, the left and right translations : $a \rightarrow ax$ and $x \rightarrow xa$ are homeomorphisms.

THEOREM 2.

Let G be a topological group and let A and B be subsets. If A or B is open then $AB$ is open.

Proof.

Since AB is a union of sets of the form Ab, $b \in B$, it is open if A is open. SimilarlyAB is a union of sets aB, $a \in A$ and is open if B is open.

COROLLARY.

Let A be a closed subset of a topological group. Then Ab and bA are closed.

Proof:

This is true because left and right translations by the constant b are homeomorphisms of G onto G.

Now, a function $f(x)$ taking a group $G_{1}$ into another group $G_{2}$ will be called a homomorphism if

*) $f(x)f(y) = f(xy)$ where $x, y \in G_{1}$

When $G_{1}$ and $G_{2}$ are topological we shall ordinarily require that f be continuous. The most useful case is where f is open as well as continuous. In many situations (see 1.26.4 and 2.13 in later blogs) continuity implies openness but this is not true in general. The set of elements going into e is a subgroup and if f is continuous it is a closed subgroup (a point in a topological group is a closed set). This is the kernel of the homomorphism.

EXAMPLE.

Let $V_{1}$ be the additive group of real numbers and G be a topological group. A continuous homomorphism $h(t)$ of $V_{1}$ into G is called a one-parameter group in G. If h is defined only on an open interval around zero satisyfying the definition of homomorphism so far as it has meaning, then $h(t)$ is called a local one-parameter group in G. If $h(t)$ is a one-parameter group, the image of $V_{1}$ may consist of e alone and then $h(t)$ is a trivial one-parameter group. If this is not the case and if for some $t_{1} \neq 0$ and $h(t_{1})=e$, then the image of $V_{1}$ is homeomorphic to a circumference. In case $h(t)=e$ only for $t =0$ the image of $V_{1}$ is a one-one image of the line which may be homeomorphism of the line or a very complicated imbedding of the line. To illustrate this let G be a torus which we obtain from the plane vector group $V_{2}$ by reducing mod one in both the x and y directions. In $V_{2}$ any line through the origin is a subgroup isomorphic to $V_{1}$ and after reduction the line $y=ax$ is mapped onto the torus G thus giving a one-parameter group in G. If a is rational the image is a simple closed curve but if a is irrational the image is everywhere dense on the torus.

1.14 PRODUCTS of Closed Sets

If A and B are subgroups of a group G, AB is not necessarily a subgroup. However if A is an invariant subgroup (that is, $g^{-1}Ag = A$ where $g \in G$) and B is a subgroup then $AB$ is a subgroup.

The product of closed subsets, even if they are subgroups, need not be closed. As an example let G be the additive group of real numbers, $H_{1}$ the subgroup of integers $\{ \pm n\}$ and $H_{2}$ the subgroup $\{ \pm n\sqrt{2}\}$. The product $H_{1}H_{2}$ is countable and a subgroup but it is not a closed set.

It will be shown later that if A is a compact invariant subgroup and B is a closed subgroup then AB is a closed subgroup (corollary of 2.1 in later blogs). (Remark: I think in I N Herstein’s language of Topics in Algebra, an “invariant subgroup” is a normal subgroup. Kindly correct me if I am mistaken).

1.15 Neighbourhoods of the identity

Let G be a topological group and U an open subset containing the identity e. We showed in 1.13 that $xU$ is open and clearly $x \in xU$. Conversely, if $x \in O$, O is open, then $U = x^{-1}O$ is an open set containing e.

If a collection of open sets $\{ U_{a}\}$ is a basis for open sets at e then every open set of G is a union of open sets of the form $x_{a}U_{a}$, where $x_{a} \in G$, $U_{a} \in \{ U_{a}\}$ and the topology of G is completely determined by the basis at e. In particular the collection $\{ xU_{a}\}$ is a basis for open sets at x so also is $\{ U_{a}x\}$

If U is a neighbourhood of e, $U^{-1}$ is a neighbourhood of e and $U \bigcap U^{-1}$ is a symmetric neighbourhood of e.

THEOREM

Let G be a topological group and U a neighbourhood of e. There exists a symmetric neighbourhood W of e such that $W^{2} \subset U$.

Proof:

Since $e.e = e$ and the product is simultaneously continuous in x and y there must exist neighbourhoods $V_{1}$ and $V_{2}$ of e such that $V_{1}V_{2} \subset U$. Define $W = V_{1} \bigcap V_{1}^{-1} \bigcap V_{2} \bigcap V_{2}^{-1}$. Then $W^{2} \subset U$ and this completes the proof.

QED.

COROLLARY.

Let G be a topological group. If $x \neq e$ there exists a neighbourhood W of e such that $W \bigcap xW$ is empty.

Proof.

Since G is a topological space there is a neighbourhood $V_{1}$ of e not containing x or there is a neighbourhood $V_{2}$ of e not containing x. In either case there is a neighbourhood of e not containing x. Let W be a symmetric neighbourhood of e, $W^{2} \subset U$. If $W \bigcap xW$ were not vacuous there would exist $w, w^{'} \in W$ with $w^{'}=w$. But this gives $x = w^{'}w^{-1} \in W^{2} \subset U$ which is false.

QED.

1.15.1

If G is a topological group then a set S of open neighbourhoods $\{ V\}$ which forms a basis at the identity e has the following properties:

(a) the intersection of all V in S is $\{ e\}$

(b) the intersection of two sets of S contains a third set of S

(c) given U in S there is a V in S such that $VV^{-1} \subset U$

(d) if U is in S and $a \in U$ then there is a V in S such that $Va \subset U$

(e) If U is in S and a is in G there is a V in S such that $aVa^{-1} \subset U$.

Conversely, a system of subsets of an abstract group having these properties may be used to determine a topology in G as will be formulated in the following theorem, the proof of which is contained in main part in remarks already made.

THEOREM.

Let G be an abstract group in which there is given a system S of subsets satisfying (a) to (e) above. If open sets in G are defined as unions of sets of the form Va, $a \in G$ then G becomes topological with S a basis for open sets at e. This is the only topology making G a topological group with S a basis at e.

1.16 COSET Spaces

Let G be a group and H a subgroup. The sets $xH$ and $yH$ where $x, y \in G$ either coincide or are mutually exclusive; and $xH = yH$ if and only if $x^{-1}y \in H$. Each set xH is called a coset of H, more specifically a left coset. Right cosets Hx, Hy will be used infrequently. We use the notation G/H for the set of all left cosets. When G is a (topological space) G/H will be made into a space (see below 1.16.2) but we speak of it as a space in the present case also, although it carries no topology at present.

If H is an invariant subgroup, that is if $x^{-1}Hx=H$ equivalently if $xH=Hx$ for any $x \in G$, then

$xHyH = xyH$ where $x, y \in G$

is a true equation in sets of elements of G. In this case, the coset space becomes a group, the factor group $G/H$.

1.16.1

DEFINITION

By the natural map T of a group G onto the coset space G/H, H being a subgroup of G, we mean the map

$T : x \rightarrow xH$, where $x \in G, xH \in G/H$

For any subset $U \subset G$ we have

$T^{-1}(T(U)) = UH \subset G$

Let G be a topological group, H a subgroup. It is useful to topologize the coset space and to do this so that the natural map T is continuous. It will become clear as we proceed that unless the group H is a closed subgroup of G, it will not be possible in general to have T continuous and G/H a topological space; for this reason only the case where H is closed will be considered.

1.16.2

DEFINITION.

Let G be a topological group and let H be a closed subgroup of G, that is a subgroup which is a closed set. By an open set in G/H we mean a set whose inverse under the natural map I is an open set in G/H.

THEOREM.

With open sets defined as above, G/H is a topological space and the map T is continuous and open. If $xH \neq yH$, there exist neighbourhoods $W_{1}$ and $W_{2}$ of xH and yH respectively such that $W_{1} \bigcap W_{2}$ is empty.

Proof:

Let U denote an arbitrary open set in G. Then U/H is open (1.13) and is the inverse of $T(U)$. Since T(U) is open in G/H, T is an open map.

Let $x, y \in G$ and $xH \neq yH$. Then $x \in yH$ and yH is closed because H is closed. There exists a neighbourhood U of e such that $Ux \bigcap yH$ is empty. Let W be open, $e \in W$ and $W^{*} \subset U$ (Theorem in 1.15 above). Then if \$latex $WxH \bigcap WyH$ is not empty we can find $w, w_{1} \in W$ and $h, h_{1} \in H$ so that $w_{1}xh = wy h_{1}$. But this leads to $w^{-1}w_{1}x = yh_{1}h^{-1}$ and implies that Ux meets yH. Therefore $WxH \bigcap WyH$ is empty. The sets Wx and Wy are open. Therefore $W_{1}=T(Wx)$ and $W_{2} = T(Wy)$ are open in G/H; $xH \in W_{1}$ $yH \in W_{2}$ and $W_{1} \bigcap W_{2}$ is empty. This is the main part of the proof and depends on the fact that H is closed. We have proved more than condition (4) of 1.1 (The T_{0} separation axiom) The remaining three conditions of 1.1 are easy to verify. The fact that T is a continuous map is stated in the definition of open set in G/H. The fact that T is open was proved in the last paragraph.

QED.

COROLLARY.

If G is a topological group, $x, y \in G$ where $x \neq y$, then there exist neighbourhoods $W_{1}$ of x and $W_{2}$ of y such that $W_{1} \bigcap W_{2}$ is empty.

Proof:

To see this it is only necessary to take $H=e$. A space in which every pair of distinct points belong to mutually exclusive open sets is called a Hausdorff space. Therefore it has been shown that a topological group G and a coset space G/H, H closed in G, are Hausdorff spaces.

COROLLARY.

Suppose that G is a topological group and H a closed invariant subgroup. Then with the customary definition of product : $(xH)(yH) = xyH$,, G/H becomes a topological group. The natural map of G onto G/H is a continuous and open homomorphism.

1.17 A FAMILY OF NEIGHBOURHOODS

Suppose that we are given a topological group G and a sequence of neighbourhoods of e: $Q_{0}, Q_{1}, \ldots$ By repeated of the following Theorem 1.15: ( Let G be a topological group and U a neighbourhood of e. There exists a symmetric neighbourhood W of e such that $W^{2} \subset U$) .We can choose a sequence of symmetric open neighbourhoods of e : $U_{0}, U_{1}, \ldots$ with $U_{0}=Q_{0}$ such that

(1) $U_{n+1}^{2} \subset U_{n}\bigcap Q_{n}$ where n=0, 1, ….

In this section we shall show how to imbed the sets $U_{n}$ in a larger family of neighbourhoods possessing a multiplicative property which generalizes (1). We shall use this family in the next section to construct a real and non-constant function which is continuous on G. In 1.22 we shall use a similar family in order to construct a metric in a metrizable group. We remark in passing that in groups which do not satisfy the first countability axiom the set: $\bigcap U_{n}$ may be of considerable interest (to be discussed in a future blog section 2.6); it is a closed group (if $x, y \in \bigcap U_{n}$ then for every n, $xy \in U_{n+1}^{2} \subset U_{n}$).

Now, for each dyadic rational $r = \frac{k}{2^{n}}$, for n=0, 1, …and $k=1, \ldots, 2^{n}$ we define an open neighbourhood $V_{r}$ of e as follows:

(2) $V_{1/2^{n}}=U_{n}$ for all n

and then using (3) and (4) alternately by induction on k.

(3) $V_{2k/2^{n+1}}=V_{k/2^{n}}$

(4) $V_{(2k+1)/2^{n+1}} = V_{1/2^{n+1}}V_{k/2^{n}}$

Each $V_{n}$ depends on the dyadic rational r, and not on the particular representation by $k/2^{n}$. The entire family has the property:

(5) $V_{1/2^{n}}V_{m/2^{n}} \subset V_{m+1/2^{n}}$ where $m+1 \leq 2^{n}$

For $m=2k$, (5) is an immediate consequence of (3) and (4). For $m=2k+1$ the left side of (5) becomes: $V_{1/2^{n}}(V_{1/2^{n}}V_{k/2^{n-1}}) \subset V_{1/2^{n-1}}V_{k/2^{n-1}}$

The right side of (5) becomes $V_{(k+1)/2^{n-1}}$. This sets up an induction on n and since (5) holds for $n=1$, we have proved that (5) is true for all n. It follows from (5) and also more directly that:

(6) $V_{r} \subset V_{r^{'}}$ if $r

1.18 COMPLETE REGULARITY

THEOREM.

Suppose that G is a topological group and that F is a closed subset of G not containing e. Then one can define on G a continuous real function f, $0 \leq f(x) \leq 1$, $f(e) =0$, $f(x)=1$ if $x \in F$.

Proof:

In virtue of the property described in the theorem, G is called a completely regular space at the point e. The theorem is due to Pontrjagin.

Set $Q_{a}=G-F$ and setting $Q_{n}=Q_{0}$ for every n, construct a family of sets $V_{r}$ as in the preceding section.

Define f(x), $x \in G$ as follows:

(1) f(x)=0 if $x \in V_{r}$ for every r

(2) f(x)=1 if $x \in V_{1}$

and in all other cases

(3) $f(x)=lub_{r}$ where $\{ r \leq 1, x \in V_{r}\}$

It is clear that e belongs to every $V_{r}$ and that $F = G - V_{1}$ and does not meet $V_{1}$ so that there remains only to prove that f is continuous; let $\epsilon>0$ be given and let n be a positive integer such that $1/2^{n}<\epsilon$.

Now suppose that $f(x) <1$ at some point $x \in G$. Then there is a pair of integers m and k such that $k >n$ (same n as above), $m < 2^{n}$ and (interpreting $V_{0}$, if it occurs, as the null set),

$x \in V_{m/2^{k}}-V_{m-1/2^{k}}$

Let y be an arbitrary element in the neighbourhood $V_{1/2^{k}}x$. Then,

$y \in V_{1/2^{k}}V_{m/2^{k}} \subset V_{(m+1)/2^{k}}$

By the choice of y, $yx^{-1} \in V_{1/2^{k}}$; therefore $xy^{-1} \in V_{1/2^{k}}$ and $x \in V_{1/2^{k}}y$. It follows from this that $y$ cannot belong to $V_{(m-2)/2^{k}}$ and this shows that

$(m-2)/2^{k} \leq f(y) \leq (m+1)2^{k}$

Concerning x we know that

$(m-1)/2^{2^{k}} \leq f(x) \leq m/2^{k}$ and we may conclude from both inequalities that

$|f(x)-f(y)| \leq 2/2^{k} \leq 1/2^{n}<\epsilon$

Suppose next that $f(y)=1$ and choose $k>n$ as before. Let y be an arbitrary element in $V_{1/2^{k}}x$. Now y cannot belong to $V_{m/2^{k}}$ with $m < 2^{k}-2$ without implying that $f(x)<1$. It follows that

$1-2/2^{k} \leq f(y) \leq 1$

and again we get $|f(x) - f(y)| \leq e$

This concludes the proof and $f(x)$ is continuous on G.

QED.

COROLLARY.

A topological group is completely regular at every point.

1.19 HOMOGENEOUS SPACES

THEOREM.

Let G be a topological group, H a closed subgroup. Each element of G determines a homeomorphism of the coset space G/H onto itself, and G becomes a group of homeomorphisms of this space (topological space); furthermore G is transitive on the space: that is, each point may be carried to any other by an element of G.

Proof

Let $a \in G$. Associate to a the mapping $T_{a}: xH \rightarrow axH$.

This is a one-one transformation of G/H onto itself with $T_{a^{-1}}$ as inverse. These transformations are open and each transformation is a homeomorphism.

Since $T_{b}T_{a}$ is given by

$T_{b}T_{a}(xH) = T_{b}(axH)=baxH = T_{ba}(xH)$.

the association of $a \in G$ and $T_{a}$ makes a group of transformations of G/H. It is clear that xH is carried to yH by $T_{a}$ with $a = yx^{-1}$. This completes the proof.

QED.

A space is called HOMOGENEOUS when a group of homeomorphisms is transitive on it. We have shown that G/H is homogeneous with G being the transitive group of homeomorphisms. This implies that the unit segment $R_{1}$, for example, cannot be the underlying space of a group or even a coset space G/H since an end point of $R_{1}$ cannot go to an interior point to a homeomorphism of $R_{1}$.

It follows from the simultaneous continuity of $ax \in G$ in the factors a and x that the image point $axH$ of $xH$ under $T_{a}$ is continuous simultaneously in the counter point xH and the element $T_{a}$ of G. This makes G an instance of what is called a topological transformation group of a space M which will be defined below.

However, we shall be principally concerned with transformation groups which are locally compact and separable, acting on spaces which are topologically locally euclidean.

1.20 LOCAL GROUPS

An open neighbourhood of the identity of a topological group when it is regarded as a space in the relative topology has some of the properties of a group. There will usually be pairs of elements for which no product element exists in the neighbourhood. A structure of this kind is called a local group and will be defined below. Local groups often arise in a natural way, especially in the case of analytic group (Lie groups of transformations) and they have been intensively studied in that form

DEFINITION.

A space G is called a local group if a product xy is defined as an element in G for some pairs x, y in G and the following conditions are satisfied:

i) there is a unique element e in G such that ex and xe are defined for each x in G and $ex = xe=x$.

ii) If x, y are in G and xy exists then there is a neighbourhood U of x and a neighbourhood V of y such that if $x^{'} \in U$ and $y^{'} \in V$ then $x^{'}y^{'}$ exists. The product xy is continuous wherever defined.

iii) The associative law holds whenever it has meaning.

iv) If $ab=e$ then $ba=e$. An element b satisfying this relation is called an inverse and is denoted by $a^{-1}$. We assume that $a^{-1}$ is unique and continuous where defined and that if it exists for an element a it exists for all elements in some neighbourhood of a. Note that $a^{-1}$ always exists in some neighbourhood of e. In fact, there exists a symmetric open neighbourhood U of e such that $U^{2}$ is defined.

The above definition is somewhat redundant.

EXAMPLE.

Any neighbourhood O of the identity oa topological group is a local group if the neighbourhood is open.

We shall call two local groups isomorphic if there is a homeomorphism between their elements which carries inverse to inverse and product to product in so far as they are defined. However, in some applications, it is natural to regard two local groups as equivalent if they belong to the same local equivalence class, that is, a neighbourhood of e in one is isomorphic to a neighbourhood of e in the other. In this book an isomorphism and preserves group operations so far as they are defined.

LEMMA.

Let G be a local group with U the symmetric open neighbourhood of e described in the definition. Given any neighbourhood V of e, $V \subset U$, there is a symmetric neighbourhood W of e, $W^{3} \subset V$. The product sets AB, BC, (AB)C, A(BC) exist for $A, B, C \subset W$ and $A(BC) = (AB)C$. The set AB is open if either A or B is open. The sets A, bA, and Ab are homeomorphic for $b \in W$. Any two points of W have homeomorphic neighbourhoods.

The proof is omitted, the details being as in 1.13, 1.14 and 1.15.

Regards,

Nalin Pithwa

# Topological Spaces and Groups : Part 1: Fast Review

Reference:

Topological Transformation Groups by Deane Montgomery and Leo Zippin, Dover Publications, Available in Amazon India.

My motivation:

We present preliminary and somewhat elementary facts of general spaces and groups. Proofs are given in considerable detail and there are examples which may be of help to a reader for whom the subject is new.

(The purpose of this blog is basically improvement in my understanding of the subject. If, by sharing, it helps some student/readers, that would be great. ) (Also, I found that reading/solving just one main text like Topology by James Munkres left me craving for more topological food: I feel I must also know how the founding fathers discovered topology)…

1.0 Introduction:

We use standard set-theoretic symbols : capitals A, B etc. for sets, $A \bigcup B$ for the union of sets (elements in one or both), and $A \bigcap B$ for the intersection (elements in both) etc.

1.1 Spaces (Topological Spaces);

The term space is sometimes used in mathematical literature in a very general sense to denote any collection whose individual objects are called points, but in topology the term space is used only when some further structure is specified for the collection. As the term will be used in this book it has a meaning which is convenient in studying topological groups. The definition is as follows:

DEFINITION: A topological space (or more simply space) is a non-empty set of points certain subsets of which are designated as open and where, moreover, these open sets are subject to the following conditions:

1. The intersection of any finite number of open sets is open.
2. The union of any number of open sets is open.
3. The empty set and the whole space are open.
4. To each pair of distinct points of space there is associated at least one open set which contains one of the points and does not contain the other.

A space is called discrete if each point is an open set.

Condition (4) is known as the $T_{0}$ separation axiomi in the terminology of Paul Alexandroff and Herr Heinz Hopf. The first three conditions define a topological space in their terminology. The designated system of open sets is the essential part of the topology, and the same set of points can become a topological space in many ways by choosing different systems of subsets designated as open.

1.2. Homeomorphisms

DEFINITION. A homeomorphism is a one-to-one relation between all points of one topological space and all points of a second which puts the open sets of the two spaces in one-one correspondence; the spaces are topologically equivalent.

The notion of homeomorphism is reflexive, symmetric, and transitive so that it is an equivalence relation in a given set of topological spaces.

EXAMPLES of topological spaces:

Let $E_{1}$ denote the set of all real numbers in its customary topology: the open intervals are the sets $\{ y: x < y < z\}$ for every $x < z$. The open sets are those which are unions of open intervals together with the empty set and the whole space.

Let $R_{1} \subset E_{1}$ denote the set of numbers in the closed interval $0 \leq y \leq 1$, where for the moment we take this subset without a topology. This set gives distinct topological spaces as follows:

1. Topologize $R_{1}$ as customarily : the open sets are the intersections of $R_{1}$ with the open sets of $E_{1}$.
2. Topologize $R_{1}$ discretely, that is, let every subset be open.
3. Topologize $R_{1}$ by the choice : open sets are the null set, the whole space and for each x of $R_{1}$ the set $\{ z, x < z \leq 1\}$
4. Topologize $R_{1}$ by the choice: the complement of any finite set is open and the empty set and whole space are open

In the sequel $R_{1}$ will denote the closed unit interval and $E_{1}$ the set of all reals in the customary topology. A set homeomorphic to $R_{1}$ is called an arc. A set homeomorphic to a circle is called a simply closed curve.

1.3 Basis

DEFINITION. A collection $\{ Q_{a}\}$ of open sets of a space is called a basis for open sets if every open set (except possibly the null set) in the space can be represented as a union of sets in $\{ Q_{a}\}$. It is called a sub-basis if every open set can be represented as a union of finite intersection of sets in $\{ Q_{a}\}$ (except possibly the null set).

A collection $\{ Q_{a}\}$ of open sets of a space S is a basis if and only for every open set Q in S and $x \in Q$, there is a $Q_{a} \in \{ Q_{a} \}$ such that $x \in Q_{a} \subset Q$.

If a collection has this property at a particular point x then the collection is called a basis at x.

If a set together with certain subsets are called a sub-basis, then another family of subsets is determined from the sub-basis by taking arbitrary unions and finite intersections. This new family (with the null set added if necessary) then satisfies conditions 1, 2, 3 for open sets of a topological space. Whether 4 will also be satisfied depends on the original family of sets.

EXAMPLE. Let $E_{1}$ denote the space of real numbers in its usual topology. For each pair of rationals $r_{1} < r_{2}$ let $\{ r_{1}, r_{2}$ denote the set of reals $r_{1}. This countable collection of open sets is a basis.

A space is said to be separable or to satisfy the second countability axiom if it has a countable basis. A space is said to satisfy the “first countability axiom” if it has a countable basis at each point.

EXAMPLE. Let S denote a topological space and let F denote a collection of real valued functions $f(x)$, where $x \in S$. If $f_{0}$ is a particular element of F then for each positive integer n let $Q(f_{0},n)=\{ f \in F: |f(x)-f_{0}(x)| < \frac{1}{n}\}$ for all x. We may topologize F by choosing the sets $Q(f_{0},n)$ for all $f_{0}$ and n as a sub-basis. The topological space so obtained has a countable basis at each point in many important cases.

1.4 Topology of subsets

Let S be a topological space, T a subset. Let $Q \bigcap T$ be called open in T or open relative to T if Q is open in S. With open sets defined in this way T becomes a topological space and the topology so defined in T is called the induced or relative topology. If S has a countable basis and $T \subset S$, then T has a countable basis in the induced topology.

DEFINITION. A subset $X \subset S$ is called closed if the complement $S - X$ is open. If $X \subset T \subset S$ then X is called closed in T when $T - X$ is open in T.

Notice that T closed in S and X closed in T implies that X is closed in S. The corresponding assertion for relatively open sets is also true.

It can be seen that finite unions and arbitrary intersections of closed sets are again closed.

DEFINITION. If $K \subset S$, the intersection of all closed subsets of S which contain K is called the closure of K and is denoted by $\overline{K}$. If K is closed, $K = \overline{K}$.

15 Continuous maps

Let S and T be topological spaces and f a map of S into T $f: S \rightarrow T$, that is, for each x in S, $y=f(x)$ is a point of T. If the inverse of each relatively open set in $f(S)$ is an open set in S, then f is called continuous. In case $f(S)=T$ then f is continuous and V open in T imply $f^{-1}(V)$ is open in S. The map is called an open map if it carries open sets to open sets.

If f is a continuous map of S onto T (that is, $f(S)=T$) and if $f^{-1}$ is also single valued and continuous, then f and $f^{-1}$ is also single valued and continuous, then f and $f^{-1}$ are homeomorphisms and S and T are homeomorphic or topologically equivalent.

EXAMPLE. The map $f(t) = \exp{2\pi it}$ is a continuous and open map of $E_{1}$ onto a circle (circumference) in the complex plane.

EXAMPLE. Let K denote the cylindrical surface, described in x ,y, z coordinates in three-space by $x^{2}+y^{2}=1$. Let $f_{1}$ denote the map of K onto $E_{1}$ given by $(x,y,z) \rightarrow (0,0,z)$, let $f_{2}$ the map $(x,y,z) \rightarrow (x,y, |z|)$ of K into K. All three maps are continuous, the first two are open and $f_{1}$ and $f_{2}$ are also closed, that is, they map closed sets into closed sets.

1.6 Topological products

The space of n real variables $(x_{1}, x_{2}, \ldots, x_{n})$ from $-\infty < x_{i} < \infty$, where $i=1,2,\ldots, n$ and the cylinder K of the preceding example are instances of topological products.

Let A denote any non-empty set of indices and suppose that to each $a \in A$ there is associated a topological space $S_{a}$. The totality of functions f defined on A such that $f(a) \in S_{a}$ for each $a \in A$ is called the product of the spaces $S_{a}$. When topologized as below it will be denoted by PROD $S_{n}$; we also use the standard symbol $\times$ thus $E \times B$ is the set of ordered pairs $(a,b)$ $e \in E$, $b \in B$.

The standard topology for this product space is defined as follows: For each positive integer n, for each choice of n indices $a_{1}, a_{2}, \ldots, a_{n}$ and for each choice of a non empty open set in $S_{a_{i}}$

$U_{a_{i}} \subset S_{a_{i}}$ for $i=1,2, \ldots, n$

consider the set of functions $f \in PROD \hspace{0.1in} S_{n}$ for which $f(a_{i}) \in U_{a_{i}}$ for $i=1,2, \ldots, n$

Let the totality of these sets be a sub-basis for the product. The resulting family of open sets satisfies the definition of space in 1.1

EXAMPLE 1.

The space $E_{n}= E_{1} \times E_{1} \times \ldots \times E_{1}$, n copies, is the space of n real variables; here $A=\{ 1, 2, 3, \ldots, n\}$ and each $S_{i}$ is homeomorphic to $E_{1}$ (1.2). Let $x_{i} \in S_{i}$. Then, $(x_{1}, x_{2}, \ldots, x_{n})$ are the co-ordinates of a point of $E_{n}$. It can be verified that the sets $U_{m}(x)$ where $m \in \mathcal{N}$, of points of $E_{n}$, whose Euclidean distance from $x = (x_{1}, x_{2}, \ldots, x_{n})$ is less than $\frac{1}{m}$ form a basis at x. The subset $R_{1} \times R_{1}\times \ldots \times R_{1}$ is an n-cell.

EXAMPLE 2.

Let A be of arbitrary cardinal power and let each $S_{n}$, $a \in A$, be homemorphic to $C_{1}$, the circumference of a circle. Then PROD $S_{n}$ is a generalized torus. If A consists of n objects, the product space is the n-dimensional torus. For n=2, we get the torus.

EXAMPLE 3.

Let $D=S_{1} \times S_{2} \times \ldots \times S_{n} \times \ldots$, where $n \in \mathcal{N}$ where each $S_{i}$ is a pair of points — conveniently regarded as the “same” pair, and designated 0 and 2. This is the Cantor Discontinuum, of Cantor Middle Third set. It is homeomorphic to the subset of the unit interval defined by the convergent series: D: $\sum { a_{n} / 3^{n}}$, where $a_{n}=0, 2$. This example will be described in another way in the next section.

THEOREM :

Let $F_{a}$ be a closed subset of the topological space $S_{a}$, and $a \in A$. Then PROD $F_{n}$ is a closed subset of PROD $S_{a}$.

Proof: The reader is requested to try. It is quite elementary.

1.7 Compactness:

DEFINITION: A topological space S is compact if every collection of open sets whose union covers S contains a finite subcollection whose union covers S.

EXAMPLE 1.

The unit interval $R_{1}$ Thus let $\{ Q\}$ denote a collection of open sets covering $R_{1}$. Let F denote the set of points $x \in R$ such that the interval $0 \leq y \leq x$ can be covered by a finite subcollection of $\{ Q \}$. Then F is not empty and is both open and closed. Hence, by the Dedekind cut postulate, or the existence of least upper bounds, or the connectedness of $R_{1}$ it follows that $F = R_{1}$ To illustrate the concept of compactness, consider the open sets $W_{n} \subset R_{1}$, $W_{n}: \frac{1}{3n}< x < \frac{1}{n}$, $n \in \mathcal{N}$. This collection does not cover $R_{1}$. Let $W_{n}$ be the union of two sets: $0 \leq x < a$ and $1-a < x \leq 1$ for some a, $0 < a < 1$. No matter how $a > 0$ is chosen, there is always some finite number of the $W_{n}$ which together with $W_{a}$ covers $R_{1}$. Of course, $R_{1}$ minus endpoints is not compact and no finite subcollection of the $W_{n}$ in this example will cover it.

THEOREM. Let S be a compact space and let $f: S \rightarrow T$ be a continuous map of S onto a topological space T Then, T is compact.

Proof:

Let $\{ O _{a}$ be a covering of T by open sets. Since f is continuous, each $f^{-1}(O_{a})$ is an open set in S. There is a finite covering of S by sets of the collection $\{ f^{-1}(O_{a})\}$ and this gives a corresponding finite covering of T by sets of $O_{a}$. This completes the proof.

COROLLARY. If f is a continuous map of S into T then $f(S)$ is a compact subset of T.

1.7.1 THEOREM

Let S be a compact space and $\{ D_{a}\}$ a collection of closed subsets such that $\bigcap_{a}D_{a}$ is empty. Then there is some finite set $D_{a_{1}}, \ldots, D_{a_{n}}$ such that $\bigcap_{i}D_{a_{i}}$ is empty.

Proof:

The complement of $\bigcap_{a}D_{a}$ is $\bigcup_{a}(S-D_{a})$; if the intersection set is empty. the union covers S. There is a finite set of indices $a_{i}$ such that $S \subset \bigcup_{i}(S - D_{a_{i}})$ and conequently $\bigcup_{i}D_{a_{i}}$ is empty for the same finite set of indices.

COROLLARY 1.

Let $D_{n}$, $n \in \mathcal{N}$ be a sequence of non empty closed subsets of the compact space S with $D_{n+1} \subset D_{n}$. Then, $\bigcap_{n}D_{n}$ is not empty.

APPLICATION:

The Cantor Middle Third Set D

From $R_{1}$, “delete” the middle third: $\frac{1}{3} < x < \frac{2}{3}$. Let $D_{1}$ denote the residue: it is a union of two closed intervals. Let $D_{2}$ denote the closed set in $D_{1}$ complementary to the union of the middle third intervals: $\frac{1}{9} < x < \frac{2}{9}$ and $\frac{7}{9} < x < \frac{8}{9}$. Continuing inductively, define $D_{n} \subset D_{n-1}$ consisting of $2^{n}$ closed mutually exclusive intervals Let $D = \bigcap_{n}$. This is homeomorphic to the space of Example 3 of 1.6.

COROLLARY 2.

A lower semi-continuous (upper semi-continuous) real-valued function on a compact space has finite glb (greatest lower bound), lub (least upper bound) and always attains these bounds at some points of space.

This follows from the preceding corollary and the fact that the set where $f(x) \leq r$ is closed, for every r (similarly, $f(x) \geq r$.

1.7.2 THEOREM

A topological space with the property: “every collection of closed subsets with empty set-intersection has a finite subcollection whose set-intersection is empty” is compact.

Proof:

The proof, like that of the Theorem of 1.7.1 is based on the duality between open and closed sets.

DEFINITION.

If a point x of a topological space S belongs to an open subset of S whose closure is compact, then S is called locally compact set at x; S i locally compact if it has this property at every point.

COROLLARY.

A closed subset of a locally compact space is locally compact in the induced topology. Similarly, a closed subset of a compact space is compact. The union of a finite number of compact subsets is compact.

Proof: HW.

A set U in a topological space is called a neighbourhood of a point z if there is an open set O such that $z \in O \subset U$; z is called an inner point of U. A set F is covered by a collection $\{ U_{i}\}$ if each point of F is an inner point of some set $U_{i}$.

1.7.3.

A space S is called a Hausdorff space if for every x, y $\in S$, $x \neq y$, there exist open sets U and V including x and y respectively such that $U \bigcap V = \phi$ where $\phi$ is the empty set; an equivalent property is the existence of a closed neighbourhood of x not meeting y.

HW: Show that a compact subset of a Hausdorff space is closed.

Lemma: Let S be a compact Hausdorff space, let F be a closed set in S, and x a point not in F. Then there is a closed neighbourhood of x, such that $W \bigcap F = \Phi$

For each $y \in F$ let $U_{y}$ be a neighbourhood of y and $W_{y}$ a neighbourhood of x, such that $U_{y} \bigcap W_{y}= \Phi$. There is a covering of F by sets $U_{y_{1}},\ldots, U_{y_{n}}$. Let $W_{x}$ be the intersection of the associated $W_{y_{i}}$ where $i = 1, 2, \ldots, n$ and let W be the closure of $W_{x}$. The union of the $U_{y_{i}}$ does not meet W. Then $W \bigcap F = \Phi$ which completes the proof.

THEOREM. Let U be a compact Hausdorff space and let $F_{n}$ be a sequence of closed subsets of U. If U is contained in the union of sets $F_{n}$, then at least one of the sets $F_{n}$ has inner points.

Proof. Take a sequence $C_{1} \supset C_{2} \supset \ldots$ of non empty compact neighbourhoods such that for each n, $(\bigcup_{1}^{n}F_{i}) \bigcap C_{n} = \Phi$. This leads to $\bigcap C_{n}=\Phi$ a contradiction. QED.

A set is nowhere dense if its closure has no inner points. A space is said to be of the second category if it cannot be expressed as the union of a countable number of nowhere dense subsets. Hence, a compact Hausdorff space is of the second category. Complete metric spaces, to be defined later, are also of the second category.

1.7.4 THEOREM:

Let S be a locally compact space. There exists a compact space $S^{*}$ and a point z in $S^{*}$ such that $S^{*}-z$ is homeomorphic to S.

Proof:

Let z denote a “new” point, not in S, and let $S^{*}$ denote the set union of S and z. If Y is a subset of S, let $Y^{*} \subset S^{*}$ denote the union of Y and z. We topologize $S^{*}$. Any open set in S is also open in $S^{*}$. In addition, if X is a compact subset of S and Y is the complement S-X, then $Y^{*}$ is open in $S^{*}$. These open sets are taken as a sub-basis for open sets in $S^{*}$.

Suppose now that we have some covering of $S^{*}$ by a family of open sets. Then z belongs to one of these open sets, say $z \in U^{*}$. The complement of $U^{*}$ is a compact subset of S . Hence, the complement is covered by a finite subset of the given covering sets, because of the compactness in S. Together with $U^{*}$, this gives a finite covering of $S^{*}$.

QED.

1.8 Tychonoff Theorem

THEOREM:

Let $S_{a}$ where $a \in \{ a \}$, be compact spaces and let P be the topological product of the $S_{a}$. Then, P is compact.

Proof:

Let $P = S_{1} \times S_{2}$ and let F denote a family of open sets of P covering P. For each point 1$x_{1}$ of $S_{1}$, the closed subset $x_{1} \times S_{2}$ of P is homeomorphic to $S_{2}$ amd is therefore compact. Each point of $x_{1} \times S_{2}$ belongs to a set in F because F is a covering. Because of the way in which a product is topologized, it follows that each point of $x_{1} \times S_{2}$ belongs to some open set $U \times V$ of P such that $U \times V$ is a subset of some set of F. It follows from its compactness that $x_{1} \times S_{2}$ is contained in the union of a finite number of sets

$U_{1} \times V_{1}, \ldots, U_{n} \times V_{n}$

each of which is a subset of some set of F. Let $U^{'} = \bigcap U_{i}$. Then, $U^{'} \times S_{2}$ is covered by a finite number of sets of F.

Since $x_{1}$ is an arbitrary point of $S_{1}$ amd $S_{1}$ is compact, there exist a finite number of open sets of $S_{1}$

$U_{1}^{'}, \ldots, U_{m}^{'}$

which cover $S_{1}$ and which are such that there is a finite number of sets of F covering $U_{1}^{'} \times S_{2}$ where $i=1, \ldots, m$. The totality of sets of F thus indicated is a finite number which covers $S_{1} \times S_{2}$. This completes the proof for the case of two factors. The case for a finite number of factors follows by a simple induction.

EXAMPLE.

Let $R_{n} = R_{1} \times R_{1} \times \ldots R_{1}$, n factors. Then, $R_{n}$ is compact and it follows that $E_{n}$ = product of n real lines is locally compact.

1.8.1

To consider the general case, let $\{ a\}$ be an arbitrary collection of at least two indices: let $S_{a}$ be compact topological spaces, let P be the topological product, and let F be a collection of open subsets of F covering P. The proof that P is compact is by contradiction. Accordingly, we shall suppose that no finite subcollection of sets of F covers P.

It was shown by Zermelo that it is possible to well-order the set of all subsets of a given set by the use of an axiom-of-choice of appropriate power, namely the cardinal number of the set of all subsets of the given set. A well-ordering of objects permits them to be inspected systematically.

Using such a well-ordering we can enlarge the given family F to a family $F^{*}$ of open sets, where $F^{*}$ has the following properties:

1. $F^{*}$ is a covering of P by open sets.
2. No finite subcollection of P by open sets.
3. If we adjoin to $F^{*}$ any open subset of P not already in $F^{*}$, then the enlarged collection does contain a finite subcollection which covers P. Of course, it is in (3) that $F^{*}$ has a property not necessarily true of F

Using this enlarged family the proof for the general case becomes similar to the proof for two factors. Let b denote an arbitrary index in $\{ a \}$ which shall be fixed temporarily, and let $P_{b}$ denote the product of all factors $S_{a}$ excepting $S_{b}$. Then P is homeomorphic to $S_{b} \times P_{b}$.

Suppose for a moment that to each point $x_{b} \times S_{b}$ there exists an open $U_{b} \subset S_{b}$ containing $x_{b}$ such that

*) $U_{b} \times P_{b} \in F^{*}$

There must then exist some finite covering of $S_{b}$ by sets $U_{b}^{1}, U_{b}^{2}, \ldots, U_{b}^{n}$ each satisfying *). The producgt P is covered by the union of $U_{b}^{i} \times P_{b}$, $i=1, \ldots, n$. This is impossible by the contradiction of $F^{*}$. Hence in each $S_{b}$ there is at least one $x_{b}$ which does not satisfy the first sentence of this paragraph.

It follows by the axiom of choice that P contains at least one point $x =$ PROD $x_{b}$ such that if $U_{b}$ is an open set in $S_{b}$ and $x_{b} \in U_{b}$ then *) is false. This holds for each coordinate $x_{b}$ of x. This implies for each coordinate $x_{b}$ of x that if $x_{b}$ is in an open set $U_{b}$ of $S_{b}$ then there is a finite collection of sets in $F^{*}$

$O_{b}^{1}, O_{b}^{2}, \ldots, O_{b}^{n_{b}}$

which together with $U_{b} \times P_{b}$ forms a covering of P.

The point x belongs to an open set $O_{x} \in F^{*}$. There is some open set contained in $O_{x}$ which contains x and is of the form

$U_{a_{1}} \times U_{a_{2}} \times \ldots U_{a_{n}} \times P_{a_{1}a_{2}a_{3}\ldots a_{n}}$

for some finite set of indices $a_{i}$ and where the last set is the product of all $S_{n}$ with the exception of $S_{a_{i}}$, where $i=1,2, \ldots, n$. For each $a_{i}$ there exists a finite collection of sets of $F^{*}$ which together with $U_{a_{i}} \times P_{a_{i}}$ covers P, say these sets are

**) O_{a_{i}}^{1}, O_{a_{i}}^{2}, \ldots, O_{a_{i}}^{n_{i}}, where $i=1, 2, \ldots, n$

Then P is covered by the union of $O_{x}$ and the sets of **). This contradiction completes the proof.

QED.

1.8.2 EXAMPLE

The infinite-dimensional torus described in 1.6 whose “dimension” equals the cardinal power of the set of indices A is compact. It is a commutative group where the addition of two points is carried out by adding the respective coordinates in each factor $S_{n} = C_{1}$ each of these factors being itself a commutative group. The group addition is continuous in the topology and this defines a topological group (1.11) In fact, this is a universal compact commutative topological group (depending on the cardinal power of the group). See for example the following paper: Discrete Abelian groups and their character groups, Ann. of Math., (2) 36 (1935) pp. 71-85.

The principal theorem of this section is due to Tychonoff (see: Uber einen Funktionenraum Math. Ann. III (1935), pp. 762-766). The present proof is dual to a proof given by Bourbaki (see: Topologie generale, Paris, 1942).

1.9 Metric Spaces

DEFINITION.

A set S of points is called a metric space if to each pair $x, y \in S$ there is associated a non-negative real number $d(x,y)$ the distance from x to y, satisfying

1. $d(x,y)=0$ if and only if $x=y$.
2. $d(x,y)=d(y,x)$
3. $d(x,y) + d(y,z) \geq d(x,z)$ where $x,y,z \in S$.

The distance function $d(x,y)$ also called the metric induces a topology in S as follows. For each $r >0$ let $S_{r}(x)$ denote the sphere of radius r, that is, the set of $y \in S$ such that $d(x,y) . Now let $S_{r}(x)$, for all positive r and all $x \in S$ constitute a basis for open sets. This choice of basis makes S a topological space. A space is called metrizable if a metric can be defined for it which induces in it the desired topology. It is clear that a metric space has a countable basis at each point x, namely $S_{r}(x)$ where r is rational.

EXAMPLE 1.

If $S_{1}$ and $S_{2}$ are metric spaces then $S_{1} \times S_{2}$ is a metric space in the metric

$d((x_{1}, x_{2}),(y_{1}, y_{2})) = max (d(x_{1},y_{1}), d(x_{2},y_{2}))$

where $x_{1},y_{1} \in S_{1}$, and $x_{2}, y_{2} \in S_{2}$. The topology determined by this metric is the same as the product topology.

EXAMPLE 2.

The set F of continuous functions defined on a compact space S with values in a metric space M becomes a metric space by defining for $f, g \in F$

$d(f,g) =$ lub $(x \in S)$ [$d_{M}(f(x), g(x))$] where $d_{M}$ is the metric in M. See corollary 2, section 1.7.1

THEOREM:

The collection of open sets of a compact metric space S has a countable basis.

Proof:

For each $n \in \mathcal{N}$ there is a covering of the space by a finite number of open sets each of diameter at most $\frac{1}{n}$. The countable collection of these sets for all n is a basis.

EXAMPLE 3:

If S is a compact metric space and $E_{1}$ denotes the real line, then $S \times E_{1}$ is a metrizable locally compact space with a countable basis for open sets.

By Example 1 above, the space is metrizable. If $\{ U_{m}\}$ and $\{ V_{m}\}$ are countable bases in S and $E_{1}$ respectively then $\{ U_{m} \times V_{m}\}$ forms a countable base in $S \times E_{1}$. If $E_{1n} = \{ x \in E_{1}, |x| \leq n\}$ then $S \times E_{1n}$ is a compact subset of $S \times E_{1}$ and any point of the product is interior to $S \times E_{1n}$ for n large enough. This proves the local compactness.

1.9.1

The following is of interest:If $(x,y)$ is a metric for a space M then the following equivalent metric:

$(x,y)^{'} = \frac{(x,y)}{1+(x,y)} \leq 1$

is a bounded metric. Properties (1) and (2) above are obviously satisfied. For (3) , one uses the fact that the function $\frac{t}{1+t}$ increases with t. Thus

$(x,y)^{'}+(y,z)^{'} \geq \frac{(x,y)}{1+(x,y)+(y,z)} + \frac{(y,z)}{1+(y,z)+(x,y)} = \frac{(x,y)+(y,z)}{1+(x,y)+(y,z)} \geq \frac{(x,z)}{1+(x,z)} = (x,z)^{'}$

This has the following consequence:

Lemma:

Let M be a space which is the union of a system $M_{a}$ where $a \in \{ a\}$ of open mutually exclusive sets. Suppose each $M_{a}$ of open mutually exclusive sets. Suppose each $M_{a}$ is a metric space and carries a metric $d_{a}$ bounded by 1. Define a function $d(x,y)$ which is equal to 2 if x and y are not in the same $M_{a}$; otherwise let d agree with the appropriate $d_{a}$. Then d is a metric for M.

Proof: HW.

1.9.2

A sequence of points $x_{n}$ in a metric space is said to converge to a point x, symbolically $x_{n} \rightarrow x$ if $\lim d(x,x_{n}) =0$. A sequence of points $x_{n}$ satisfies the Cauchy convergence criterion if when $\epsilon >0$ is given there is an N such that for $m,n >N$ $d(x_{n},x_{m})<\epsilon$. A metric space is called complete if every sequence of points satisfying the Cauchy criterion converges to a point of the space. A subset of a space is called dense (everywhere dense) in the space if every point of space is a limit of some sequence of points of the subsets.

To be continued in next blog,

Cheers,

Nalin Pithwa

# Is Math really abstract? I N Herstein answers…

Reference: Chapter 1: Abstract Algebra Third Edition, I. N. Herstein, Prentice Hall International Edition:

For many readers/students of pure mathematics, such a book will be their first contact with abstract mathematics. The subject to be discussed is usually called “abstract algebra,” but the difficulties that the reader may encounter are not so much due to the “algebra” part as they are to the “abstract” part.

On seeing some area of abstract mathematics for the first time,be it in analysis, topology, or what not, there seems to be a common reaction for the novice. This can best be described by a feeling of being adrift, of not having something solid to hang on to. This is not too surprising, for while many of the ideas are fundamentally quite simple, they are subtle and seem to elude one’s grasp the first time around. One way to mitigate this feeling of limbo, or asking oneself “What is the point of all this?” is to take the concept at hand and see what it says in particular concrete cases. In other words, the best road to good understanding of the notions introduced is to look at examples. This is true in all of mathematics.

Can one, with a few strokes, quickly describe the essence, purpose, and background for abstract algebra, for example?

We start with some collection of objects S and endow this collection with an algebraic structure by assuming that we can combine, in one or several ways (usually two) elements of this set S to obtain, once more, elements of this set S. These ways of combining elements of S we call operations on S. Then we try to condition or regulate the nature of S by imposing rules on how these operations behave on S. These rules are usually called axioms defining the particular structure on S. These axioms are for us to define, but the choice made comes, historically in mathematics from noticing that there are many concrete mathematical systems that satisfy these rules or axioms. In algebra, we study algebraic objects or structures called groups, rings, fields.

Of course, one could try many sets of axioms to define new structures. What would we require of such a structure? Certainly we would want that the axioms be consistent, that is, that we should not be led to some nonsensical contradiction computing within the framework of the allowable things the axioms permit us to do. But that is not enough. We can easily set up such algebraic structures by imposing a set of rules on a set S that lead to a pathological or weird system. Furthermore, there may be very few examples of something obeying the rules we have laid down.

Time has shown that certain structures defined by “axioms” play an important role in mathematics (and other areas as well) and that certain others are of no interest. The ones we mentioned earlier, namely, groups, rings, fields, and vector spaces have stood the test of time.

A word about the use of “axioms.” In everyday language, “an axiom means a self-evident truth”. But we are not using every day language; we are dealing with mathematics. An axiom is not a universal truth — but one of several rules spelling out a given mathematical structure. The axiom is true in the system we are studying because we forced it to be true by “force” or “our choice” or “hypothesis”. It is a licence, in that particular structure to do certain things.

We return to something we said earlier about the reaction that many students have on their first encounter with this kind of algebra, namely, a lack of feeling that the material is something they can get their teeth into. Do not be discouraged if the initial exposure leaves you in a bit of a fog.Stick with it, try to understand what a given concept says and most importantly, look at particular, concrete examples of the concept under discussion.

Follow the same approach in linear algebra, analysis and topology.

Cheers, cheers, cheers,

Nalin Pithwa