# VI Convergence. Open and Closed Sets.

Reference: Introductory Real Analysis by Kolmogorov and Fomin.

Reference: Introduction to Analysis by Rosenlicht.

Reference: Analysis by Walter Rudin.

Sec 6.1 Closure of a set. Limit points.

By the open sphere (or open ball) $S(x_{0},r)$ in a metric space R we mean the set of points $x \in R$ satisfying the inequality $\rho(x_{0},r)

($\rho$ is the metric of R.)

Note: Any confusion between “sphere” meant in the sense of spherical surface and “sphere” meant in the sense of a solid sphere (or ball) will always be avoided by judicious use of the adjectives “open” or “closed.”

The fixed point $x_{0}$ is called the center of the sphere and the number r is called its radius. By the closed sphere (or closed ball) $S[x_{0},r]$ with center $x_{0}$ and radius r we mean the set of points $x \in R$ satisfying the inequality

$\rho(x_{0},x) \leq r$

An open sphere of radius $\epsilon$ with center $x_{0}$ will also be called an $\epsilon$neighbourhood of $x_{0}$ denoted by $O_{\epsilon}(x_{0})$.

A point $x \in R$ is called a contact point of a set $M \subset R$ if every neighbourhood of x contains at least one point of M. The set of all contact points of a set M is denoted by $[M]$ and is called the closure of M. Obviously, $M \subset [M]$, since every point of M is a contact point of M. By the closure operator in a metric space R, we mean the mapping of R carrying each set $M \subset R$ into its closure $[M]$.

Theorem 1:

The closure operator has the following properties:

1. If $M \subset N$, then $[M] \subset [N]$
2. $[[M]] = [M]$
3. $[M \bigcup N] = [M] \bigcup [N]$
4. $[\phi] = \phi$.

Proof I:

Property I is obvious.

Proof of property 2:

Let $x \in [[M]]$. Then, any given neighbourhood $O_{\epsilon}(x)$ contains a point $x_{1} \in [M]$. Consider the sphere $O_{\epsilon_{1}}(x_{1})$ of radius

$\epsilon_{1} = \epsilon - \rho(x,x_{1})$

PS: It helps to draw an illustrative diagram to understand this proof.

Clearly, $O_{\epsilon_{1}}(x_{1})$ is contained in $O_{\epsilon}(x)$. In fact, if $z \in O_{\epsilon_{1}}(x_{1})$ then $\rho(z,x_{1}) < \epsilon_{1}$ and hence, since $\rho(x,x_{1}) = \epsilon - \epsilon_{1}$, it follows from the triangle inequality that

$\rho(z,x) < \epsilon_{1} + (\epsilon - \epsilon_{1}) = \epsilon$,

That is, $z \in O_{\epsilon}(x)$. Since $x_{1} \in [M]$, there is a point $x_{2} \in M$ in $O_{\epsilon_{1}}(x)$. But then $x_{2} \in O_{\epsilon}(x)$ and hence, $x \in [M]$, since $O_{\epsilon}(x)$ is an arbitrary neighbourhood of x. Therefore, $[[M]] \subset [M]$. But, obviously $[M] \subset [[M]]$ and hence, $[[M]] = [M]$, as required.

Proof of property 3:

Let $x \in [M \bigcup N]$ and suppose $x \notin [M] \bigcup [N]$. Then, $x \notin [M]$ and $x \notin [N]$. But then there exist neighbourhoods $O_{\epsilon_{1}}(x)$ and $O_{\epsilon_{2}}(x)$ such that $O_{\epsilon_{1}}(x)$ contains no points of M while $O_{\epsilon_{1}}(x)$ contains no points of N. It follows that the neighbourhood $O_{\epsilon}(x)$, where $\epsilon=\min (\epsilon_{1}, \epsilon_{2})$, contains no points of either M or N, and hence no points of $M \bigcup N$, contrary to the assumption that $x \in [M \bigcup N]$. Therefore, $x \in [M] \bigcup [N]$, and hence,

$[M \bigcup N] = [M] \bigcup [N]$…call this I

since x is an arbitrary point of $[M \bigcup N]$. On the other hand, since $M \subset M \bigcup N$ and $N \subset M \bigcup N$, it follows that from property (i) that $[M] \subset [M \bigcup N]$ and $[N] \subset [M \bigcup N]$. But then,

$[M] \bigcup [N] \subset [M \bigcup N]$

which together with (i) implies that $[M \bigcup N] = [M] \bigcup [N]$.

Proof of property 4:

We observe that given any $M \subset R$,

$[M] = [M \bigcup \phi] = [M] \bigcup [\phi]$ by property (iii).

It follows that $[\phi] \subset [M]$. But this is possible for arbitrary M only if $[\phi] = \phi$. (Alternatively, the set with no elements can have no contact points!). QED.

Definition:

A point $x \in R$ is called a limit point of a set $M \subset R$ if every neighbourhood of x contains infinitely many points of M. The limit point may or may not belong to M. For example, if M is the set of rational numbers in the interval $[0,1]$, then every point of $[0,1]$, rational or irrational, is a limit point of M. (It is said that the rationals are dense in R; we will discuss this in detail later).

A point x belonging to a set M is called an isolated point of M if there is a (“sufficiently small”) neighbourhood of x containing no points of M other than x itself.

6.2 Convergence and limits.

A sequence of points $\{x_{n}, n \in N \}$ in a metric space R is said to converge to a point $x \in R$ if every neighbourhood $O_{\epsilon}(x)$ of x contains all points $x_{n}$ starting from a certain index (more exactly, if, given any $\epsilon >0$, there is an integer $N_{\epsilon}$ such that $O_{\epsilon}(x)$ contains all points $x_{n}$ with $n > N_{\epsilon}$). The point x is called the limit of the sequence $\{ x_{n}\}$, and we write $x_{n} \rightarrow x$ (as $n \rightarrow \infty$). Clearly, $\{ x_{n}\}$ converges to x if and only if

$\lim_{n \rightarrow \infty} \rho(x,x_{n}) = 0$.

It is an immediate consequence of the definition of a limit of a sequence that

(1) No sequence can have two distinct limits.

(2) If a sequence $\{ x_{n}\}$ converges to a point x, then so does every subsequence of $\{ x_{n}\}$.

(Details are filled in here: I believe it is helpful to write little proofs of little details…I call it playing with small pearls of math…: )

Proof of (1) : Let, if possible, a sequence $\{ x_{n}\}$ have two distinct limit points, namely, $x_{0}$ and $x_{0}^{'}$. Then, there exists a neigbourhood or open ball $O_{\epsilon/2}(x)$ which contains all points of the given sequence after some $N_{0} \in \mathscr{N}$. So also there exists a neighbourhood (again of radius $\epsilon/2$ about $x_{0}^{'}$ such that it contains all points of the given sequence after some natural number $N_{0}^{'} \in \mathscr{N}$.

But, we know that $\rho(x_{0},x_{0}^{'}) \leq \rho(x_{0},x) + \rho(x_{0}^{'},x) \leq \epsilon/2+\epsilon/2=\epsilon$. Which in turn means that the two “distinct limit points” lie in the same open ball. Hence, they cannot be distinct. Hence, a sequence can have only one limit point, and it is unique. QED.

Proof of (2):

Definition of subsequence: Given a sequence $\{ x_{n}\}$, where $n \in \mathscr{N}$, consider a strictly increasing sequence of positive integers (that is, $n_{1}, n_{2}, n_{3}, \ldots$ are positive integers and $n_{1}) : then the sequence $x_{n_{1}}, x_{n_{2}}, x_{n_{3}}, \ldots$ is called a subsequence of the original sequence $\{ x_{n}\}$.

Now, the claim is that: if a given sequence $\{ x_{n}\}$ converges to a limit point, so does every subsequence of the given sequence: —-

So, it is given that $\rho(x_{n},x) < \epsilon$ when $n > N$ for some natural number $N \in \mathscr{N}$. Since $n_{m} \geq m$ ( here m is some dummy variable, some appropriate positive integer) for all positive integers m, we have $\rho(x, x_{n_{m}})<\epsilon$ whenever $m > N$. This means that $\rho(x_{n_{m}},x)<\epsilon$, which is what we wanted to prove. QED.

THEOREM 2:

A necessary and sufficient condition for a point x to be a contact point of a set M is that there exists a sequence $\{ x_{n}\}$ of points of M converging to x.

PROOF 2:

The condition is necessary since if x is a contact point of M, then every neighbourhood $O_{\frac{1}{n}}(x)$ contains at least one point $x_{n} \in M$, and these points form a sequence $\{ x_{n}\}$ converging to M. The sufficiency is obvious. QED.

THEOREM 2′:

A necessary and sufficient condition for a point x to be a limit point of a set M is that there exists a sequence $\{ x_{n}\}$ of distinct points of M converging to x.

Proof 2′:

Clearly, if x is a limit point of M, then the points $x_{n} \in O_{\frac{1}{n}}(x) \bigcap M$ figuring in the proof of Theorem 2 (just above) can be chosen to be distinct. This proves the necessity and the sufficiency is again obvious. QED.

6.3 Dense subsets, Separable spaces:

Let A and B be two subsets of a metric space R. Then, A is said to be dense in B if $[A] \supset B$. In particular, A is said to be everywhere dense (in R) if $[A] = R$. A set A is said to be nowhere dense if it is dense in no (open) sphere at all. (Note; it is w.r.t. to the metric function defined in the space R.) (Famous example: The rationals are dense in $\Re$).

Example 1:

The set of all rational points is dense in the real line $\Re$.

proof 1:

We want to prove that $[Q] \supset \Re$. Clearly, as $Q \subset R$, the set of rationals are dense in rationals themselves. We only need to show that any irrational number is also a limit point of a sequence of rationals. For example, this is crystal clear: consider the sequence of decimal numbers (finitely many decimals) that we get by extracting square root of 2 by long division. The same applies to any other surd or irrational number. Hence, we can always find a rational sequence of numbers converging to an irrational number. Hence, the rationals are dense in the reals. QED.

Example 2:

The set of all points $x = (x_{1}, x_{2}, x_{3}, \ldots, x_{n})$ with rational co-ordinates is dense in each of the spaces $R^{n}$, $R_{0}^{n}$, $R_{1}^{n}$ introduced in the following examples (in earlier blog, and presented here again for handy reference):

2a’) the set of all ordered n-tuples $x=(x_{1}, x_{2}, x_{3}, \ldots, x_{n})$ of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ with distance $\rho(x,y) = \sqrt{\Sigma_{k=1}^{n}(x_{k}-y_{k})^{2}}$ is a metric space denoted by $R^{n}$ and called n-dimensional Euclidean space (or, simply Euclidean n-space).

proof 2a’:

same proof as example 1. QED.

2b) the set of all ordered n-tuples $x=(x_{1}, x_{2}, x_{3}, \ldots, x_{n})$ of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ with distance $\rho_{1}(x,y) = \Sigma_{k=1}^{n}|x_{k}-y_{k}|$ is a metric space denoted by $R_{1}^{n}$.

proof 2b:

same proof as example 1. QED.

2c) The set of all ordered n-tuples $x=(x_{1}, x_{2}, x_{3}, \ldots, x_{n})$ of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ with distance $\rho_{0}(x,y) = \max_{1 \leq k \leq n}|x_{k}-y_{k}|$ is a metric space denoted by $R_{0}^{n}$.

proof 2c:

once again same proof as example 1 with minor differences. QED.

Example 3:

The set of all points $x = (x_{1}, x_{2}, x_{3}, \ldots, x_{k}, \ldots)$ with only finitely many non-zero co-ordinates, each a rational number, is dense in the space $l_{2}$ introduced in earlier blog (and presented here again for handy reference):

Definition of $l_{2}$ metric space: Let $l_{2}$ be the set of all infinite sequences $x = (x_{1}, x_{2}, x_{3}, \ldots, x_{k}, \ldots)$ of real numbers $x_{1}, x_{2}, \ldots, x_{k}, \ldots$ satisfying the convergence condition $\Sigma_{k=1}^{\infty} x_{k}^{2}< \infty$, where distance between points is defined by $\rho(x,y) = \sqrt{\Sigma_{k=1}^{\infty}(x_{k}-y_{k})^{2}}$.

Again, some detailed steps show that the  essence of this proof is also the fact that the rationals are dense in the reals.

Example 4: So, also, similarly, the set of all polynomials with rational coefficients is dense in both spaces $C_{[a,b]}$ and $C_{[a,b]}^{2}$. Let me reproduce the definitions of these spaces: The set $C_{[a,b]}$ of all continuous functions defined on the closed interval $[a,b]$ with distance $\rho(f,g)= \max_{a \leq t \leq b}|f(t)-g(t)|$ is a metric space. The set of all functions continuous on the interval $[a,b]$ with distance function defined by $\rho(x,y) = (\int_{a}^{b}[x(t)-y(t)]^{2}dt)^{\frac{1}{2}}$ is also a metric space denoted by $C_{[a,b]}^{2}$.

Definition:

A metric space is said to be separable if it has a countable everywhere dense subset.

Example 5:

The spaces $\Re^{1}$, $\Re^{n}$, $\Re_{0}^{n}$, $l_{2}$, $C_{[a,b]}$ and $C_{[a,b]}^{2}$ are all separable since the sets in Examples 1 to 4 above are all countable. (this can be easily verified).

Example 6:

The “discrete space” M described in Example 1 (reproduced here again) contains a countable everywhere dense subset and hence is separable if and only it is itself a countable set, since clearly $[M]=M$ in this case. (Put $\rho(x,y)=1$, when $x=y$ and $\rho(x,y)=0$ if $x \neq y$. Then, this space M is called the discrete metric space).

Example 7:

There is no countable everywhere dense set in the space m of all bounded sequences, introduced in an earlier blog ( To help recall, we present it here again: Consider the set of all bounded infinite sequences of real numbers $x = (x_{1}, x_{2}, \ldots, x_{n}, \ldots)$, and let $\rho(x,y) = \sup_{k}|x_{k}-y_{k}|$. This gives a metric space which we denote by m. The fact that it has the three properties of a metric is almost obvious.)

In fact, consider the set E of all sequences consisting exclusively of zeros and ones. Clearly, E has the power of continuum since there is a one-to-one correspondence between E and the set of all subsets of the set $\mathcal{Z_{+}} = \{1,2, \ldots, n, \ldots \}$ (what is the correspondence here? answer:) Now, the distance between any two points of E equals 1. Suppose we surround each point of E by an open sphere of radius $\frac{1}{2}$, thereby obtaining an uncountably infinite family of pairwise disjoint spheres. Then, if some set M is everywhere dense in m, there must be at least one point of M in each of the spheres. It follows that M cannot be countable and hence, that m cannot be separable.

Closed Sets:

We say that a subset M of a metric space R is closed if it coincides with its own closure, that is, if $[M]=M$. In other words, a set is called closed if it contains all its limit points (we will solve this problem as an exercise).

Example 1:

The empty set $\phi$ and the whole space R are closed sets.

Example 2:

Every closed interval $[a,b]$ on the real line is a closed set.

Example 3:

Every closed sphere in a metric space is a closed set. In particular, the set of all functions f in the space $C_{[a,b]}$ such that $|f(t)|\leq K$ (where K is a constant) is closed.

Example 4:

The set of all functions f in $C_{[a,b]}$ such that $|f(t)| (an open sphere) is not closed. The closure of the set is the closed sphere in the preceding example.

Example 5:

Any set consisting of a finite number of points is closed.

Theorem 3:

The intersection of arbitrary number of closed sets is closed. The union of a finite number of closed sets is closed.

Proof 3:

Given arbitrary sets $F_{\alpha}$ indexed by a parameter $\alpha$, let x be a limit point of the intersection $F = \bigcap_{\alpha}F_{\alpha}$

Then, any neighbourhood $O_{\epsilon}(x)$ contains infinitely many points of F, and hence, infinitely many points of each $F_{\alpha}$. Therefore, x is a limit point of each $F_{\alpha}$ and hence belongs to each $F_{\alpha}$, since the sets $F_{\alpha}$ are all closed. It follows that $x \in F_{\alpha}$ and hence that F itself is closed.

Next, let $F = \bigcup_{k=1}^{\infty}F_{k}$ be the union of a finite number of closed sets $F_{k}$, and suppose x does not belong to F. Then, x does not belong to any of the sets $F_{k}$, and hence, x cannot be a limit point of any of them. But then, for every k, there is a neighbourhood $O_{\epsilon_{k}}(x)$ containing no more than a finite number of points of $F_{k}$. Choosing

$\epsilon = \min (\epsilon_{1}, \epsilon_{2}, \ldots, \epsilon_{n})$

we get a neighbourhoood $O_{\epsilon}(x)$ containing no more than a finite number of points of F, so that x cannot be a limit point of F. This proves that a point $x \in F$ cannot be a limit point of F. Therefore, F is closed. QED.

6.5 Open sets.

A point x is called an interior point of a set M if x has a neighbourhood $O_{\epsilon}(x) \subset M$, that is, a neighbourhood consisting entirely of points of M. A set is said to be open if all points are interior points (with respect to that metric).

Example 1:

Every open interval $(a,b)$ on the real line is an open set. In fact, if $a < x < b$, choose $\epsilon = \min(x-a, b-x)$. Then, clearly $O_{\epsilon}(x) \subset (a,b)$.

Example 2:

Every open sphere $S(a,r)$ in a metric space is an open set. In fact, $x \in S(a,r)$ implies that $\rho(a,x). Hence, choosing $\epsilon = r - \rho(a,x)$ we have $O_{\epsilon}(x)=S(x,\epsilon) \subset S(a,r)$.

Example 3:

Let M be the set of of all functions f in $C_{[a,b]}$ such that $f(t) < g(t)$, where g is a fixed function in $C_{[a,b]}$. Then, M is an open subset of $C_{[a,b]}$.

Theorem 4:

A subset M of a metric space R is open if and only its complement $R-M$ is closed.

Proof of theorem 4:

If M is open, then every point $x \in M$ has a neighbourhood (entirely) contained in M. Therefore, no point $x \in M$ can be a contact point of $R - M$. In other words, if x is a contact point of $R-M$ then $x \in R-M$, that is, $R-M$ is closed.

Conversely, if $R-M$ is closed, then any point $x \in M$ must have a neighbourhood contained in M; since, otherwise every neighbourhood of x would contain points of $R-M$, that is, x would be a contact point of $R-M$, not in $R-M$. Therefore, M is open.

QED.

corollary to theorem 4:

The empty set $\phi$ and the whole space R are open sets.

Proof:

An immediate consequence of Theorem 4 and Example 1 in the previous section.

QED.

THEOREM 5:

The union of an arbitrary number of open sets is open. The intersection of a finite number of open sets is open.

Proof of theorem 5:

this is the dual of theorem 3. The proof is an immediate consequence of theorem 4 and DeMorgan laws. QED.

6.6 Open and closed sets on the real line:

The structure of open and closed sets in a given metric space can be quite complicated. This is true even for open and closed sets in a Euclidean space of two or more dimensions. In the one-dimensional case, however, it is an easy matter to give a complete description of all open sets (and, hence, of all closed sets).

Theorem 6:

Every open set G on the real line is the union of a finite or countable system of pairwise  disjoint open intervals.

Proof of theorem 6:

Let x be an arbitrary point of G. By the definition of an open set, there is at least one open interval containing x and contained in G. Let $I_{x}$ be the union of all such open intervals. Then, as we now show,

$I_{x}$ is itself an open interval.

proof:

In fact, let $a = \inf {I_{x}}$ and $b = \sup {I_{x}}$ where we allow the cases $-\infty=a$ and $+\infty=b$.

Then, obviously, $I_{x} \subset (a,b)$….call this II.

Moreover, let us assume that y is an arbitrary point of $(a,b)$ distinct from x, where, to be explicit, we assume that $a. Then, there is a point $y^{'}\in I_{x}$ such that $a < y^{'}. (HW: think, why!) Hence, G contains an open interval containing the points $y^{'}$ and $x$. But, then this interval also contains y, that is, $y \in I_{x}$. (The case $y>x$ is treated similarly. )

Moreover, $x \in I_{x}$, by hypothesis. It hence follows that $I_{x} \supset (a,b)$, and hence, by (II) that $I_{x}=(a,b)$. Thus, $I_{x}$ is itself an open interval, as asserted, in fact the open interval $(a,b)$.

Next, we want to show that $I_{x}$ is a union of countable, pairwise, disjoint open intervals. Let us proceed as follows:

By its very construction, the interval $(a,b)$ is contained in G and is not a subset of a larger interval contained in G. Moreover, it is clear that two intervals $I_{x}$ and $I_{x^{'}}$ corresponding to distinct points x and $x^{'}$ either coincide or else are disjoint (otherwise, $I_{x}$ and $I_{x^{'}}$ would both be contained in a larger interval $I_{x} \bigcup I_{x^{'}} = I \subset G$. There are no more than countably many such pairwise disjoint intervals $I_{x}$. In fact, choosing an arbitrary rational point in each $I_{x}$ we establish a one-to-one correspondence between the intervals $I_{x}$ and a subset of the rational numbers. Finally, it is obvious that

$G = \bigcup_{x}I_{x}$.

QED.

COROLLARY TO PROOF 6:

Every closed set on the real line can be obtained by deleting a finite or countable system of pairwise disjoint intervals from the line.

EXAMPLE 1:

Every closed interval $[a,b]$ is a closed set (here, a and b are necessarily finite).

EXAMPLE 2:

Every single element set $\{ x_{k}\}$ is closed.

EXAMPLE 3:

The union of a finite number of closed intervals and single element sets is a closed set.

EXAMPLE 4 (The CANTOR Set):

A more interesting example of a closed set on the real line can be constructed as follows:

Delete the open interval $(\frac{1}{3},\frac{2}{3})$ from the closed interval $F_{0}=[0,1]$ and let $F_{1}$ denote the remaining closed set, consisting of two closed intervals. Then, delete the open intervals $(\frac{1}{9}, \frac{2}{9})$ and $(\frac{7}{9}, \frac{8}{9})$ from $F_{1}$ and let $F_{2}$ denote the remaining closed set consisting of four closed intervals. Then, delete the “middle third” from each of these four intervals, getting a new closed set $F_{3}$, and so on. Continuing this process indefinitely, we get a sequence of closed sets $F_{n}$ such that

$F_{0} \supset F_{1} \supset F_{2} \supset \ldots \supset F_{n} \supset \ldots$

(such a sequence is said to be decreasing). The intersection

$F = \bigcap_{n=0}^{\infty}F_{n}$

of all these sets is called the CANTOR SET.

Clearly, F is closed, and is obtained from the unit interval $[0,1]$ by deleting a countable number of open intervals. In fact, at the nth stage of construction, we delete $2^{n-1}$ intervals, each of length $\frac{1}{3^{n}}$.

To describe the structure of the set $F_{n}$ we first note that F contains the points, 0,1, $\frac{1}{3}$, $\frac{2}{3}$, $\frac{1}{9}$, $\frac{2}{9}$, $\frac{7}{9}$, $\frac{8}{9}$, $\ldots$,….call this III.

that is, the end points of the deleted intervals (together with the points 0 and 1)

However, F contains many other points.

In fact, given any $x \in [0,1]$, suppose we write x in ternary notation, representing x as a series :

$x=\frac{a_{1}}{3}+\frac{a_{2}}{3^{2}}+\frac{a_{3}}{3^{3}}+\ldots + \frac{a_{n}}{3^{n}}+\ldots$…call this IV.

where each of the numbers $a_{1}, a_{2}, \ldots$ is 0, 1 or 2 in radix-3. Then, it is easy to see that x belongs to F if and only if x has a representation (IV) such that none of the numbers $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ equals 1 (you need to think about this as HW!!)

Remarkably enough, the set F has the power of the continuum, that is, there are as many points in F as in the whole interval $[0,1]$, despite the fact that the sum of the lengths of the deleted intervals equals

$\frac{1}{3} + \frac{2}{3^{2}} + \frac{2^{2}}{3^{3}} + \ldots =1$

To see this, we associate a new point

$y=\frac{b_{1}}{2} + \frac{b_{2}}{2^{2}}+\ldots + \frac{b_{n}}{2^{n}}+\ldots$

with each point IV, where

$b_{n}=0$, if $a_{n}=0$; $b_{n}=1$; if $a_{n}=2$.

In this way, we set up a one-to-one correspondence between F and the whole interval $[0,1]$. It follows that F has the power of the continuum, as asserted. Let $A_{1}$ be the set of points III. Then,

$F = A_{1} \bigcup A_{2}$ where the set $A_{2}=F-A_{1}$ is uncountable, since $A_{1}$ is countable and F itself is not. The points of $A_{1}$ are often called “points of F of the first kind,” while those of $A_{2}$ are called “points of the second kind.”

QED.

# Solutions to System of Sets:

Reference: Introductory Real Analysis, Kolmogorov and Fomin. Dover Publications. Translated and edited by Richard A. Silverman:

Problem 1:

Let X be an uncountable set, and let $\mathscr{R}$ be the ring consisting of all finite subsets of X and their complements. Is $\mathscr{R}$ a $\sigma$-ring?

Solution 1:

Recall the definition: A ring of sets is called $\sigma$-ring if it contains the union $S= \bigcup_{n=1}^{\infty}A_{n}$ whenever it contains the sets $A_{1}, A_{2}, \ldots, A_{n}, \ldots$.

There are countably many finite subsets of X (given uncountable set) and their complements (also especially) so that the union of all these is the whole set X. Hence, X is a $\sigma$-ring as it is closed under countably many set unions. QED.

Problem 2:

Are open intervals Borel sets?

Solution 2:

Recall the following:

(1) Borel sets or B-sets are the subsets of the real line belonging to the minimal B-algebra  generated by the set of all closed intervals $[\alpha, \beta]$.

Problem 3:But a B-algebra is also a ring and is closed under differences/complements. Open intervals are complements of closed intervals. And, they belong hence to B-algebra. Note we still have to answer the question whether open sets are Borel sets and so we have to check if they are contained in a minimal B-algebra. As per the following theorem: Given any non-empty system of sets $\mathscr{S}$, there is a unique irreducible (minimal) B-algebra $\mathscr{B}(\mathscr{P})$ generated by the system $\mathscr{S}$ or the Borel closure of $\mathscr{P}$. So, yes indeed open sets are Borel sets. QED.

Problem 3A:

Let $y=f(x)$ be a function defined on a set M and taking values in a set N. Let $\mathscr{M}$ be a system of subsets of M, and let $f(\mathscr{M})$ denote the system of all images $f(A)$ of sets $A \in \mathscr{M}$. Moreover, let $\mathscr{N}$ be a system of subsets of N, and let $f^{-1}(\mathscr{N})$ denote the system of all preimages $f^{-1}(B)$ of sets $B \in \mathscr{N}$. Prove that:

i) If $\mathscr{N}$ is a ring, so is $f^{-1}(\mathscr{N})$.

ii) If $\mathscr{N}$ is an algebra, so is $f^{-1}(\mathscr{N})$

iii) If $\mathscr{N}$ is a B-algebra, so is $f^{-1}(\mathscr{N})$

iv) $\mathscr{R}(f^{-1}(\mathscr{N})) = f^{-1}(\mathscr{R}(\mathscr{N}))$ where $\mathscr{R}$ stands for a ring.

v) $\mathscr{B}(f^{-1}(\mathscr{N})) = f^{-1}(\mathscr{B}(\mathscr{N}))$ where $\mathscr{B}$ is irreducible or minimum Borel algebra generated by a non-empty system of sets $\mathscr{S}$.

Problem 3B:

Prove the following:

i) If $\mathscr{M}$ is a ring, so is $f(\mathscr{M})$.

ii) If $\mathscr{M}$ is an algebra, so is $f(\mathscr{M})$.

iii) If $\mathscr{M}$ is Borel-algebra, so is $f(\mathscr{M})$.

iv) $\mathscr{R}(f(\mathscr{M}))=f(\mathscr{R}(\mathscr{M}))$

v) $\mathscr{B}(f(M))=f(\mathscr{B}(\mathscr{M}))$ where $\mathscr{B}$ is irreducible minimum Borel algebra generated by a non-empty system of sets $\mathscr{S}$.

Solution 3A:

Part 1: Given mapping $f: M \rightarrow N$, that is, $y=f(x)$, Let $Y_{1}, Y_{2}, Y_{3}, \ldots$ be subsets of N and let $X_{1}, X_{2}, X_{3}, \ldots$ be subsets of M. Then, as it is given that $\mathscr{N}$ is a ring then :

$Y_{1} \bigcap Y_{2} \in \mathscr{N}$

$Y_{1}\bigcup Y_{2} \in \mathscr{N}$

$Y_{1}-Y_{2} \in \mathscr{N}$

$Y_{1} \triangle Y_{2} = (Y_{1}-Y_{2}) \bigcup (Y_{2}-Y_{1}) \in \mathscr{N}$

The above implies:

$f^{-1}(Y_{1}\bigcap Y_{2}) \in f^{-1}(\mathscr{N})$, that is, $f^{-1}(Y_{1}) \bigcap f^{-1}(Y_{2}) \in f^{-1}(\mathscr{N})$, that is, $X_{1} \bigcap X_{2} \in \mathscr{M}$.

Also, $Y_{1} \bigcup Y_{2} \in \mathscr{N}$ so that $f^{-1}(Y_{1} \bigcup Y_{2}) \in f^{-1}(\mathscr{N})$, that is, $f^{-1}(Y_{1}) \bigcup f^{-1}(Y_{2}) \in f^{-1}(\mathscr{N})$, that is, we conclude $X_{1} \bigcup X_{2} \in \mathscr{M}$.

Also, $Y_{1}-Y_{2} \in \mathscr{N}$ as $\mathscr{N}$ is a ring; so that $f^{-1}(Y_{1}-Y_{2}) \in f^{-1}(\mathscr{N})$ so that for some $X_{0}$, which is $f^{-1}(Y_{1}-Y_{0}) \longrightarrow f^{-1}(Y_{1}) - f^{-1}(Y_{0}) \in \mathscr{M}$

So, we have shown that the system of sets $\mathscr{M}$ is closed under set union, set intersection, set difference, and hence, also set symmetric difference. Hence, it is is a ring. QED.

Part ii:

We have already shown that $f^{-1}(\mathscr{N})$ is a ring, now $\mathscr{N}$ is also an algebra meaning a ring of sets with a unit element. Let it contain a unit E $\in \mathscr{N}$ such that if $Y_{0}$ is any set belonging to $\mathscr{N}$, then $Y_{0} \bigcap E = Y_{0}$. so that $f^{-1}(Y_{0} \bigcap E) = f^{-1}(Y_{0})$, hence, $f^{-1}(Y_{0}) \bigcap f^{-1}(E) = f^{-1}(Y_{0})$, which in turn implies that $f^{-1}(E)$ is a unit of $f^{-1}(\mathscr{N})$. Hence, $f^{-1}(\mathscr{N})$ is a ring of sets with a unit element, or in other words, an algebra of sets.

Part iii:

To prove that: if $\mathscr{N}$ is a Borel algebra, so is $f^{-1}(\mathscr{N})$. From parts i and ii above, $f^{-1}(\mathscr{N})$ is a ring with a unit or in other words, $f^{-1}(\mathscr{N})$ is an algebra. Moreover, if it contains $\bigcup_{k=1}^{\infty}Y_{k}$ whenever it contains the sets $Y_{1}, Y_{2}, Y_{3}, \ldots, Y_{n}, \ldots$ all of which are members of $\mathscr{N}$.

But for an arbitrary (finite or infinite) collection of sets:

$f^{-1}(\bigcup_{\alpha}A_{\alpha}) = \bigcup_{\alpha}f^{-1}(A_{\alpha})$

Now, $\bigcup_{k=1}^{\infty}Y_{k} \in \mathscr{N}$

hence $f^{-1}(\bigcup_{k=1}^{\infty}Y_{k}) \in f^{-1}(\mathscr{N})$

Consequently, $\bigcup_{k=1}^{\infty}f^{-1}(Y_{k}) \in f^{-1}(\mathscr{N})$, which in turn implies that $\mathscr{N}$ is a Borel algebra also. QED.

Part iv and part v:

Using parts i and iii above, these can be very easily proved.

Solutions 3b:

i) Prove: If $\mathscr{M}$ is a ring, so is $f(\mathscr{M})$. Note that $f: M \rightarrow N$, $y=f(x)$. Proof (i): As $\mathscr{M}$ is a system of subsets of M. and it is a ring, so if $A_{1} \in \mathscr{M}$, and $A_{2} \in \mathscr{M}$, it implies that $A_{1} \bigcup A_{2} \in \mathscr{M}$; $A_{1} \bigcap A_{2} \in \mathscr{M}$; $(A_{1}-A_{2}) \in \mathscr{M}$. We know that $f(A_{1}\bigcup A_{2}) = f(A_{1}) \bigcup f(A_{2}) \in f(\mathscr{M})$ and $f(A_{1} \bigcap A_{2}) \subset f(A_{1}) \bigcap f(A_{2}) \in f(\mathscr{M})$. And, quite obviously, $f(A_{1}-A_{2}) = f(A_{1}) - f(A_{2}) \in f(\mathscr{M})$ so that it is true that if $\mathscr{M}$ is a ring, then $f(\mathscr{M})$ is also a ring.

ii) Check if it is true or false: if $\mathscr{M}$ is an algebra, so is $f(\mathscr{M})$. Argument: As $\mathscr{M}$ is an algebra, it is already a ring. But additionally, $\mathscr{M}$ contains an element E such that $A_{1} \bigcap E= A_{1}$ (whenever $A_{1} \in \mathscr{M}$ but $f(A_{1} \bigcap E) \subset f(A_{1}) \bigcap f(E) \in f(\mathscr{M})$ but it need not be true that $f(E)$ is a unit element of $f(\mathscr{M})$. So, we can say that in case $\mathscr{M}$ is an algebra, then $f(\mathscr{M})$ may or may not be an algebra. QED.

iii) Check if it is true or false: If $\mathscr{M}$ is a Borel algebra, then so also $f(\mathscr{M})$ is also a Borel algebra. Argument: As in previous case, $f(\mathscr{M})$ may not have a unit element. So, $f(\mathscr{M})$ need not be a Borel algebra. QED.

iv) $\mathscr{R}(f(\mathscr{M})) = f(\mathscr{R}(\mathscr{M}))$. Check if this is true or false. Argument: Let $f(A_{1}) \in f(\mathscr{M})$, $f(A_{2}) \in \mathscr{M}$. By definition of a ring, $f(A_{1}) \bigcup f(A_{2}) \in f(\mathscr{M})$, $f(A_{1}) \bigcap f(A_{2}) \in f(\mathscr{M})$, $f(A_{1}) - f(A_{2}) \in f(\mathscr{M})$, hence we get the following: $f(A_{1} \bigcup A_{2}) \subset f(\mathscr{M})$; $f(A_{1} \bigcap A_{2}) \subset f(A_{1}) \bigcap f(A_{2}) \in f(\mathscr{M})$ and $f(A_{1}-A_{2}) = f(A_{1}) - f(A_{2}) \in f(\mathscr{M})$. This in turn means that $A_{1} \bigcup A_{2} \in \mathscr{M}$, $A_{1} \bigcap A_{2} \in \mathscr{M}$ and $A_{1} - A_{2} \in \mathscr{M}$. Hence, $LHS \subset RHS$. Similarly, we can show that $RHS \subset LHS$. QED.

v) $\mathscr{B}(f(\mathscr(M))) = f(\mathscr{B}(\mathscr{M}))$. Check if this is true or false. (We will discuss this later; meanwhile, if you wish, please try as homework and let me know).

Cheers,

Nalin Pithwa.

# Introductory Real Analysis: Exercise 3 solutions: related to set theory

Problem 1:

Exhibit both a partial ordering and a simple ordering on the set of complex numbers.

Solution I:

Recall the definition of a partial ordering: A binary relation R on a set M is said to be a partial ordering (and the set M itself is said to be partially ordered) if:

(i) R is reflexive (aRa for every $a \in M$) (ii) R is transitive (aRb and bRc together imply aRc) (iii) aRb and bRa together imply $a=b$ (anti symmetric).

Recall the definition of a simple ordering or totally ordered or (just) ordered set: a set M is said to be ordered if it is a partially ordered set and if, given any two distinct elements $a, b \in M$, either $a < b$ or $b < a$.

Let us consider the following relation R on the set of complex numbers, C:

consider any two elements $z_{1}, z_{2} \in C$ such that $z_{1}=x_{1}+i y_{1}$ and $z_{2} = x_{2}=iy_{2}$. We define a relation $z_{1}Rz_{2}$ if and only if $x_{1} \leq x_{2}$ and $y_{1} \leq y_{2}$. Clearly, this is both a partial ordering on C as well as total ordering on C.

Now consider the following relation K on the set of complex numbers:

We define the relation K as: $z_{1}Kz_{2}$ if and only if $\frac{x_{1}}{y_{1}} = \frac{x_{2}}{y_{2}}$ where $x_{1}, y_{1}, x_{2}, y_{2} \in \Re$. Clearly, if any one of $y_{1}$ or $y_{2}$ is zero, then the complex numbers $z_{1}$ and $z_{2}$ are non-comparable. So this could be a partial ordering but not a total ordering on the set of complex numbers.

Problem 2:

What is the minimal element of the set of all subsets of a given set X partially ordered by set inclusion. What is the maximal element?

Solution 2:

Recall the definition: An element “a” of a partially ordered set is said to be maximal if $aRb$ implies $b=a$ and minimal if $bRa$ implies b=a.

Note that in the above definition, element “a” is our given/chosen element to be compared with other elements of the set. So, clearly, if the partial ordering is set inclusion, the minimal element is the null set $\phi$ and the maximal element is the given set X itself.

Problem 3:

A partially ordered set M is said to be a directed set if, given any two elements $a, b \in M$, there is an element $c \in M$ such that $a \leq c$ and $b \leq c$. Are the following partially ordered sets directed sets also ?

PS: in the above we use the notation $aRb$ or equivalently, $a \leq b$ to mean one and the same thing.

Question 3a:  Any set M can be trivially partially ordered by setting $a \leq b$ if and only if $a=b$.

Answer 3a: Clearly, this is a trivial directed set.

Question 3b: Let M be the set of all continuous functions $f, g, \ldots$ defined in a closed interval $[\alpha, \beta]$. Then, we get a partial ordering by setting $f \leq g$ if and only if $f(t) \leq g(t)$ for every $t \in [\alpha, \beta]$. Is this also a directed set ?

Answer 3b: We need to choose two arbitrary elements say $f_{1}$ and $f_{2}$ of the given set of continuous functions on $[\alpha, \beta]$. Our desired “element” could be $f_{3}= |f_{1}(t)|+|f_{2}(t)|$ again defined on $[\alpha, \beta]$. Then, the given set becomes a directed set.

Note that the function $y=f(x) = |x|$ is continuous at zero also (but not differentiable at zero).

Question 3c:

Set inclusion.

Answer 3: Clearly, given any two arbitrary subsets, both are contained always in the given big set M or X. So, this is again a directed set.

Question 3d: The set of all integers greater than 1 is partially ordered if $a \leq b$ means that “b is divisble by a.”

Answer 3: So let us consider two arbitrary positive integers, both greater than 1, call them a and b. We want to know if there exist a positive integer, also greater than 1, call it c such that $a \leq c$ and $b \leq c$? That is, “c is divisible by a” and “c is divisible by b”. Clearly, c can be least common multiple of a and b. So, yes, this is also a directed set.

Problem 4:

By the greatest lower bound of two elements a and b of a partially ordered set M, we mean an element $c \in M$ such that $c \leq a$ and $c \leq b$ and there is no element $d \in M$ such that $c < d \leq a$, $d \leq b$. Similarly, by the least upper bound of a and b, we mean an element $c \in M$ such that $a \leq c$, $b \leq c$, and there is no element $d \in M$ such that $a \leq d < c$ and $b \leq d$. By a lattice is meant a partially ordered set any two elements of which have both a greatest lower bound and a least upper bound. Prove that the set of all subsets of a given set M, partially ordered by set inclusion, is a lattice. What is the set theoretic meaning of the greatest lower bound and least upper bound of two elements of this set ?

Solution 4:

It can be easily checked that the null set is the greatest lower bound and the set M itself if the least upper bound. (PS: these are the minimal and maximal elements of this set M respectively.)

Problem 5:

Prove that an order-preserving mapping of one ordered set onto another is automatically an isomorphism.

Solution 5:

By definition.

Problem 6:

Prove that ordered sums and products of ordered sets are associative, that is, prove that if $M_{1}, M_{2}$ and $M_{3}$ are ordered sets, then

$(M_{1}+M_{2})+M_{3}=M_{1}+(M_{2}+M_{3})$

$(M_{1}.M_{2}).M_{3} = M_{1}.(M_{2}.M_{3})$

PS: comment: this allows us to drop the parentheses in writing ordered sums and products.

where the operations are as defined below:

Let $M_{1}$ and $M_{2}$ be two ordered sets of type $\theta_{1}$ and $\theta_{2}$ respectively. Then, we can introduce an ordering in the union $M_{1} \bigcup M_{2}$ of the two sets by assuming that: (i) a and b have the same ordering as in $M_{1}$ if $a, b \in M_{1}$ (ii) a and b have the same orderiing as in $M_{2}$ if $a, b \in M_{2}$ (iii) $a \leq b$ if $a \in M_{1}$ and $b \in M_{2}$.

In this case/example, the ordered sum or union of three ordered sets $M_{1}, M_{2}, M_{3}$ would mean that any two elements a, b would have the same ordering if $a, b \in M_{1}$, or $a, b \in M_{2}$ or $a,b \in M_{3}$. Also, comparing with ordered sum of two ordered sets, we can say that $a \leq b$ if and only if $a \in M_{1}, b \in M_{2}$ or $a \in M_{1}, b \in M_{3}$, or $a \in M_{2}, b \in M_{3}$. In such a case, the ordered sum is also associative.

The operation of ordered product of two ordered sets is defined as follows: $M_{1}. M_{2}$ is the set of all pairs $(a,b)$ where $a \in M_{1}$ and $b \in M_{2}$ ordered in such a way that (i) $(a_{1}, b_{1}) < (a_{2}, b_{2})$ if $b_{1} < b_{2}$ for arbitrary $a_{1}, a_{2}$, and (ii) $(a_{1},b) < (a_{2}, b)$ if $a_{1} < a_{2}$.

So, if we have three ordered sets $M_{1}, M_{2}, M_{3}$, we can define their ordered product $M_{1}.M_{2}.M_{3}$ as consisting of all ordered triples such that (i) $(a_{1}, b_{1}, c_{1}) < (a_{2}, b_{2}, c_{2}) < (a_{3}, b_{3}, c_{3})$ if and only if $b_{1} and $c_{1} < c_{2} < c_{3}$ for arbitrary $a_{1}, a_{2}, a_{3}$, and (ii) $(a_{1}, b, c) < (a_{2}, b, c) < (a_{3}, b, c)$ if $a_{1}; $(a, b_{1}, c) < (a, b_{2}, c) < (a, b_{3}, c)$ if $b_{1}; $(a,b, c_{1})< (a,b,c_{2}) < (a,b, c_{3})$ of $c_{1} < c_{2} < c_{3}$. In such a case, the product is well-defined and is also associative. (any comments ??)

Problem 7:

Construct well-ordered sets with ordinals

$\omega + n$, $\omega + \omega$, $\omega + \omega + n$, $\omega + \omega + \omega$, $\ldots$

Show that the sets are all countable.

Solution 7:

$\omega + n$: $\{ 1,2,3,\ldots, a_{1}, a_{2}, \ldots, a_{n}\}$. This is countable because there can be bijection from the set $Z^{+}$ to the given set: $1 \rightarrow a_{1}$, $2 \rightarrow a_{2}$, $\ldots$, $n \rightarrow a_{n}$, $n+1 \rightarrow 1$, $n+2 \rightarrow 2$, $\ldots$

$\omega + \omega$: $\{1,2,3, \ldots, 1,2,3, \ldots \}$, this is also countable as it can be put in one-to-one correspondence with $Z^{+}$.

$\omega + \omega + n$; $\{ 1,2, 3, \ldots, 1, 2, 3, \ldots, a_{1}, a_{2}, a_{3}, \ldots, a_{n}\}$. This is also countable in the same manner as the first example.

$\omega + \omega + \omega$: $\{ 1,2, 3, \ldots, 1,2, 3, \ldots, 1,2,3, \ldots\}$: this can be put in one-to-one correspondence with $Z^{+}$

Problem 8:

Construct well-ordered sets with ordinals

$\omega.n$, $\omega^{2}$, $\omega^{2}.n$, $\omega^{3}$, $\ldots$.

Show that the sets are all countable.

Solution 8:

First let us recall the definition: An ordered set M is said to be well-ordered if every non empty subset A of M has a smallest (or, first) element, that is, an element $\mu$ such that $\mu < a$ for every $a \in A$.

Ans 1: $\omega.n$ is set : $M= \{(1,a_{1}), (2,a_{2}), (3, a_{3}), \ldots, (n, a_{n}), (n+1, a_{1}), (n+2, a_{2}), \ldots \}$. This set is ordered clearly by the definition of ordered product of two ordered sets; and this is well-ordered if we define the first element to be $(1, a_{1})$.

Ans 2: $\omega^{2} = \omega. \omega$: this set can be constructed as $M= \{ (1,1), (1,2),(1,3), \ldots, (2,1), (2,2), (2,3), \ldots, \}$ and we can define the first element as $(1,1)$ so that it becomes well-ordered under the usual definition of product of two ordered sets.

Ans 3: $\omega^{2}.n = \omega. \omega. n$: this set can be constructed as follows as an ordered product of three ordered sets: $M = \{ (1,1,a_{1}), (1,1,a_{2}), (1,1,a_{3}), \ldots, (1,1,a_{n}), (1,2,a_{1}), (1,2,a_{2}), (1,2,a_{3}), \ldots, (1, 2, a_{n}), \ldots\}$. We see that this is also well-ordered if we define the first element to be $(1,1,a_{1})$ and “order or list or count” them as shown. (any comments ? )

Ans 4: $\omega^{3}= \omega.\omega.\omega$: this can be well-ordered as above with clearly the first element to be $(1,1,1)$. (any comments?)

PS: the explicit listing shows that all the above sets are clearly countable.

More later,

Cheers,

Nalin Pithwa

# Metric Spaces: Tutorial problems: Solutions: Part I

Reference: Introductory Real Analysis by Kolmogorov and Fomin.

Reference: Analysis by Walter Rudin.

Reference: Introduction to Analysis by Rosenlicht.

Reference: Topology by Hocking and Young.

The problems proposed were as follows:

1. Given a metric space $(X,\rho)$, prove that (1a) $|\rho(x,z)-\rho(y,u)| \leq \rho(x,y) + \rho(z,u)$ where $x,y,z,u \in X$; (1b) $|\rho(x,z) -\rho(y,z)| \leq \rho(x,y)$ where $x,y,z \in X$.
2. Verify that $(\Sigma_{k=1}^{n}a_{k}b_{k})^{2} = \Sigma_{k=1}^{n}a_{k}^{2} \times \Sigma_{k=1}^{n}b_{k}^{2} - \frac{1}{2}\Sigma_{i=1}^{n}\Sigma_{j=1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}$. Deduce the Cauchy-Schwarz inequality from the above identity. (refer second last blog from here). For convenience, we reproduce it here again: consider the set of all ordered n-tuples such that $x=(x_{1}, x_{2}, \ldots, x_{n})$ and $y=(y_{1}, y_{2}, \ldots, y_{n})$ and $z=(z_{1}, z_{2}, \ldots, z_{n})$ and let $a_{k}=x_{k}-y_{k}$, $b_{k}=y_{k}-z_{k}$ where $k=1,2, \ldots, n$: then Cauchy Schwarz inequality tells us that : $(\Sigma_{k=1}^{n}a_{k}b_{k})^{2} \leq \Sigma_{k=1}^{n}a_{k}^{2}\Sigma_{k=1}^{n}b_{k}^{2}$
3. Verify that $(\int_{a}^{b}x(t)y(t)dt)^{2} = \int_{a}^{b}x^{2}(t)dt \int_{a}^{b}y^{2}(t)dt -\frac{1}{2}\int_{a}^{b}\int_{a}^{b}(x(s)y(t)-x(t)y(s))^{2}dsdt$.

Deduce Schwarz’s inequality from this identity. (refer second last blog from here). To aid the memory, we present Schwarz’s inequality below: $(\int_{a}^{b}x(t)y(t))^{2} \leq \int_{a}^{b}x^{2}(t)dt \int_{a}^{b}y^{2}(t)dt$

4. Identify the fallacy in example 10, if $p \leq 1$? (refer second last blog from here). Hint: Show that Minkowski’s inequality fails for $p \leq 1$. To help recall, we are reproducing Minkowski’s inequality below: Consider again the set of all ordered n-tuples, and let $x=(x_{1}, x_{2}, \ldots, x_{n})$, and $y=(y_{1}, y_{2}, \ldots, y_{n})$ and $z=(z_{1}, z_{2}, \ldots, z_{n})$ and again further let $a_{k}=x_{k}-y_{k}$ and $b_{k}=y_{k}-z_{k}$, where $k=1,2, \ldots, n$ then Minkowski’s inequality tells us that: $(\Sigma_{k=1}^{n}|a_{k}+b_{k}|^{p})^{\frac{1}{p}} \leq (\Sigma_{k=1}^{n}|a_{k}|^{p})^{\frac{1}{p}} + (\Sigma_{k=1}^{n}|b_{k}|^{p})^{\frac{1}{p}}$

5. Prove that the following metric : Consider two points $x=(x_{1}, x_{2}, \ldots, x_{n})$ and $y=(y_{1}, y_{2}, \ldots, y_{n})$ with the metric function: $\rho_{0}(x,y) = \max_{1 \leq k \leq n}|x_{k}-y_{k}|$

is the limiting case of the following metric: consider the set of all ordered n-tuples $x = (x_{1}, x_{2}, \ldots, x_{n})$ of real numbers with the metric function now defined by: $\rho_{p}(x,y) = (\Sigma_{k=1}^{\infty}|x_{k}-y_{k}|^{p})^{\frac{1}{p}}$

in the sense that

$\rho_{0}(x,y) = \max_{1 \leq k \leq n}|x_{k}-y_{k}| = \lim_{p \rightarrow \infty}(\Sigma_{k=1}^{\infty}|x_{k}-y_{k}|^{p})^{\frac{1}{p}}$

6. Starting from the following inequality: $ab \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}$, deduce Holder’s integral inequality as given by:

$\int_{a}^{b} x(t)y(t)dt \leq (\int_{a}^{b}|x(t)|^{p}dt)^{\frac{1}{p}}(\int_{a}^{b}|y(t)|^{q})^{\frac{1}{q}}$ where $\frac{1}{p} + \frac{1}{q} =1$ valid for any function $x(t)$ and $y(t)$ such that the integrals on the right exist.

7.  Use Holder’s integral inequality to prove Minkowski’s integral inequality:

$(\int_{a}^{b}|x(t)+y(t)|^{p}dt)^{\frac{1}{p}} \leq (\int_{a}^{b}|x(t)|^{p}dt)^{\frac{1}{p}} + (\int_{a}^{b}|y(t)|^{p}dt)^{\frac{1}{p}}$

where $p \geq 1$.

8. Exhibit an isometry between the spaces $C_{[0,1]}$ and $C_{[1,2]}$.

9. Verify that the following is a metric space: all bounded infinite sequences $x=(x_{1}, x_{2}, x_{3}, \ldots)$ of elements of $\Re$, with $\rho(x,y) = lub \{ |x_{1}-y_{1}|,|x_{2}-y_{2}|,|x_{3}-y_{3}|, \ldots\}$

10. Verify that $(\Re \times \Re, \rho)$ is a metric space where $\rho((x,y),(x^{'}, y^{'})) = |y|+|y^{'}|+|x-x^{'}|$ when $x \neq x^{'}$ and is $|y-y^{'}|$ when $x=x^{'}$. Illustrate by diagrams in the plane $E^{2}$ or $R^{2}$ what the open balls of this metric space are.

11. Show that a one-to-one transformation $f: S \rightarrow T$ of a space S onto a space T is a homeomorphism if and only if both f and $f^{'}$ are continuous.

The solutions are as follows:

Problem 1:

1. Given a metric space $(X,\rho)$, prove that (1a) $|\rho(x,z)-\rho(y,u)| \leq \rho(x,y) + \rho(z,u)$ where $x,y,z,u \in X$; (1b) $|\rho(x,z) -\rho(y,z)| \leq \rho(x,y)$ where $x,y,z \in X$.

Solution 1:

Part I: TPT: $|\rho(x,z)-\rho(y,u)| \leq \rho(x,y) + \rho(z,u)$

Proof: We know that: $\rho(x,z) \leq \rho(x,u)+\rho(z,u)$, which also means that, $\rho(x,z) \leq \rho(x,y) + \rho(y,u) + \rho(z,u)$, that is, $\rho(x,z)-\rho(y,u) \leq \rho(x,y)+\rho(z,u)$. On the other hand, we also know that $\rho(y,u) \leq \rho(y,z) + \rho(z,u)$, which also means that, $\rho(y,u) \leq \rho(x,y) + \rho(x,z) + \rho(z,u)$, that is, $-\rho(x,z) + \rho(y,u) \leq \rho(x,y) + \rho(z,u)$. Combining the two smallish results, we get $|\rho(x,z)-\rho(y,u)| \leq \rho(x,y) + \rho(z,u)$. QED.

Part II: TPT: $|\rho(x,z)-\rho(y,z)| \leq \rho(x,y)$. Proof : We know that $\rho(x,z) \leq \rho(x,y)+\rho(y,z)$, that is, $\rho(x,z) - \rho(y,z) \leq \rho(x,y)$. On the other hand, we know that $\rho(y,z) \leq \rho(y,z)$, that is, $\rho(y,z) \leq \rho(x,y) +\rho(x,z)$, that is, $-\rho(x,z)+\rho(y,z) \leq \rho(x,y)$. Combining the two smallish results, we get $|\rho(x,z) -\rho(y,z)| \leq \rho(x,y)$ where $x,y,z \in X$. QED.

PS: Now that the detailed proof is presented, try to interpret the above two results “geometrically.”

Problem 2 was:

Verify that $(\Sigma_{k=1}^{n}a_{k}b_{k})^{2} = \Sigma_{k=1}^{n}a_{k}^{2} \times \Sigma_{k=1}^{n}b_{k}^{2} - \frac{1}{2}\Sigma_{i=1}^{n}\Sigma_{j=1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}$. Deduce the Cauchy-Schwarz inequality from the above identity. (refer second last blog from here). For convenience, we reproduce it here again: consider the set of all ordered n-tuples such that $x=(x_{1}, x_{2}, \ldots, x_{n})$ and $y=(y_{1}, y_{2}, \ldots, y_{n})$ and $z=(z_{1}, z_{2}, \ldots, z_{n})$ and let $a_{k}=x_{k}-y_{k}$, $b_{k}=y_{k}-z_{k}$ where $k=1,2, \ldots, n$: then Cauchy Schwarz inequality tells us that : $(\Sigma_{k=1}^{n}a_{k}b_{k})^{2} \leq \Sigma_{k=1}^{n}a_{k}^{2}\Sigma_{k=1}^{n}b_{k}^{2}$.

Proof of problem 2: We present a proof based on the principle of mathematical induction. Let us check the proposition for $n=1$: Then, $LHS=(a_{1}b_{1})^{2}=a_{1}^{2}b_{1}^{2} = RHS = a_{1}^{2}b_{1}^{2}-\frac{1}{2} \times 0$. So now let the proposition be true for $m=n, m \in N, m >1$. Then, the following holds true:

$(\Sigma_{k=1}^{m}a_{k}b_{k})^{2} = \Sigma_{k=1}^{m}a_{k}^{2}\Sigma_{k=1}^{m}b_{k}^{2} -\frac{1}{2}\Sigma_{i=1}^{m}\Sigma_{j=1}^{m}(a_{i}b_{j}-a_{j}b_{i})^{2}$.

Now, we want to prove the proposition for $n=m+1$, that is, we need to prove that:

$(\Sigma_{k=1}^{m+1}a_{k}b_{k})^{2}=\Sigma_{k=1}^{m+1}a_{k}^{2}\Sigma_{k=1}^{m+1}b_{k}^{2} - \frac{1}{2}\Sigma_{i=1}^{m+1}\Sigma_{j=1}^{m+1}(a_{i}b_{j}-a_{j}b_{i})^{2}$.

Now, let us see what we need really; we need to add some(same) terms to both sides of the proposition for $n=m$ and manipulate to derive the proposition for $n=m+1$:

Further notice that

$(\Sigma_{k=1}^{m+1}a_{k}b_{k})^{2}=(a_{1}b_{1}+a_{2}b_{2}+ \ldots+a_{m}b_{m}+a_{m+1}b_{m+1})^{2}=\Sigma_{k=1}^{m}a_{k}^{2}b_{k}^{2}+a_{m+1}^{2}b_{m+1}^{2}+2(a_{1}b_{1})(a_{2}b_{2}+a_{3}b_{3}+\ldots+a_{m}b_{m}+a_{m+1}b_{m+1})+2(a_{2}b_{2})(a_{3}b_{3}+a_{4}b_{4}+\ldots+a_{m}b_{m}+a_{m+1}b_{m+1}) + 2a_{3}b_{3}(a_{4}b_{4}+a_{5}b_{5}+ \ldots + a_{m+1}b_{m=1})+ \ldots + 2a_{m}b_{m}a_{m+1}b_{m+1}$

Comparing the algebraic statements for proposition for $n=m$ and $n=m+1$, we can clearly see which terms are to be added to both sides of the proposition for $n=m$,

(please fill in the few missing last details…a good deal of algebraic manipulation…some concentrated effort …:-))

From this the Cauchy Schwarz inequality follows naturally.

PS: I have no idea how to solve question 3. If any reader helps, I would be obliged.

PS: I have yet to try the other questions.

Cheers,

Nalin Pithwa

# Metric Spaces: some additional stuff

Reference: Introductory Real Analysis by Kolmogorov and Fomin, Dover Pub.

Reference: Analysis by Walter Rudin

Section 5.2: Continuous mappings and homeomorphisms. isometric spaces:

Let f be a mapping of one metric space X into another metric space Y, so that f associates an element $y=f(x) \in Y$ with each element $x \in X$. Then, f is said to be continuous at the point $x_{0}$ if, given any $\epsilon >0$, there exists a $\delta >0$ such that

$\rho^{'}(f(x),f(x_{0})) < \epsilon$ whenever $\rho(x,x_{0}) < \delta$.

In the above, the metric $\rho$ is in X and the metric $\rho^{'}$ is in Y. The mapping f is said to be continuous on X if it is continuous at every point in domain X.

Remarks:

This definition reduces to the usual definition of continuity familiar from calculus if X and Y are both numerical sets, that is, if f is a real function defined on some subset of this real line.

Given two metric spaces X and Y, let f be one-to-one mapping of X onto X, and suppose $f$ and $f^{-1}$ are both continuous. Then, f is called a homeomorphic mapping, or simply a homeomorphism between X and Y. Two spaces X and Y are said to be homeomorphic if there exists a homeomorphism between them.

Example :

are said to be isometric to each other. The function $y=f(x) = \frac{2}{\pi}\arctan{x}$

establishes a homeomorphism between the whole real line $(-\infty, \infty)$ and the open interval $(-1,1)$.

DEFINITION 2:

A one-to-one mapping f of one metric space $R=(X,\rho)$ onto another metric space $R^{'} = (X, \rho^{'})$ is said to be an isometric mapping (or simply, an isometry) if

$\rho(x_{1}, x_{2}) = \rho^{'}(f(x_{1}), f(x_{2}))$

for all $x_{1}, x_{2} \in R$. Correspondingly, the spaces R and $R^{'}$ are said to be isometric to each other.

Thus, if $R$ and $R^{'}$ are isometric, the “metric relations” between the elements of R are the same as those between the elements of $R^{'}$, that is, R and $R^{'}$ differ only in the explicit nature of their elements (this distinction is unimportant from the viewpoint of metric space theory). From now on, we will not distinguish between isometric spaces regarding them simply as identical.

Remark:

We will discuss continuity and homeomorphism from a more general viewpoint later.

The next blog poses some exercise problems.

Regards,

Nalin Pithwa.

# II Metric Spaces:

Reference: Introductory Real Analysis Kolmogorov and Fomin, translated by Richard A Silverman, Dover Publications.

Reference: Analysis by Walter Rudin, Third Edition.

Chapter II Metric Spaces:

V: Basic Concepts:

Section 5.1: Definitions and examples:

One of the most important operations in mathematical analysis is the taking of limits. Here what matters is not so much as the algebraic nature of the real numbers (that is, the fact that real numbers form a field), but rather the fact that distance from one point to another on the real line(or, in two or three dimensional space) is well-defined and has certain properties. Roughly speaking, a metric space is a set equipped with a distance (or, ‘metric’) which has these same properties. More exactly, we have:

Definition 1:

By a metric space is meant a pair $(X, \rho)$ consisting of a set X and a distance $\rho$, that is, a single valued, nonnegative, real function $\rho(x,y)$ defined for all $x, y \in X$ which has the following three properties:

1. $\rho(x,y)=0$ if and only if $x=y$
2. Symmetry: $\rho(x,y)=\rho(y,x)$
3. Triangle Inequality: $\rho(x,z) \leq \rho(x,y)+\rho(y,z)$

We will often refer to the set X as a “space” and its elements x, y, …, as “points.” Metric spaces are usually denoted by a single letter, like

$R=(X,\rho)$

or, even by the same letter X as used for the underlying space, in cases where there is no possibility of confusion.

Example 1:

Setting $\rho(x,y)=0$, if $x=y$ and $\rho(x,y)=1$, when $x \neq y$, where x and y are elements of an arbitrary set X, we obviously get a metric space, which might be called a “discrete space” or a “space of isolated points.”

Check: does this satisfy all the three axioms of a metric space: clearly, the first axiom is true. So, also the second axiom is true because $\rho(x,y)=\rho(y,x)$ is zero when $x = y$ and is 1 when $x \neq y$ or $y \neq x$.

Now we have to check: $\rho(x,z) \leq \rho(x,y) + \rho(y,z)$. Case I: if $x=z$, LHS is zero and again RHS could be zero or one depending on y and z. In all cases, the inequality holds. Case II: If $x \neq z$, then LHS is 1. Now, $x \neq y$, then $\rho(x,y)$ is 1 and depending on z, $\rho(y,z)$ is zero or 1. So, we get LHS = RHS = 1 or LHS=1 less than RHS, which is 2.

So, yes indeed this is a well-defined metric function.

Some remarks: To think further:  Suppose we are given the following function : $f(x)=1$ when $x \in \mathscr{Q}$ and $f(x)=0$ when $x \in \mathscr{Q^{'}}$. Can what can we say about this function with respect to the above discrete space ? What are the limit points of such a function ? Is such a function continuous (if so, at which points) in this metric space? Is it dense in this metric space?

Example 2:

The set of all real numbers with distance $\rho(x-y) = |x-y|$ is a metric space, which we denote by $R^{1}$ —- one dimensional real line.

Check: is this a well-defined metric ? Clearly, axiom 1 holds true because $|x-x|=0$ and axiom 2 is true because $\rho(x-y) = |x-y| = |-(x-y)|=|y-x|=\rho(y-x)$. Now, we need to check: $\rho(x,z) \leq \rho(x,y) + \rho(y,z)$. Here, LHS is $|x-z|=|x-y+y-z| = |(x-y) + (y-z)| \leq |x-y| + |y-z| = \rho(x,y) + \rho(y,z)$ where we have used the triangle inequality. So, axiom 3 holds true.

So, this is indeed a well-defined metric.

Example 3:

The set of all ordered n-tuples $x=(x_{1}, x_{2}, \ldots, x_{n})$ of real numbers $x_{1}, x_{2}, \ldots, x_{n}$ with distance

$\rho(x,y)= \sqrt{\Sigma_{k=1}^{n}(x_{k}-y_{k})^{2}}$ ….call this relation I

is a metric space denoted by $R^{n}$ and called n-dimensional Euclidean space (or, simply Euclidean n-space). The distance (I) obviously satisfies axioms 1 and 2 of definition of a metric. Moreover, it can be seen that (I) satisfies the third axiom also:

In fact, let $x=(x_{1}, x_{2}, \ldots, x_{n})$, $y=(y_{1}, y_{2}, y_{3}, \ldots, y_{n})$, $z=(z_{1}, z_{2}, \ldots, z_{n})$ be three points in $R^{n}$.

Futher, let $a_{k}=x_{k}-y_{k}$ and $b_{k}=y_{k}-z_{k}$ when $k=1,2,\ldots, n$.

Then, the triangle inequality takes the form:

$\sqrt{\Sigma_{k=1}^{n}(x_{k}-z_{k})^{2}} \leq \sqrt{\Sigma_{k=1}^{n}(x_{k}-y_{k})^{2}} + \sqrt{\Sigma_{k=1}^{n}(y_{k}-z_{k})^{2}}$….let us call this Relation II.

Or equivalently,

$\sqrt{\Sigma_{k=1}^{n}(a_{k}+b_{k})^{2}} \leq \sqrt{\Sigma_{k=1}^{n}a_{k}^{2}} + \sqrt{\Sigma_{k=1}^{n}b_{k}^{2}}$….call this as relation II’

It follows from the Cauchy-Schwarz inequality that $(\Sigma_{k=1}^{n}a_{k}b_{k})^{2} \leq \Sigma_{k=1}^{n}a_{k}^{2} \times \Sigma_{k=1}^{n}b_{k}^{2}$ …call this as relation III.

so that we have now,

$\Sigma_{k=1}^{n}(a_{k}+b_{k})^{2} = \Sigma_{k=1}^{n}a_{k}^{2} + 2 \Sigma_{k=1}^{n}a_{k}b_{k} + \Sigma_{k=1}^{n}b_{k}^{2} \leq \Sigma_{k=1}^{n}a_{k}^{2}+2\sqrt{\Sigma_{k=1}^{n}a_{k}^{2}\times \Sigma_{k=1}^{n}b_{k}^{2}} + \Sigma_{k=1}^{n}b_{k}^{2} = (\sqrt{\Sigma_{k=1}^{n}a_{k}^{2}} + \sqrt{\Sigma_{k=1}^{n}}b_{k}^{2})^{2}$.

Taking square roots, we get II’ and hence, II.

QED.

Example 4:

Take the same set of ordered tuples as in preceding example $x = (x_{1}, x_{2}, \ldots, x_{n})$, but this time define the distance function by $\rho_{1}(x,y)= \Sigma_{k=1}^{n}|x_{k}-y_{k}|$….call this as relation IV.

It is clear that this is also a well-defined metric function.

Check: Axiom 1 is obvious. So, also axiom 2 because $|a-b|=|b-a|$. Axiom 3 holds true because the following general inequality holds true: $|\pm x_{1} \pm x_{2} \ldots \pm x_{n}| \leq |x_{1}|+|x_{2}|+ \ldots + |x_{n}|$.

Example 5:

Take the same set as in the previous two examples, but now let us define the distance to between two points $x=(x_{1},x_{2}, \ldots, x_{n})$ and $y=(y_{1}, y_{2}, \ldots, y_{n})$ to be $\rho_{0}(x,y)= \max_{1 \leq k \leq n}|x_{k}-y_{k}|$….call this V.

This is also a well-defined metric function.

This space, denoted by $R_{0}^{n}$ is often as usual the Euclidean space $R^{n}$.

Remark: The last three examples show that it is sometimes important to use a different notation for a metric space than for the underlying set of points in the space, since the latter can be “metrized” in a variety of different ways.

Example 6:

The set $C_{[a,b]}$ of all continuous functions defined on the closed interval $[a,b]$ with distance $\rho(f,g)=-\max_{a \leq t \leq b}|f(t)-g(t)|$…call this VI. This is a metric space of great importance in analysis.

Let us verify so:

*****

****

***

This metric space and the underlying set of points in the space will both be denoted by $C_{[a,b]}$. Instead of $C_{[0,1]}$ we just write C. A space like $C[a,b]$ is often called a “function space” to emphasize that its elements are functions.

Example 7:

Let $I_{2}$ be the set of all “infinite” sequences : $x=(x_{1}, x_{2}, \ldots, x_{k}, \ldots)$ of real numbers $x_{1}, x_{2}, \ldots, x_{k}, \ldots$ satisfying the convergence condition:

$\Sigma_{k=1}^{\infty}x_{k}^{2} \leq \infty$

Note: the infinite sequence with the general term $x_{k}$ can be written as $\{ x_{k}\}$ or simply as $x_{1}, x_{2}, \ldots, x_{k}, \ldots$  (this notation is familiar from calculus). It can also be written in “point notation” as $x=(x_{1}, x_{2}, \ldots, , x_{k}, \ldots)$, that is, as an “ordered $\infty-$tuple” generalizing the notion of an ordered n-tuple. (In writing $x_{k}$) we have another use of curly brackets, but the context will always prevent any confusion between the sequence $x_{k}$  and the set whose only element is $x_{k}$).

Where distance between points is defined by

$\rho(x,y)=\sqrt{\Sigma_{k=1}^{\infty}(x_{k}-y_{k}^{2})}$…call this VII.

Clearly, VII makes sense for all $x, y \in l_{2}$ since it follows from the elementary inequality

$(x_{k} \pm y_{k})^{2}\leq 2(x_{k}^{2}+y_{k}^{2})$ that the convergence of the two series $\Sigma_{k=1}^{n}x_{k}^{\infty}$ and $\Sigma_{k=1}^{\infty}y_{k}^{2}$ also implies the convergence of the series $\Sigma_{k=1}^{\infty}(x_{k}-y_{k})^{2}$.

At the same time, we find that if the points $(x_{1}, x_{2}, \ldots, x_{k}, \ldots)$ and $(y_{1}, y_{2}, \ldots, y_{k}, \ldots)$ both belong to $l_{2}$, then so does the point:

$(x_{1}+y_{1}, x_{2}+y_{2}, \ldots, x_{k}+y_{k}, \ldots)$

(since the lim of a sum of two sequences is the sum of the individual limits)

The function VII obviously has the first two defining properties of a distance. To verify the triangle inequality, which takes the form:

$\sqrt{\Sigma_{k=1}^{\infty}(x_{k}-z_{k})^{2}} \leq \sqrt{\Sigma_{k=1}^{\infty}(x_{k}-y_{k})^{2}} + \sqrt{\Sigma_{k=1}^{\infty}(y_{k}-z_{k})^{2}}$ ….call this VIII.

for the metric VII, we first note that all three series converge, for the reason just given. Moreover, the inequality:

$\sqrt{\Sigma_{k=1}^{n}(x_{k}-z_{k})^{2}} \leq \sqrt{\Sigma_{k=1}^{n}(x_{k}-y_{k})^{2}} + \sqrt{\Sigma_{k=1}^{n}(y_{k}-z_{k})^{2}}$…call this IX, holds for all n, (as proved in Example 3 above). Taking the limit as $n \rightarrow \infty$ in IX, we get VIII, thereby satisfying the triangle inequality in $l_{2}$. Therefore, $l_{2}$ is a metric space.

Example 8:

As in Example 6, consider the set of all functions continuous on the interval $[a,b]$, but now let us define the metric by the formula:

$\rho(x,y) = (\int_{a}^{b}[x(t)-y(t)]^{2}dt)^{\frac{1}{2}}$….call this X.

The resulting metric space will be denoted by $C_{[a,b]}^{2}$. The first two axioms of the metric clearly hold, and the fact that X satisfies the triangle inequality is an immediate consequence of the following Schwarz’s inequality:

$(\int_{a}^{b}x(t)y(tdt))^{2} \leq \int_{a}^{b}x^{2}(t)dt \times \int_{a}^{b}y^{2}(t)dt$

(see Problem 3 in the exercises below), by the continuous analogue of the argument given in example 3 above.

Example 9:

Next consider the set of all bounded infinite sequences of real numbers $x=(x_{1},x_{2}, \ldots, x_{k}, \ldots)$ and let

$\rho(x,y) = \sup_{k} {|x_{k}-y_{k}|}$….call this XII.

This gives a metric space which we denote by m. The fact that XII satisfies axioms 1 and 2 of a metric space is obvious by the definition of a supremum.

Axiom 3 can be verified as follows:

***

***

Example 10:

As in example 3, consider the set of all ordered n-tuples, $x=(x_{1}, x_{2}, \ldots, x_{k}, \ldots)$, but now let the metric be given by the more general formula as follows:

$\rho_{p}(x,y)=(\Sigma_{k=1}^{n}|x_{k}-y_{k}|^{p})^{\frac{1}{p}}$….call this XIII.

where p is a fixed real number greater than or equal to 1. (Examples 3 and 4 correspond to the cases $p=2$ and $p=1$, respectively.) This gives a metric space, which we denote by $R_{p}^{n}$.

It is obvious that $\rho_{p}(x,y) = 0$ if and only if $x=y$.

It is obvious that $\rho_{p}(x,y)=\rho_{p}(y,x)$.

But, verification of the third axiom of the definition of a metric (XIII) (that is, the triangle inequality) requires a little work as follows:

Let $x=(x_{1}, x_{2}, \ldots, x_{n})$, $y=(y_{1}, y_{2}, \ldots, y_{n})$, $z=(z_{1}, z_{2}, \ldots, z_{n})$ be three points in $R_{p}^{n}$, and let:

$a_{k}=x_{k}-y_{k}$, $b_{k}=y_{k}-z_{k}$ for $k=1,2,\ldots, n$ just as in example 3. Then, the triangle inequality

$\rho_{p}(x,z) \leq \rho_{p}(x,y)+\rho_{p}(y,z)$

takes the form of Minkowski’s inequality:

$(\Sigma_{k=1}^{\infty}|a_{k}+b_{k}|^{p})^{\frac{1}{p}} \leq (\Sigma_{k=1}^{\infty}|a_{k}|^{p})^{\frac{1}{p}} + (\Sigma_{k=1}^{\infty}|b_{k}|^{p})^{\frac{1}{p}}$

Call the above inequality as XIV in the current blog article.

PS: I think the proof of the Minkowski inequality can be found in any standard text on Inequalities by B. J. Venkatachala, for example;; or, in wikipedia.

The above inequality holds true clearly for $p=1$, and hence, we assume the case $p \geq 1$.

The proof of XIV in turn again is based on Holder’s inequality:

$\Sigma_{k=1}^{n}|a_{k}b_{k}| \leq (\Sigma_{k=1}^{n}|a_{k}|^{p})^{\frac{1}{p}} (\Sigma_{k=1}^{n}|b_{k}|^{q})^{\frac{1}{q}}$

Call the above as inequality XV.

Where the numbers $p>1$, $q >1$ satisfy the condition:

$\frac{1}{p} + \frac{1}{q} =1$….call this as XVI.

We begin by observing that the inequality XV is homogeneous, that is, if it holds for any two points $(a_{1}, a_{2}, \ldots, a_{n})$ and $(b_{1}, b_{2}, \ldots, b_{n})$ then it holds for the two points $(\lambda a_{1}, \lambda a_{2}, \ldots, \lambda a_{n})$ and $(\mu b_{1}, \mu b_{2}, \ldots, \mu b_{n})$ where $\lambda$ and $\mu$ are any two real numbers. Therefore, we need only prove XV for the following case:

$\Sigma_{k=1}^{n}|a_{k}|^{p}=\Sigma_{k=1}^{n}|b_{k}|^{p}$….call this relation XVII.

Thus, assuming that XVII holds, we now have to prove that: $\Sigma_{k=1}^{n}|a_{k}b_{k}| \leq 1$…call this XVIII.

Consider the two areas $S_{1}$ and $S_{2}$, associated with the curve defined in the $\xi\eta$-plane and given by the equation:

$\eta = \xi^{p-1}$, or equivalently by the equation:

$\xi = \eta^{q-1}$

Then, clearly $S_{1}=\int_{0}^{a}\xi^{p-1}d\xi = \frac{a^{p}}{p}$, and $S_{2}=\int_{0}^{b}\eta^{q-1}d\eta = \frac{b^{q}}{q}$

Moreover, it is apparent (if we draw the figure suitably) that $S_{1}+S_{2} \geq ab$ for arbitrary positive a and b. It follows that $ab \leq \frac{a^{p}}{q} + \frac{b^{q}}{q}$…call this relation (19 or XIX).

Setting $a = |a_{k}|$, $b=|b_{k}|$, summing over k from 1 to n, and taking account of (16, or XVI) and (17, or XVII), we get the desired inequality (18, or XVIII). This proves Holder’s inequality (15 or XV). Note that (15 or XV) reduces to Schwarz’s inequality if $p=2$.

It is now an easy matter to prove Minkowski’s inequality (14 or XIV), starting from the identity

$(|a|+|b|)^{p} = (|a|+|b|)^{p-1}|a|+(|a|+|b|)^{p-1}|b|$.

In fact, putting $a=a_{k}$, $b=b_{k}$ and summing over k from 1 to n, we obtain

$\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p}=\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p-1}|a_{k}|+\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p-1}|b_{k}|$

Next, we apply Holder’s inequality (15 or XV) to both sums on the right, bearing in mind that $(p-1)q=p$:

$\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p} \leq (\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p})^{\frac{1}{q}}([\Sigma_{k=1}^{n}|a_{k}|^{p}]^{\frac{1}{p}}+[\Sigma_{k=1}^{n}|b_{k}|^{p}]^{\frac{1}{p}})$

Dividing both sides of this inequality by

$(\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p})^{\frac{1}{a}}$

we get

$(\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p})^{\frac{1}{p}} \leq (\Sigma_{k=1}^{n}|a_{k}|^{p})^{\frac{1}{p}} + (\Sigma_{k=1}^{n}|b_{k}|^{p})^{\frac{1}{p}}$

which immediately implies (14 of XIV), thereby proving the triangle inequality in $R_{p}^{n}$.

QED.

Example 11:

Finally, let $l_{p}$ be the set of all infinite sequences $x = (x_{1}, x_{2}, \ldots, x_{k}, \ldots)$ of real numbers satisfying the convergence condition

$\Sigma_{k=1}^{n}x_{k}^{p} < \infty$ for some fixed number $p \geq 1$, where distance between points is defined by

$\rho(x,y) = (\Sigma_{k=1}^{\infty}|x_{k}-y_{k}|^{p})^{\frac{1}{p}}$….call this (20 or XX)

(the case $p=2$ has already been considered in Example 7). It follows from Minkowski’s inequality (14 or XIV) that

$(\Sigma_{k=1}^{n}|x_{k}-y_{k}|^{p})^{\frac{1}{p}} \leq (\Sigma_{k=1}^{n}|x_{k}|^{p})^{\frac{1}{p}} + (\Sigma_{k=1}^{n}|y_{k}|^{p})^{\frac{1}{p}}$….call this (21 or XXI) for any n.

Since the series $\Sigma_{k=1}^{\infty}|x_{k}|^{p}$, and $\Sigma_{k=1}^{\infty}|y_{k}|^{p}$ converge, by hypothesis, we can take the limit as $n \rightarrow \infty$ in (21 or XXI) obtaining

$(\Sigma_{k=1}^{\infty}|x_{k}-y_{k}|^{p})^{\frac{1}{p}} \leq (\Sigma_{k=1}^{\infty}|x_{k}|^{p})^{\frac{1}{p}} + (\Sigma_{k=1}^{\infty}|y_{k}|^{p})^{\frac{1}{p}} < \infty$.

This shows that (20 or XX) actually makes sense for arbitrary $x, y \in l_{p}$. At the same time, we have verified that the triangle inequality holds in $l_{p}$ (the other two properties of a metric space are obviously satisfied). Therefore, $l_{p}$ is a metric space.

QED.

Remarks:

If $R = (X, \rho)$ is a metric space and M is any subset of X, then obviously $R^{*} = (M, \rho)$ is again a metric space, called a subspace of the original metric space R. This device gives us infinitely more examples of metric spaces.

Example 12:

For $x, y \in R^{1}$, define:

12a) $\rho(x,y) = (x-y)^{2}$

12b) $\rho(x,y) = \sqrt{|x-y|}$

12c) $\rho(x,y) = |x^{2}-y^{2}|$

12d) $\rho(x,y) = |x-2y|$

12e) $\rho(x,y) = \frac{|x-y|}{1+|x-y|}$

Determine for each of these, whether it is a metric or not.

Solution 12a:

Axioms 1 and 2 are clearly satisfied. We have to verify if the following holds true:

$(x-z)^{2} \leq (x-y)^{2} + (y-z)^{2}$. Clearly, RHS is $x^{2} + 2y^{2}+ z^{2} -2xy -2yz$ whereas LHS is $x^{2}+y^{2}-2xy$ so it may not always be true that LHS is lesser than or equal to RHS.

Hence, this is not a metric function.

Solution 12b:

this also satisfies the first two axioms of the definition of a metric.

So, we have to verify if the following is true:

$\rho(x,z) \leq \rho(x,y) + \rho(y,z)$, that is, TPT:

$\sqrt{|x-z|} \leq \sqrt{|x-y|} + \sqrt{|y-z|}$.

Consider the following:

$|x-z|=|x-y+y-z| \leq |x-y|+|y-z|$

Also, $(\sqrt{|x-y|} + \sqrt{|y-z|})^{2} = |x-y|+|y-z|+2\sqrt{|x-y|\times|y-z|}$

So the third axiom holds true in this case. So, the given function is a metric.

Solution 12c:

Once again, the first two axioms clearly hold.

We have to verify if the following holds true:

$\rho(x,z) \leq \rho(x,y) = \rho(y,z)$, that is, to prove that:

$|x^{2}-z^{2}| \leq |x^{2}-y^{2}| + |y^{2}-z^{2}|$, which is obviously true by triangle inequality of real numbers. So, the given function is a metric.

Solution 12d:

Clearly, the first two axioms do not hold. It can easily be checked that the third axiom also does not hold. So, the given function is not a metric.

Solution 12e:

The first axiom holds true.

To check the second axiom consider and compare:

$\rho(x,y) = \frac{|x-y|}{1+|x-y|}$ whereas $\rho(y,x) = \frac{|y-x|}{1+|y-x|}=\frac{|x-y|}{1+|x-y|} = \rho(x,y)$ clearly again.

To verify the third axiom, we have to check if the following is true:

$\rho(x,z) \leq \rho(x,y) + \rho(y,z)$, that is, to prove that:

$\frac{|x-z|}{1+|x-z|} \leq \frac{|x-y|}{1+|x-y|} + \frac{|y-z|}{1+|y-z|}$. A little algebraic work shows that this is aot always possible. Hence, the given function is not a metric.

# Exercises based on system of sets

Reference: Introductory Real Analysis, Kolomogorov and Fomin, Dover Publications, Translated and edited by Richard A. Silverman:

Problem 1:

Let X be an uncountable set, and $\mathscr{R}$ be the ring consisting of all finite subsets of X and their complements. Is $\mathscr{R}$ a $\sigma$ -ring also?

Problem 2:

Are open intervals Borel sets ?

Problem 3:

Let $y=f(x)$ be a function defined on a set M and taking values in a set N. Let $\mathscr{M}$ be a system of subsets of M, and let $f(\mathscr{M})$ denote the system of all images $f(A)$ of sets $A \in \mathscr{M}$. Moreover, let $\mathscr{N}$ be a system of subsets of N, and let $f^{-1}(\mathscr{N})$ denote the system of all preimages of $f^{-1}(B)$ of sets $B \in \mathscr{N}$. Prove that

(i) If $\mathscr{N}$ is a ring, so is $f^{-1}(\mathscr{N})$

(ii) If $\mathscr{N}$ is an algebra, so is $f^{-1}()\mathscr{N}$

(iii) If $\mathscr{N}$ is a borel algebra, then so is $f^{-1}(\mathscr{N})$

(iv) $\mathscr{R}(f^{-1}(\mathscr{N}))=f^{-1}(\mathscr{R}(\mathscr{N}))$

(v) $\mathscr{R}(f^{-1}(\mathscr{N}))=f^{-1}(\mathscr{R}(\mathscr{N}))$.

Which of these assertions remain true if $\mathscr{N}$ os replaced by $\mathscr{M}$ and $f^{-1}$ by f?

Regards,

Nalin Pithwa