Month: July 2022

Connectedness and separated, some basics only

Posted on by Nalin Pithwa

Reference: Topology by Hocking and Young, Dover Publications, Inc. (available Amazon India also)

Definition: A topological space is separated if it is the union of two disjoint, non-empty open sets.

Definition: A topological space is connected if it is not separated.

It is obvious that both these properties are invariant under homeomorphisms.

Lemma 1-10: A space is separated if and only if it is the union of two disjoint, nonempty closed sets.

Lemma 1-11: A space is connected if and only if the only sets which are both open and closed in it are the space itself and the empty set.

Note: A subset X of a space S is separated if there exist two open sets U and V of S such that U \bigcap X and V \bigcap S are disjoint and nonempty and such that U \bigcap V \supset S. We cannot assume that U and V are disjoint in S, however, Consider a space S consisting of three points a, b and c with the open sets being S, \phi, a \bigcup c and a \bigcup b. Then, b \bigcup c is not a connected subset of S, but there are no disjoint open sets in S, one containing b and the other containing c.

Theorem 1-13: A subset X of a space S is connected if and only if there do not exist two non-empty subsets A and B of X such that X = A \bigcup B and such that (\overline{A} \bigcap B) \bigcup (A \bigcap \overline{B}) is empty.

Theorem 1.-14: Suppose that C is a connected subset of a space S and that \{ C_{\alpha}\} is a collection of connected subsets of S, each of which intersects C. Then, S^{'} = C \bigcup (\bigcup_{\alpha}C_{\alpha}) is connected.

Cheers,

Nalin Pithwa

Various equivalent definitions of continuity

Posted on by Nalin Pithwa

Reference: Topology by Hocking and Young, Dover Publications Inc., NY (available in Amazon India also)

  1. A transformation f: S \rightarrow T is continuous provided that if p is a limit point of a subset X of S, then f(p) is a limit point of f(X) or a point of f(X).
  2. A transformation f: S \rightarrow T is continuous provided that if p is a point of \overline{X} (where X is a subset of S), then f(p) is a point of \overline{f(X)}.
  3. A real valued function y=f(x) defined on an interval [a,b] is continuous provided that if a \leq x_{0} \leq b and \epsilon>0, then there is a number \delta >0 such that if |x-x_{0}|<\delta, x \in [a,b] then |f(x)-f(x_{0})|<\epsilon.
  4. Let f: S \rightarrow T be a transformation of the space S into the space T. A necessary and sufficient condition that f be continuous is that if O is any open subset of T, then its inverse image f^{-1}(O) is open in S.
  5. A necessary and sufficient condition that the transformation f:: S \rightarrow T of the space S into the space T be continuous is that if x is a point of S, and V is an open subset of T containing f(x), then there is an open set U in S containing x and such that f(U) lies in V.
  6. (A rewording of the above theorem for metric spaces strongly resembles that classic definition of continuity in analysis) Let f: M \rightarrow N be a transformation of the metric space M with metric d and the metric space N with metric \rho. A necessary and sufficient condition that f be continuous is that if given \epsilon>0 and given x \in M, then there exists a \delta>0 such that if d(x,y)<\delta, then \rho(f(x),f(y))<\epsilon.
  7. PS: A transformation f: S \rightarrow T of the space S into the space T is said to be interior if f is continuous and if the image of every open subset of S is open in T.
  8. PS: Some writers discuss transformations that carry open sets into open sets that are not necessarily continuous. Such transformations are usually called open.

Cheers,

Nalin Pithwa

Every separable metric space has a countable basis

Posted on by Nalin Pithwa

Reference: Topology by Hocking and Young, Dover Publications Inc., NY. (available in Amazon India also)

….As an example of the topological power of a metric, we give the following result. First, a set X of points in a space S is said to be dense in S if every point of S is a point or a limit point of X, that is, if S = \overline{X}. A space is separable if it has a countable dense subset. For instance, E^{n}, the Euclidean n-space is separable since the set of all points whose coordinates are all rational is countable and dense.

Theorem 1-5: Every separable metric space has a countable basis.

Proof of Theorem 1-5: Let M be a metric space with metric d(x,y) and having a countable dense subset X = \{ x_{i} \}. For each rational number r>0 and each integer i>0, there is a spherical neighbourhood S(x_{i},r) and the set \mathcal{B} of all these is countable. We will show that \mathcal{B} is a basis. (Recall definition of a basis: A subcollection \mathcal{B} of open sets of a topological space S is a basis for S if and only if every open set in S is a union of elements of \mathcal{B}) (A countable basis for a space S is a basis that contains only countably many sets).Let p be any point of M and let O be an open set containing p. Then there is a positive number \epsilon such that S(p, \epsilon) is contained in O, by definition. There is a point x_{i} of X such that d(x_{i},p) < \epsilon/3 since X is dense. Let r be a rational number satisfying \epsilon/3<r<2\epsilon/3 and consider S(x_{i},r). Certainly S(x_{i},r) contains p and if y is any point of S(x_{i},r), then d(y,p) \leq d(y,x_{i}) + d(x_{i},p) < \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon.

Thus y is an element of S(p, \epsilon) and so S(x_{i},r) is an element of \mathcal{B} that contains p and lies in O. It follows that O is a union of elements of \mathcal{B} and that \mathcal{B} is a basis for the topology of M.

QED.

Important remarks about the above theorem:

Without the assumption of metricity, Theorem 1-5 is not true. In E^{n}, the Euclidean n-space, consider the set P of all points (x,y) with y \geq 0. Let a basis for P consist of (1) all interiors of circles of P but not touching the x-axis and (2) the union of a point on the x-axis and the interior of a circle tangent from above to the x-axis at that point. The set of points in P both of whose coordinates are rational is both countable and dense in P. But no element of the basis just defined contains two points on the x-axis. If there were a countable basis \mathcal{B} for P, then each basis element above would be a union of elements of \mathcal{B}. This would imply that there is a subcollection of \mathcal{B} such that each point of the x-axis lies in one and only one element of that subcollection. This contradicts the fact the real numbers are uncountable. QED.

Cheers,

Nalin Pithwa