Ex: 1-7, relation between homeomorphism and continuity

Reference: Exercise: 1-7. Chapter 1. Topology by Hocking and Young.

Prove 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^{-1} are continuous.

Proof:

Consider the following defintion:

Definition 1-1: A real point p is a limit point of a set X of real numbers if for any positive number \epsilon there exists a point x \in X such that 0 < |p-x| < \epsilon.

Definition of homeomorphism: A homeomorphism of S onto T is a one-to-one transformation f: S \rightarrow T which is onto and such that a point p is a limit point of a subset X of S if and only if f(p) is a limit point of f(X).

But, the above two definitions when combined mean the following: at least for the case of a Euclidean space: 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.

The above sub-case settles the proof for the Euclidean space E^{1}.

Now, for the more general transformation f: S \rightarrow T, consider definition 1-1 above and the following two theorems (both being equivalent definitions of continuous functions):

Theorem 1-6: 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.

Next, theorem 1-7: A necesssary 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 such that f(U) lies in V.

Clearly, the above settles the claim for a general transformation which is one-to-onto that it is a homeomorphism if and only if both f and f^{-1} are continuous.

QED.

Your comments are welcome !

Regards,

Nalin Pithwa.

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