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

Prove that a one-to-one transformation of a space S onto a space T is a homeomorphism if and only if both f and 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 there exists a point such that .

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

But, the above two definitions when combined mean the following: at least for the case of a Euclidean space: A real valued function defined on an interval is continuous provided that if and , then there is a number such that if , x in , then .

The above sub-case settles the proof for the Euclidean space .

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

Theorem 1-6: Let 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 is open in S.

Next, theorem 1-7: A necesssary and sufficient condition that the transformation 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 , then there is an open set, U in S containing x such that 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 are continuous.

QED.

Your comments are welcome !

Regards,

Nalin Pithwa.

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