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.