# Proof that retract of Hausdorff space is closed

andrewkirk
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I think it is explained in Harald Hanche-Olsen's comment in reply to his own answer. It is the second comment to that answer, and dated Mar 6 '13 at 7:07. Except, I think he should have written ##y## instead of ##x## in it because he is referring to an arbitrary element of U, and ##x## has been fixed.

To flesh it out a bit.
1. We chose U and V to be disjoint nbds of x and a, which the Hausdorff property allows us to do.
2. We know ##S\equiv r^{-1}(V\cap A)\cap U## is open because it is the intersection of two open sets. The first set is open because it is the pre-image of a set that is open in A under a continuous map r.
3. Let ##y## be an arbitrary element of S. Then ##r(y)\in V## and ##y\in U## so, since ##U,V## are disjoint, we have ##r(y)\neq y##.
4. Hence ##y\notin A## since every element of A maps to itself under ##r##.
5. Hence S is an open nbd of ##x## that is disjoint from A.
6. Since ##x## was chosen as an arbitrary point in ##X-A##, that means that every element of ##X-A## has an open nbd in ##X-A##.
7. Hence ##X-A## is open. Hence ##A## is closed.

PsychonautQQ