This Question gives the following argument for why the retract of a Hausdorff space is closed:

Proof. Let $x∉A$ and $a=r(x)∈A$. Since $X$ is Hausdorff, $x$ and $a$ have disjoint neighborhoods $U$ and $V$, respectively. Then $r^{−1}(V∩A)∩U$ is a neighborhood of $x$ disjoint from $A$. (*) Hence, $A$ is closed.

It looks to me like this presumes that every $x∉(X-A)$ has an neighborhood, $U$, that doesn't intersect $A$. Otherwise $r^{−1}(V∩A)∩U$ could be in $A$, and the proof wouldn't work. It seems to me though that the only way every $U$ could not intersect $A$ is if we've already assumed that $A$ is closed because otherwise there would be a point in the boundary of $A$, but not in $A$, and every neighborhood of this point would intersect $A$. And so we wouldn't be able to find a neighborhood, $U$, of this point that doesn't intersect $A$.

So does this proof actually work and I've missed something?

  • $\begingroup$ Use $ symbols for Latex code. a -1 superindex: 'r^{-1}' $\endgroup$ – MyUserIsThis May 6 '13 at 6:27
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    $\begingroup$ Welcome to math.SE. For some basic information about writing math (including superscripts) at this site see e.g. here, here, here and here. $\endgroup$ – Julian Kuelshammer May 6 '13 at 6:43

We need to show that for each $x \in X \setminus A$ there is an open neighbourhood $U$ of $x$ which is disjoint from $A$ (this then implies that $X \setminus A$ is open, and so $A$ is closed). Below is the proof spelled out in quite a bit of detail.

Now given any $x \in X \setminus A$, since $r$ is a retraction onto $A$ we have that $r(x) \in A$, and so in particular $r(x) \neq x$. Since $X$ is Hausdorff we know that there are disjoint open sets $V , W$ about $x$ and $r(x)$, respectively. Since $r$ is continuous, it follows that $r^{-1} [ W ]$ is open, and since $r(x) \in W$ we have that $x \in r^{-1} [ W ]$; so $U = V \cap r^{-1} [ W ]$ is also an open neighbourhood of $x$.

Can $U$ intersect $A$? Well, if $y \in U$, then it must be that $r(y) \in W$, and since $V$ and $W$ are disjoint we have that $r(y) \notin V$, but since $y \in V$ it follows that $y \neq r(y)$, and so $y \notin A$. Therefore $U$ is an open neighbourhood of $x$ which is disjoint from $A$.

Notice that above we have only used the assumption that $r$ is a retraction onto $A$ (so $r$ is continuous, and $f(x) \in A$ for all $x \in X$, and $r(a) = a$ for all $a \in A)$, and that $X$ is Hausdorff (so different points can be separated by disjoint open neighbourhoods). What we do is construct from these assumptions an open neighbourhood of $x$ disjoint from $A$.

  • $\begingroup$ Excellent explanation! +1 $\endgroup$ – Error 404 Feb 25 '18 at 2:47

Here we are proving $A$ is closed by showing that $X/A$ is open. Let $x \in X/A$, we find neighborhood of $x$, $r^{-1}(V\cap A) \cap U$ which lies completely in $X/A$. Now $A \cap r^{-1}(V\cap A) \cap U= \phi$, since if $x \in r^{-1}(V\cap A) \cap U$, we have $x \in U$ and $r(x) \in V$. Now $U$ and $V$ are disjoint so $x \neq r(x)$ hence $x $ does not lies in $A$ [because $\forall x \in A$ we have $r(x)=x$] . (I have given the same argument as in the answer for the question in the link.)

  • $\begingroup$ This doesn't work if x is in the closure of A but not in A. We don't know if r−1(V∩A)∩U lies completely in X/A, because it wont if x is in the closure of A but not in A. Every U which contains such an x will intersect A. $\endgroup$ – David May 6 '13 at 6:52
  • $\begingroup$ Please explain: We don't know if $r^{−1}(V\cap A)\cap U$ lies completely in X/A, because it wont if x is in the closure of A but not in A. I have shown that $r^{−1}(V\cap A)\cap U$ and $A$ are disjoint which says that $ r^{−1}(V\cap A)\cap U \subset X/A$. Is there anything that I am missing here? $\endgroup$ – Girish May 6 '13 at 7:35

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