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Can I please receive help/feedback on my proof? Thank you for your time and help!

Suppose $X$ is a topological space, $A\subseteq X, f: X\to A$ is continuous and for each $a\in A, f(a) = a.$ Prove that $A$ is closed in $X.$

$\textbf{Solution:}$ Let $(x,y)$ be a Hausdorff space and let $f$ be a continuous mapping of $X$ into itself. Then the set $A= \{x\colon f(x) = x\}$ is closed. We shall show $X\setminus A$ is open. If $X\setminus A = \emptyset$ then it is clearly open. So let $X\setminus A \ne \emptyset$. Let $a$ be an arbitrary point of $X\setminus A$. Then $f(a) \ne a.$ Since $X$ is a Hausdorff space and $f(a), a$ are distinct points of $X$, there exist disjoint open sets $G$ and $H$ such that $f(a) \in G$ and $a\in H$. As $f$ is continuous, $f^{-1}(G)$ is an open set containing $a$. So $f^{-1}(G)\cap H$ is an open set containing $a$.

We claim $f^{-1}(G) \cap H \subseteq X\setminus A.$ Let $z\in f^{-1}(G) \cap H.$ Then $f(z) \in G, z\in H.$ Since $G\cap H = \emptyset, f(z) \ne z.$ So $z\notin A$, i.e. $z\in X\setminus A.$ Thus for each $a\in X\setminus A,$ there exists an open set $f^{-1}(G)\cap H$ such that $a\in f^{-1}(G)\cap H \subseteq X\setminus A.$ Hence, $X\setminus A$ is in a neighborhood of each of its points and so it is open.

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    $\begingroup$ Did you miss the Hausdorff requirement in you statement? Also, what does your first sentence mean, what is $(x,y)$? $\endgroup$ May 7, 2020 at 5:39
  • $\begingroup$ @G.Chiusole I meant to say $X$, sorry about that confusion $\endgroup$ May 7, 2020 at 13:41

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The proof looks fine. Note that this generally goes under the fact that the retract of a Hausdorff space is closed. For a general topological space this is not true however. Pick for example a space with a point that is not closed and then project onto that point.

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    $\begingroup$ That a point is open does not prevent it from being closed. However, with $X$ the two-point space with indiscrete topology and $A$ a one-point subspace, we do have a counterexamaple $\endgroup$ May 7, 2020 at 5:49
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    $\begingroup$ @HagenvonEitzen oh I now noticed I should have worded that better. What I meant was “a point which is open and also not closed”, and not “a point which is open, and thereby not closed”. I’ll edit that. Thx $\endgroup$ May 7, 2020 at 5:51
  • $\begingroup$ Thank you for the feedback G. Chiusole $\endgroup$ May 7, 2020 at 13:38
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Another way to see this fact (retracts are closed in Hausdorff spaces), is to use nets: suppose $x \in \overline{A}$, then there is a net $a_d, d \in D$ from $A$ converging to $x$. By continuity of $f$, $\lim_d f(a_d) = f(x)$, but because $f$ is the identity on $A$, we also have $\lim_d f(a_d) = \lim_d a_d = x$ and as limits of nets are unique in Hausdorff spaces, we conclude that $f(x)=x$ and so $x \in A$ in particular, showing $\overline{A} \subseteq A$, or $A$ is closed.

Your proof works too, though. But if nets are a tool you know about, then the above proof feels quite natural (at least to me). In metric spaces we could use sequences and have a rigorous proof this way too. In general spaces we do need nets, but their use is similar to the way we use sequences in analysis, e.g.

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  • $\begingroup$ Thank you, Henno for the feedback $\endgroup$ May 7, 2020 at 13:39

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