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Let $X$ be Hausdorff and $A \subset X$. Let $r:X \to A$ be a retraction. Show that $A$ is closed.

Let $x \in X \setminus A$ and let $a=r(x) \in A$, then as $X$ is Hausdorff there exists disjoint neighborhoods $U_x$ and $U_a$ respectively. Now I need to show that I can find an open set around $x$ that's disjoint from $A$ to finish the proof. I don't have the intuition on how to do this. Since the retraction is continuous I have the preimages at my disposal, but I don't know how to use them. Here is a bad sketch I tried to come up of the situation. I have a feeling that it's misleading me.

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what sort of preimages I could consider?

Herman Himmerloos
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    Suppose $p : X \to A$ is continuous, $i : A \to X$ is the inclusion and $p \circ i = \mathrm{id}$. Consider the map $(i \circ p, \mathrm{id}) : X \to X \times X$. $A$ is the preimage of the diagonal and since $X$ is Hausdorff, the diagonal is closed. – Julian Quast Jan 28 '22 at 23:33

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We'll show that $X\setminus A$ is open.

Pick a point $x\in X\setminus A$. Since $r(x)\neq x$, $X$ is Hausdorff and $r$ is continuous, there exists a neighborhood $N_x$ of $x$ such that $\forall y\in N_x$ $f(y)\neq y$. This means $N_x\subset X\setminus A$, ie $X\setminus A$ is a neighborhood of each of its points, and therefore is open.

No-one
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