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Let $X$ be a Hausdorff space, and $f:X \to X$ continuous. Then the following set is closed in $X$: $$B=\{x \in X;f(x)=x\}$$

I saw that question that says to consider $g:X \to X \times X$ with $g(x)=(x,f(x))$, and I could do it by arguing that $g(B)=\Delta = \{(x,x);x \in X\}$ and showing that $\Delta$ is closed. But that argument depends on $g$ being a continuous function. How can I show that?

Gea5th
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    A function into a product is continuous iff both components are – Alessandro Codenotti Apr 21 '21 at 22:39
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    You can show directly from the definition that if $f_1:X\to Y_1$ and $f_2:X\to Y_2$ are continuous, then $g:X\to Y_1\times Y_2$ (given the product topology) defined by $g(x):= (f_1(x),f_2(x))$ is continuous (actually its even an if and only if statement) – peek-a-boo Apr 21 '21 at 22:39
  • More generally, if $Y$ is Hausdorff and $f:X\to Y,,g:X\to Y$ are continuous, then $S={x\in X: f(x)\ne g(x)}$ is open in $X$. (For your Q, let $Y=X$ and $g=id_X$.) For if $x\in S$ let $V', V''$ be disjoint open subsets of $Y$ with $f(x)\in V'$ and $g(x)\in V''.$ There are open subsets $U', U''$ of $X,$ both containing $x$, with $f[U']\subset V'$ and $g[U'']\subset V''.$ Now if $z\in U'\cap U''$ then $f(z)\in V'$ and $g(z)\in V'',$ but $V', V'' $ are disjoint so $f(z)\ne g(z).$ So the open set $U'\cap U''$ contains $x$ and is a subset of $S$. – DanielWainfleet Apr 22 '21 at 09:00

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You can also give a direct proof: suppose that $p \notin B$ so that $f(p) \neq p$. As $X$ is Hausdorff, there are open sets $U_p, U_{f(p)}$ such that $p \in U_p, f(p) \in U_{f(p)}, U_p \cap U_{f(p)} = \emptyset$. As $f$ is continuous, $f^{-1}[U_{f(p)}]$ is an open neighbourhood of $p$ as well and we define $O_p = U_p \cap f^{-1}[U_{f(p)}]$, which is thus an open neighbourhood of $p$ too. One easily checks that $O_p \cap B = \emptyset$ and so as $p \notin B$ was arbitrary, $B$ is closed.

$g$ in your question is continuous because $\pi_1 \circ g = 1_X$ and $\pi_2 \circ g = f$ are both continuous, using the universal property for continuity of maps into a product.

Henno Brandsma
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  • For a counter-example when $X$ is $not$ Hausdorff, let $X={0,1}$ where ${0}$ is open but ${1}$ is $not$ open, and $f(0)=f(1)=0.$ – DanielWainfleet Apr 22 '21 at 09:09
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    @DanielWainfleet Any bijection $f$ on $\Bbb Z$ when we give $\Bbb Z$ the cofinite topology is continuous. We can give $f$ any fixed point set we like, including infinite (non-closed) ones; this gives sharper ($T_1$ instead of $T_0$) counter-examples.. – Henno Brandsma Apr 22 '21 at 09:21