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Suppose $X$ is a Hausdorff space and $f:X \rightarrow X$ a continuous function. Prove that the set $\{x \in X \mid f(x)=x\}$ is closed in $X$.

I've already proved this proposition:

Let $X,Y$ be topological spaces with $Y$ Hausdorff, and let $f,g:X \rightarrow Y$ be continuous maps. Then the set

$$ \{x \in X \mid f(x)=g(x)\} $$

is closed in $X$.

My question is whether I can use this proposition to prove the first statement?

I think I can, because if we let $g:X \rightarrow X$ be the identity map, then $g$ is continuous aswell, and the codomain of $f$ and $g$ is obviously Hausdorff, so the conditions in the above proposition seems to be satisfied.

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    $\begingroup$ Yes, taking $g$ the identity function is a particular case. If you think it will be needed add a word about why the identity is continuous. $\endgroup$ – conditionalMethod Dec 5 '19 at 20:51
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    $\begingroup$ Yes... I dk what method you used to prove the general result about $X$ and $Y$ but for me it seems easiest to prove that $\{x\in X: f(x)\ne g(x)\}$ is open in $X$. A useful corollary, for continuous $f,g$ from $X$ to Hausdorff $Y,$ is that if $f,g$ agree on a dense subset of $X,$ then $f=g,$ which can put an upper limit on the cardinal of the set of continuous $f:X\to Y$, which is applied in the proof of the Jones Lemma about (some) non-normal spaces. A familiar special case, with the standard topology on $\Bbb R,$ is that if the continuous real functions $f,g$ agree on $\Bbb R$ then $f=g. $ $\endgroup$ – DanielWainfleet Dec 6 '19 at 8:00
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As has been noted in a comment, the answer is “yes”.

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    $\begingroup$ In general, you will have that the preimage by a continuous function of a point (closed subset by Haussdorff condition) is closed. Since $f$ is continuos, the function $g(x)=f(x)-x$ is also continuos, therefore $g^{-1}(0)$ is closed. You just need to use that the preimage by continuous functions of a closed subset is closed. $\endgroup$ – Senna Dec 5 '19 at 21:27
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    $\begingroup$ @Senna What does $f(x) - x$ mean in a general Hausdorff space? The correct general argument is this: If $X$ is Hausdorff, then the diagonal $\Delta = \{(x,x)\mid x\in X\}$ is a closed subset of $X\times X$. Given a continuous function $f\colon X\to X$, we have a continuous function $(\text{id}_X,f)\colon X\to X\times X$, by $x\mapsto (x,f(x))$, and $\{x\in X\mid f(x) = x\}$ is the preimage $(\text{id}_X,f)^{-1}(\Delta)$, hence it is closed. Note that we really need the Hausdorff condition to get that $\Delta$ is closed. The condition that points are closed ($T_1$) is not enough. $\endgroup$ – Alex Kruckman Dec 5 '19 at 21:43
  • $\begingroup$ For a counterexample in the $T_1$ case, let $X = \mathbb{N}$ with the cofinite topology. This is $T_1$ but not Hausdorff. Let $f\colon X\to X$ be the function which swaps $0$ and $1$ and fixes every other element. Then $f$ is continuous. But $\{x\in X\mid f(x) = x\} = \mathbb{N}\setminus \{0,1\}$, which is not closed. $\endgroup$ – Alex Kruckman Dec 5 '19 at 21:46
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As an alternative proof for the general case (with both $f$ and $g$, and yes, we can of course take $g=\textrm{id}_X$ to derive the first from the second, as identities are always continuous), we can use nets: if $(x_i)_{i \in I}$ is a net in $X$ converging to some $x \in X$ and all $x_i, i \in I$ are in $C:=\{x\mid f(x)=g(x)\}$ then we know that for all $i$, $f(x_i)=g(x_i)$ by definition of $C$ and so, as $f$ and $g$ are continuous:

$$\lim_i f(x_i) = f(\lim_i x_i) = f(x) \text{ and } \lim_i g(x_i)=g(\lim_i x_i)=g(x)$$ and as the nets $(f(x_i))_i$ and $(g(x_i))_i$ in $Y$ are the same by hypothesis and $Y$ is Hausdorff so that limits of nets are unique: $f(x)=g(x)$ and so $x \in C$ as well.

So nets from $C$ can only converge to members of $C$, which implies $C$ is closed.

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