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Let $X,Y$ be a pair of Hausdorff spaces. Let $f,g \in C(X,Y)$. Is it guaranteed that $\{x \in X: f(x)=g(x)\}$ is a closed set? If not, is it guaranteed for some reasonably wide family of Hausdorff spaces?

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Yes, $Y$ is Hausdorff iff $\Delta(Y) = \{(y,y): y \in Y\}$ is closed in $Y \times Y$ (in the product topology).

If $f,g:X \to Y$ are continuous then $F: X \to Y \times Y$ defined by $F(x) = (f(x), g(x))$ is also continuous, as $\pi_1 \circ F =f$ and $\pi_2 \circ F= g$ are continuous, so

$$\{x: f(x) = g(x)\}= \{x: F(x) \in \Delta(Y)\} = F^{-1}[\Delta(Y)]$$

is closed in $X$ as the inverse image of a closed set under a continuous map.

Hausdorffness of $X$ is not needed, that of $Y$ is essential.

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Yes, it is guaranteed if $Y$ is a Hausdorff space, because if $Y$ is such a space, then $D=\bigl\{(y,y)\,|\,x\in Y\bigr\}$ is a closed subset of $Y\times Y$. But$$\bigl\{x\in X\,|\,f(x)=g(x)\bigr\}=\varphi^{-1}(D),$$where $\varphi$ is the map$$\begin{array}{rccc}\varphi\colon&X&\longrightarrow&Y\times Y\\&x&\mapsto&\bigl(f(x),g(x)\bigr).\end{array}$$

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  • $\begingroup$ Thank you very much! Please do not take offense, but although your answer is great, I'll accept Henno Brandsmas one instead, just because I've learned that "hausdorffness" is a (somewhat) legit word <3 $\endgroup$ – Michał Zapała Nov 24 '18 at 22:06
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    $\begingroup$ The choice of an accepted answer is strictly a matter of individual taste and therefore no offense is, or could be, taken. $\endgroup$ – José Carlos Santos Nov 24 '18 at 22:08
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    $\begingroup$ @MichałZapała We have typed essentially the same answers simultaneously, indeed. It's Hausdorffness with capital H BTW. Named after a mathematician.. $\endgroup$ – Henno Brandsma Nov 24 '18 at 22:08

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