Are continuous functions between Hausdorff spaces always equal on a closed set?

Question

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?

Answer 1

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.

Answer 2

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$$\begin{array}{rccc}\varphi\colon&X&\longrightarrow&Y\times Y\\&x&\mapsto&\bigl(f(x),g(x)\bigr).\end{array}$$