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

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?

## 2 Answers

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.

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

• 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 – Michał Zapała Nov 24 '18 at 22:06
• The choice of an accepted answer is strictly a matter of individual taste and therefore no offense is, or could be, taken. – José Carlos Santos Nov 24 '18 at 22:08
• @MichałZapała We have typed essentially the same answers simultaneously, indeed. It's Hausdorffness with capital H BTW. Named after a mathematician.. – Henno Brandsma Nov 24 '18 at 22:08