# Hausdorffness of a space

We know that if $X$ and $Y$ are two topological spaces with $Y$ Hausdorff, then for any two continuous functions $f:X\to Y$ and $g:X\to Y$, $\{x\in X: f(x)=g(x)\}$ is a closed set.

Is the converse true? Let us suppose that $Y$ be a topological space and for any topological space $X$ and for any continuous functions $f:X\to Y$ and $g:X\to Y$, $\{x\in X: f(x)=g(x)\}$ is a closed set in $X$. I want to prove that $Y$ is Hausdorff. Is it really possible? Any hint will be very much helpful.

• Hint: Product, projection. – Daniel Fischer Oct 15 '13 at 12:34

I think we might as well spill the beans here. Let $f = \pi_1: X \times X \to X$, and let $g = \pi_2: X \times X \to X$.
As a hint I give you my favorite proof of the direction that you are not interested in: Let $\Delta=\{(y,y)\in Y\times Y\colon y\in Y\}$. We know that $Y$ is Hausdorff if and only if $\Delta$ is closed in $Y\times Y$. Hence $\{x\in X: f(x)=g(x)\}=(f,g)^{-1}[\Delta]$ is closed.