The following statement is given:
$X$ is a topological space and $Y$ is a Hausdorff space, then $f:X\to Y$ is countinous iff $\mathcal{G}(f)$ is closed in $X\times Y$, when $\mathcal{G}(f) := \{(x,y)\in X\times Y: f(x) = y\}$
Note: The statement was given as true, to be proven, as part of an assignment. Also, no additional condition was given for $X$, $Y$.
One direction of statement (the graph of a a continuous function to Hausdorff space is closed in the product), is certainly true. However, the opposite direction seems false (although it does hold, if $Y$ is compact). The following is an attempt to provide a counterexample.
Assume $X=Y=\mathbb{R}$ with the standard topology, and define: $$f:X\to Y \\ f(x)=\begin{cases}\frac{1}{x},&x\neq 0\\0,&x=0\end{cases} $$ Every point $x\in (X\times Y)\setminus \mathcal{G}(f)$ has some nbhd that is a subset of $(X\times Y)\setminus \mathcal{G}(f)$, so $(X\times Y)\setminus \mathcal{G}(f)$ is union of such nbhds and thus is an open set, implying $\mathcal{G}(f)$ is closed.
The interval $(-1,1)$ is an open subset of $Y=\mathbb{R}$, but: $$f^{-1}((-1,1)) = (-\infty,-1)\cup(1,\infty)\cup \{0\}$$ is not an open subset of $X=\mathbb{R}$, so $f$ is not continuous.
Is this a proper disproof, or is there some mistake?
