graph theorems for topological spaces

We wish to show the relation between continuous maps and their graphs is closer that it may look. Recall, that if $f:X\to Y$ is a function between sets, then the set $\mathrm{\Gamma }(f)=\{(x,f(x))\in X\times Y\}$ is called the graph of $f$.

Proposition 1. If $f:X\to Y$ is a continuous map between topological spaces such that $Y$ is Hausdorff, then the graph $\mathrm{\Gamma }(f)$ is a closed subset of $X\times Y$ in product topology.

Proof. Indeed, we will show, that $Z=(X\times Y)\backslash \backslash \mathrm{\Gamma }(f)$ is open. Let $(x,y)\in Z$. Then $f(x)\ne y$ and thus (since $Y$ is Hausdorff) there exist open subsetes $V_1,V_2\subseteq Y$ such that $f(x)\in V_1$, $y\in V_2$ and $V_1\cap V_2=\mathrm{\varnothing }$. Since $f$ is continuous, then $U=f^{-1}(V_1)$ is open in $X$.

Note, that the condition $V_1\cap V_2=\mathrm{\varnothing }$ implies, that $f(U)\cap V_2=\mathrm{\varnothing }$. Therefore $U\times V_2$ is a subset of $Z$. On the other hand this subset is open (since it is a product of two open sets) in product topology and $(x,y)\in U\times V_2$. This shows, that every point in $Z$ belongs to $Z$ together with a small neighbourhood, which completes the proof. $\mathrm{\square }$

Unfortunetly, the converse of this theorem is not true as we will see later. Nevertheless we can achieve similar result, if we assume a bit more about spaces:

Proposition 2. Let $f:X\to Y$ be a function, where $X,Y$ are Hausdorff spaces with $Y$ compact. If $\mathrm{\Gamma }(f)$ is a closed subset of $X\times Y$ in product topology, then $f$ is continuous.

Proof. Let $F\subseteq Y$ be a closed set. We will show that $f^{-1}(F)$ is also closed. Consider projections

$${\pi }_Y:X\times Y\to Y;{\pi }_X:X\times Y\to X.$$

They are both continuous and thus ${\pi }_Y^{-1}(F)$ is closed in $X\times Y$. Since $\mathrm{\Gamma }(f)$ is also closed, then

$$Z={\pi }_Y^{-1}(F)\cap \mathrm{\Gamma }(f)$$

is closed in $X\times Y$. It is well known, that since $Y$ is compact, then ${\pi }_X$ is a closed map (this is easily seen to be equivalent to the tube lemma). Furthermore it is easy to see, that ${\pi }_X(Z)=f^{-1}(F)$ and the proof is complete. $\mathrm{\square }$

Counterexample. Let $ℝ$ denote the set of reals (with standard topology). Consider function $f:ℝ\to ℝ$ given by $f(x)=1/x$ and $f(0)=0$. It is obvious, that $f$ is discontinuous at $x=0$, but also it can be easily checked, that $\mathrm{\Gamma }(f)$ is closed in ${ℝ}^2$. Note, that $ℝ$ is not compact.