# 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 $\Gamma(f)=\{(x,f(x))\in X\times Y\}$ is called the graph of $f$.

Proof. Indeed, we will show, that $Z=(X\times Y)\backslash\Gamma(f)$ is open. Let $(x,y)\in Z$. Then $f(x)\neq y$ 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 $V_{1}\cap V_{2}=\emptyset$. Since $f$ is continuous  , then $U=f^{-1}(V_{1})$ is open in $X$.

Note, that the condition $V_{1}\cap V_{2}=\emptyset$ implies, that $f(U)\cap V_{2}=\emptyset$. 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. $\square$

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 $\Gamma(f)$ 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 $\Gamma(f)$ is also closed, then

 $Z=\pi_{Y}^{-1}(F)\cap\Gamma(f)$

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. $\square$

Let $\mathbb{R}$ denote the set of reals (with standard topology). Consider function $f:\mathbb{R}\to\mathbb{R}$ 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 $\Gamma(f)$ is closed in $\mathbb{R}^{2}$. Note, that $\mathbb{R}$ is not compact.

Title graph theorems for topological spaces GraphTheoremsForTopologicalSpaces 2013-03-22 19:15:09 2013-03-22 19:15:09 joking (16130) joking (16130) 7 joking (16130) Theorem msc 54C05 msc 26A15