# 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 $Z=(X\times Y)\backslash \mathrm{\Gamma}(f)$ 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 $\mathbb{R}$ 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 $\mathrm{\Gamma}(f)$ is closed in ${\mathbb{R}}^{2}$. Note, that $\mathbb{R}$ is not compact.

Title | graph theorems for topological spaces |
---|---|

Canonical name | GraphTheoremsForTopologicalSpaces |

Date of creation | 2013-03-22 19:15:09 |

Last modified on | 2013-03-22 19:15:09 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 7 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 54C05 |

Classification | msc 26A15 |