# Closed graph theorem( metric spaces) [duplicate]

So what I want to prove is the following:

If $f:X \to Y$ is a continuous function and $X$ and $Y$ are metric spaces, then the graph of $f$ is closed.

All the proofs that I’ve encountered in the internet involve some topological concepts that I’m not aware of. I was looking for a proof which would involve only concepts from metric spaces. Any suggestions would be really helpful.

## marked as duplicate by Siminore, José Carlos Santos, Jimmy R., Robert Soupe, Rhys SteeleMar 26 '18 at 15:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Well, any reasonable proof must make use of some ideas like neighborhoods or convergent sequences. What kind of proof are you looking for? – Siminore Mar 26 '18 at 11:36
• I’m good with convergent sequences and I’ve quite reasonable understanding of most of the concepts in metric spaces. – John Mitchell Mar 26 '18 at 11:59

## 1 Answer

Take a look at a sequence $(x_n,y_n) \in \text{graph}(f)$ converging in $X\times Y$ to some $(x,y)$. Since we look at the graph of f, we have $y_n=f(x_n)$. We have to prove, that $(x_n,y_n)$ converges in $\text{graph}(f)$. Since $x_n\rightarrow x$ and f is continuous, $y_n=f(x_n)\rightarrow f(x)=y$, thus the limit of our sequence is $(x,y)=(x,f(x))$ and our sequence converges in $\text{graph}(f)$. Thus $\text{graph}(f)$ is closed.