0
$\begingroup$

Possible Duplicate:
If the graph of a function $f: A \rightarrow \mathbb R$ is compact, is $f$ continuous where $A$ is a compact metric space?

I was wondering about whether I can use the proposition above in the following manner

So if I have two metric spaces, say A and B and $f: A\rightarrow B$ and the graph of $f$, say $G$ is compact. Then I know that $G\subseteq A \times f(A) \text{ where $f(B)$} \subseteq B$ is compact. Since $G$ is compact, for every sequence, I have a convergent subsequence $(a_{n_k},b_{n_k})$ that converges to an ordered pair $(a,b)$ in $G$.

Now $(a_{n_k},ba_{n_k})$ converges iff $a_{n_k} \rightarrow a \in A$ and $b{n_k} \rightarrow b \in f(A)\subseteq B$. $(a_{n_k})$ is a convergent subsequence of $(a_n)$ in $A$. So $A$ is compact. Similarly $f(A)$ is compact. Since the image of $f$ is compact and the domain is compact, $f$ is continuous.

Does this make sense?

  • 0
    @ZhenLin: That’s exactly what the OP is trying to prove, and it’s been thoroughly dealt with in the answers and comments for the previous question, of which it is a part.2012-11-09

0 Answers 0