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Prove that $f: A \to B$ is continuous iff its graph is compact where $A$ is compact and $A,B$ are metric spaces.

My attempt: I have already proved it. But somehow i am not satisfied with my proof. Implies part is Ok. But converse part i want to prove f is continuous by using $f(\overline C) \subset \overline{ f(C)}$.

For this i took $C\subset E$ arbitrary. Let $T=\{(x, f(x)):\, x \in C\}$, $F=\{(x, f(x)): x \in \overline{C}\}$. Now $\overline T$ is closed in graph(f), hece it is compact. I cant proceed further. Please give me hint.

BePure
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  • No, absolutely not. Two problems are totally different – BePure Dec 03 '20 at 18:59
  • Apologies for the confusion, but your title should describe the question stated in the body of the text. I was misled. – Seokbin Lee Dec 03 '20 at 19:02
  • My title already described my question. Ponder over body and title. I don't think it misleading. Infact i choose it to attract all those peoples who are expert in metric spaces. – BePure Dec 03 '20 at 19:05
  • Its Okay, No need to apologise. It happens sometimes. – BePure Dec 03 '20 at 19:06
  • This question has been answered here: https://math.stackexchange.com/questions/440676/graph-of-continuous-function-from-compact-space-is-compact – Oreomair Dec 03 '20 at 19:18
  • You already have the assumption that $A$ is compact, so $f(\bar{A}) = f(A) \subset \overline{f(A)}$. – angryavian Dec 03 '20 at 19:29

2 Answers2

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I'll try proving what the title states when $f$ is continuous.

If $x\in f(\overline{A})$ then there exists $a\in \overline{A}$ such that $f(a)=x$, now since it is in the closure, we have $$\forall U \mbox{ open set containing } a \mbox{, } U\cap A\not=\emptyset$$ Now, taken an open set $V$ containing $x$, since $f$ is continuous, $f(V)^{-1}$ is a an open set containing $a$, so as we saw $A\cap f(V)^{-1} \not=\emptyset$.

Now, taking it's immage, $f(\emptyset)\not=f(A\cap f(V)^{-1}) \subseteq f(A)\cap V$, so every $V$ intersects $f(A)$: by definition $x$ is in the closure of $f(A)$

I have show that $x\in f(\overline{A}) \Rightarrow x\in \overline{f(A)}$; this means $f(\overline{A})\subseteq \overline{f(A)}$.

Measure me
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As noted in the comments, I don't think what you're asking to prove will actually solve the desired question. So I will use sequences to prove your question. So suppose your graph is compact. Note that since we are dealing with Metric spaces then we have the Bolzano-Weierstrass property (in the sense that every sequence in our graph has a convergent subsequence).

Let $C \subset f(A)$ be closed and let $x_n \in f^{-1}(C)$ with $x_n \rightarrow x \in A$. Since our graph is compact then the sequence $(x_n, f(x_n))$ has a convergent subsequence in the graph of $f$, i.e. $(x_{n_k}, f(x_{n_k})) \rightarrow (x,y)$, which must be in the graph of $f$ since our graph is closed. But then $(x,y) = (x,f(x))$, and $f(x_{n_k}) \rightarrow f(x)$ so then $f(x) \in C$. Finally, this means that $x \in f^{-1}(C)$, thus $f^{-1}(C)$ is closed and $f$ is continuous.

Oreomair
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  • Are Xn from A? If so how can you let that "let xn∈f−1(C) ". It may be from outside of f inverse C. – BePure Dec 04 '20 at 01:58
  • @BePure certainly the $x_n$ are all in $A$ as it is the domain. The point is that I make it such that the sequence is in $f^{-1}(C)$. One equivalency of a closed set is that if you have a convergent sequence in that set, then the value it converges to is also in that set. So by proving that statement then we have have proven that $f^{-1}(C)$ is closed and thus $f$ is continuous. – Oreomair Dec 04 '20 at 02:02