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Let $\text{Dom}(f)=E$ be compact and $G_f=\{(x,f(x)): x\in E\}$ the graph of $f$.

$G_f$ is compact iff $f$ is continuous on $E$.

Proof: $\Rightarrow$ Let $f$ is not continuous at some point $x_0\in E$. Then $\exists \{x_n\}$ on $E$ such that $x_n\to x_0$ but $f(x_n)\nrightarrow f(x_0)$. It means that exists some subsequence $\{n_k\}$ such that $|f(x_{n_k})-f(x_0)|\geqslant \varepsilon_0$ for some $\varepsilon_0>0$. We'll define function $g:x\mapsto (x,f(x))$. Then $g(x_{n_k})=(x_{n_k},f(x_{n_k}))$ is sequence in $G_f$ which is compact. Thus this sequence has limit point in $G_f$ namely $(p,f(p))$. Hence $\exists \{k_j\}$ such that $g(x_{n_{k_j}})\to (p,f(p)).$ Hence $p=x_0$ and $f(x_{n_{k_j}})\to f(x_0)$ and we get contradiction. Right?

$\Leftarrow$ Let $\{G_{\alpha}\}$ be an open cover of $G_f$ i.e. $G_f\subset \cup _{\alpha}G_{\alpha}$. Consider "projections" of $G_{\alpha}$ to $E$ we define $F_{\alpha}:=\{x\in E: (x,f(x))\in G_{\alpha}\}.$ It's clear that $E\subset \cup _{\alpha}F_{\alpha}$. Also we define function $g:x\mapsto (x,f(x))$ and $g$ is continouos on $E\times f(E)$ since all his components are continuous. Then $g^{-1}(G_{\alpha})=F_{\alpha}$ is open for any $\alpha$. Since $E$ is compact then $\exists$ $\alpha_1, \cdots, \alpha_n$ such that $E\subset F_{\alpha_1}\cup \cdots\cup F_{\alpha_n}$. Then $g(E)\subset g(F_{\alpha_1})\cup\cdots \cup g(F_{\alpha_n})=G_{\alpha_1}\cup\cdots\cup G_{\alpha_n}.$ But $g(E)=G_f$.

Sorry if my topic is repeated but I solved it myself and canyone check my proof please?

I would be very grateful.

RFZ
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  • Are you working with $E \subseteq \mathbb{R}$ explicitly? You are attempting to use "sequentially compact iff compact", which is only true in metric spaces. – Patrick Stevens Oct 05 '15 at 07:02
  • @PatrickStevens, No. I'm considering the general case. – RFZ Oct 05 '15 at 07:19
  • By the way it's problem from baby Rudin. And I don know what is sequentially compact. I think that I has little problem in first part of proof. Right? I'll edit it soon. – RFZ Oct 05 '15 at 07:21
  • @RFZ Since $RFZ$ is compact, it is sequentially compact. Thus, there given $(x_n,f(x_n))$, there is a subsequence converging to $(x,f(x)) \in G_f$. This implies $f$ is continuous directly. – Anthony Peter Oct 05 '15 at 08:06
  • @RFZ Can you define "compact", then, please? – Patrick Stevens Oct 05 '15 at 08:08
  • @PatrickStevens, I thought on this about day but no any good results. That's my sketch: Let $f$ is not continuous then exists some sequence ${x_n}$ on $E$ such that $x_n\to x_0$ but $f(x_n)\nrightarrow f(x_0)$. We'll define function $g:x\mapsto (x,f(x))$ and ${g(x_n)}{n=1}^{\infty}$$={(x_n,f(x_n))}{n=1}^{\infty}$ is sequence in compact $G_f$. Hence it has some limit point in $G_f$, namely $(p,f(p))$. It's easy to see that $p=x_0$. Hence $\exists {n_k}$ such that $f(x_{n_k})\to f(x_0)$. – RFZ Oct 05 '15 at 17:56
  • But I have the last question. How to use compactness of $E$ to get contradiction? Can you help me with this last part of proof? – RFZ Oct 05 '15 at 17:56
  • Please define "compact", then. Do you mean "every open cover has a finite subcover"? – Patrick Stevens Oct 05 '15 at 18:07
  • @PatrickStevens, Dear! I think that I solved problem. I will write my solution above and can you check it please? But I think that I nowhere used the compactness of $E$. Hmmm – RFZ Oct 05 '15 at 20:37
  • @PatrickStevens, Yes I mean that definition. Also was theorem in baby Rudin which states that every infinite subset $E$ of compact set $K$ has limit point in $K$. – RFZ Oct 05 '15 at 21:19
  • @PatrickStevens, what can you say about my above proof? – RFZ Oct 06 '15 at 07:28

1 Answers1

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Forwards direction

You said in the comments that your definition of "compact" was "every open cover has a finite subcover", and that you have never encountered sequential compactness in general metric spaces. Therefore, you're going to have to prove that if the space is compact, then it has the property that every sequence has a convergent subsequence. Proofs of this are widely available. The step "this sequence has a limit point" relies on this theorem.

Backwards direction

"$g$ is continuous on $E \times f(E)$" is badly phrased: you mean "$g$ is continuous as a function $E \to E \times f(E)$".

Otherwise, I think it's fine.

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Patrick Stevens
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  • I'm not rely that "the product of two compact spaces is compact". We have sequence $(x_n,f(x_n))$ in $G_f$ which is compact. Thus it has limit point $(p,f(p))$ in $G_f$. It means that there is subsequence that converges to this $(p,f(p))$. Right? – RFZ Oct 06 '15 at 08:50
  • True. I'll edit that out. – Patrick Stevens Oct 06 '15 at 08:54