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Let $X$ be a compact subset of $\mathbb{R}$. Let $f$ be a real-valued function on $X$. Prove that $f$ is continuous if and only if {$(x,f(x));x\in X$} is a compact subset of $\mathbb{R}^2$

I need some help and hints on this one.

1 Answers1

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If $f$ is continuous, then also the map $g:X\rightarrow\mathbb{R}^2, x\mapsto (x,f(x))$ is continuous. Therefore $g(X)=\{(x,f(x))\,:\,x\in X\}$ is compact.

On the other hand, let $g(X)$ be compact and $x_n\rightarrow x$ be a convergent sequence. We show that $f(x_n)$ converges to $f(x)$. Since the graph is compact, $f(x_n)$ has a convergent subsequence, i.e. $f(x_{n_j})\rightarrow y$. That is $(x_{n_j}, f(x_{n_j}))\rightarrow (x, y)$. The graph is closed. That is, the limit of every convergent sequence in $g(X)$ is again in $g(X)$. Therefore $(x,y)\in g(X)$, i.e. $y=f(x)$.

Since this is true for every convergent subsequence, we showed that $f(x)$ is the only accumulation point of $f(x_n)$, i.e. $f(x_n)$ converges to $f(x)$.

This is also known as the closed graph theorem.

J.R.
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  • You wrote: it is enough to ask, that the graph is closed. Isn't $X=[-1,1]$, and $f(x)=\frac1x$, $f(0)=0$ a counterexample? This function is not continuous, but it has closed graph. – Martin Sleziak Feb 17 '14 at 15:20
  • BTW one implication from this question seems to already have answer here: http://math.stackexchange.com/questions/231797/if-the-graph-of-a-function-f-a-rightarrow-mathbb-r-is-compact-is-f-conti – Martin Sleziak Feb 17 '14 at 15:21
  • Hmm thanks for the help, however i wonder, if a topological space is limit point compact, does this mean that every point of the space is a limit point? –  Feb 17 '14 at 17:31
  • @MartinSleziak: Good point, I edited the proof. – J.R. Feb 17 '14 at 17:56
  • Could you please explain why we get $y=f(x)$ because the graph is closed? –  Feb 17 '14 at 19:26
  • @George, edited. – J.R. Feb 17 '14 at 19:52