Please can you help with this exercise.
Let $F$ be a metric space such that for any metric space $E$, any mapping of $E$ into $F$ whose graph is closed in $E\times F$ is continuous. Show that $F$ is compact.
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Note that the converse is also true. – Julien Jun 23 '13 at 04:30
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related – leo Jun 23 '13 at 19:16
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Recall that for metric spaces, sequential compactness is equivalent to compactness, so it suffices to show that $F$ is sequentially compact. Suppose otherwise, so we have some sequence $x_n$ with no limit points. Define $f:\{1/n:n\in\mathbb N\}\cup \{0\}\to F$ by letting $f(0)$ be any point and $f(1/n)=x_n$. Then the graph of $f$ is closed, since each point in the graph is isolated (show this). Thus $f$ is continuous, so $$\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty} f(1/n)=f\left(\lim\limits_{n\to\infty}\frac1n\right)=f(0)$$ contradicting the fact that $x_n$ has no limit points.
Alex Becker
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I have tried to prove that the graph of the function is closed but I have not achieved, could helpme? – Roiner Segura Cubero Jun 26 '13 at 18:54
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@RoinerSeguraCubero Recall that a set consisting entirely of isolated points is closed. To see that each point $x_i$ is isolated, note that since $(x_n)$ has no convergent subsequence, we have some $\epsilon>0$ such that $d(x_i,x_j)>\epsilon$ for all $j\ne i$. Thus $d((1/i,x_i),(1/j,x_j))> d(x_i,x_j)>\epsilon$, so each point $(1/i,x_i)$ in the graph is isolated. – Alex Becker Jun 26 '13 at 19:35
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Alex as you could use the following result to prove my exercise. Let $E$ be a compact metric space, $F$ a metric space, $A$ a closed subset of $E\times F$. Show that the projection of $A$ into $F$ is a closed. – Roiner Segura Cubero Jun 26 '13 at 20:19
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@RoinerSeguraCubero No, that implication goes the wrong way. You know that the projection of $A$ is closed in this case, and you want to show that $A$ itself is closed. This holds in this case but not always. – Alex Becker Jun 27 '13 at 01:29