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This is the analogue of closed graph theorem for compact space

Suppose that $X$ and $K$ are metric spaces, that $K$ is compact, and that the graph of $f: X \rightarrow K$ is a closed subset of $X \times K$. Prove that $f$ is continuous.

Because metric space is first countable, all we need is to prove that if $x = \lim_{n \rightarrow \infty} x_{n}$, then $f(x) = \lim_{n \rightarrow \infty} f(x_{n})$. Suppose $x = \lim_{n \rightarrow \infty} x_{n}$, then because $K$ is compact, sequence $\{f(x_{n})\}$ has a subsequence $\{f(x_{n_{i}})\}$ converge. Because the graph of $f$ is closed, we must have $\lim_{n \rightarrow \infty}\{f(x_{n_{i}})\} = f(x)$. All we need to prove now is $\{f(x_{n})\}$ converge to $f(x)$, but I can't prove that. Can any one help me. I really appreciate. Thanks

le duc quang
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2 Answers2

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Suppose it weren't so. Then you'd have a subsequence $(x_{n_k})$ with $d(f(x_{n_k}), f(x)) \geqslant \varepsilon$ for some $\varepsilon > 0$.

By your argument, $f(x_{n_k})$ has a convergent subsequence $f(x_{n_{k_m}})$. Since the graph is closed, as you said, we must have $f(x_{n_{k_m}}) \to f(x)$. But that contradicts the assumption that $d(f(x_{n_k}), f(x)) \geqslant \varepsilon$ for all $k$.

Daniel Fischer
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This is an opportunity to use the following useful lemma.

Let $X$ be space, $(x_n)_n$ a sequence in $X$, $y\in X$. If every subsequence of $(x_n)$ has a subsequence that converges to $y$, then $(x_n)$ converges to $y$.

The lemma is easily proved by contraposition.

For a proof without sequences that therefore applies more generally, first show that for compact $K$ the projection $\pi\colon X\times K\to X$ maps closed subsets to closed subsets, then use $$f^{-1}[A]=\pi[(X\times A)\cap{\rm graph}\,f] .$$

Carsten S
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