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The space $X=C(\mathbb{R})=\{f:\mathbb{R}\to\mathbb{C}: f \text{ is continuous}\}$ is metric (not normed) and a Frechet space. I want to show that this space does not satisfy the Heine-Borel property (which means that any closed and bounded subset of $X$ is compact)

I feel like the collection $\{\exp(2\pi inx):n\in\mathbb{N}\}$ is a suitable candidate for a counterexample, since it is clearly bounded. How can I show that this set is closed, but not compact in the topology generated by the metric?

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    What metric are you using?2012-10-08
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    Or if it’s easier to describe, what countable family of seminorms are you using?2012-10-08
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    If you show the sequence has no convergent subsequence, this will show it is closed (because it has no limit points) and not compact (by Bolzano-Weierstrass).2012-10-08
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    I’m going to guess that $\|f\|_k=\sup_{x\in[-k,k]}|f(x)|$.2012-10-08

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