# How to prove that $C(\mathbb{R})$ does not have the Heine-Borel property

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? – Chris Eagle Oct 8 '12 at 17:08

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Or if it’s easier to describe, what countable family of seminorms are you using? – Brian M. Scott Oct 8 '12 at 17:16
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). – Nate Eldredge Oct 8 '12 at 17:25
I’m going to guess that $\|f\|_k=\sup_{x\in[-k,k]}|f(x)|$. – Brian M. Scott Oct 8 '12 at 17:26

## 1 Answer

I’m going to guess that $\|f\|_k=\sup_{x\in[-k,k]}|f(x)|$. If $m<n$, then $$\exp(2\pi inx)-\exp(2\pi imx)=\exp(2\pi imx)\Big(\exp(2\pi i(n-m)x)-1\Big)\;,$$ which at $x=\frac1{2(n-m)}$ is $-2\exp\left(\frac{\pi im}{n-m}\right)$. Let $f_n(x)=\exp(2\pi inx)$; then $\|f_n-f_m\|_k=2$ for all $k\in\Bbb Z^+$, and $\|f_n-f_m\|_0=0$, so $$d(f_n,f_m)=\sum_{k\ge 1}\frac{\|f_n-f_m\|_k}{1+\|f_n-f_m\|_k}2^{-k}=\frac23\sum_{k\ge 1}2^{-k}=\frac23\;.$$ Clearly $\langle f_n:n\in\Bbb N\rangle$ can have no Cauchy subsequence and hence no convergent subsequence.

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It is simpler to just pick any non-zero function with support contained in $[0,1]$ and consider the set of its integer translates. – Mariano Suárez-Alvarez♦ Oct 8 '12 at 18:08
@MarianoSuárez-Alvarez This is boring and not challenging :) Brian M. Scott (+1). – Norbert Oct 8 '12 at 18:59
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