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From Bolzano-Weirerstrass we can demonstrate that in a normed vector space $E$ of finite dimension, every bounded sequence admits a limit point.

What are some counterexamples in infinite dimension? Does there exist a counterexample in every infinite dimensional normed space?

I believe this one works: Let $E$ be the space of sequences of real numbers with finite support, equipped with the norm $\| (a_k)_{k \in \mathbb{N} } \|=\sup |a_k|$. Then take define the sequence $(s_n)$ as follows: $s_n$ is the sequence whose $n$th term is $1$ and every other term is $0$. Then $(s_n)$ is bounded, and we can easily show that it has no limit point.

Joonas Ilmavirta
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math_lover
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3 Answers3

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Yes. A normed vector space satisfies the Bolzano-Weierstrass property (i.e. any bounded sequence has a convergent subsequence) if and only if it is of finite dimension. This means there is a counterexample in any infinite dimensional normed vector space.

I only found the german Wikipedia article for this result (Riesz's compactness theorem), but it uses Riesz's lemma to proof it. I think the proof of the relevant direction might work something like this:

Proof.

Let $X$ be an infinite dimensional normed vector space.

Choose some $s_0\in X$ with $\|s_0\|=1$. For any $i\in \Bbb N$ we can define $U_i:=\mathrm{span}(s_0,...,s_{i-1})\subseteq X$ as a closed proper subspace. By the lemma of Riesz there exists $s_{i}\in X- U_i$ with $\|s_i\|=1$ and

$$d(s_{i},U_i):=\inf_{s\in U_i}\|s-s_{i}\|\ge 1/2.$$

The so built sequence is bounded, but no subsequence can be Cauchy, hence not convergent. $\square$

M. Winter
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Consider $X=L^2[0,2\pi]$ and $E=\{\sin nx:n\in\mathbb N\}$.

$E$ does not have a limit point in $X$. Assume otherwise that there is a subsequence $$\sin {n_k}x\to f(x)\in L^2[0,2\pi].$$ Then we would have that $$ \int_0^{2\pi} (\sin n_kx)^2\,dx- \int_0^{2\pi} f(x)\,\sin n_kx\,dx\to 0. $$ But $\int_0^{2\pi} f(x)\,\sin n_kx\,dx\to 0$, while $\int_0^{2\pi} (\sin n_kx)^2\,dx=\pi$.

Yiorgos S. Smyrlis
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Yes, that's the obvious counter example. You don't need to require finite support, when using supremum norm you only need the sequences to be bounded. Let the $j$th sequence be $k\to \delta_{jk}$ where $\delta_{jk}$ is the Kronecker delta. Every element in the sequence has norm $1$ and the distance between any two elements is $1$ so clearly this is a bounded sequence that and no subsequence is Cauchy and therefore not convergent.

skyking
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