compactness of a general normed vector space

Question

Let $(V,||.||)$ be a normed vector space and suppose $S:=\{s\in V:||s||=1\}$ is sequentially compact then would $S^1:=\{s\in V:||s||\leq1\}$ be compact as well?

Is there a way of proving this or are there counterexample to this? I know that $S$ would be complete and totally bounded but I am not sure how to further that argument.

Thanks!

Answer 1

Define $f:[0,1]\times S \to S^{1}$ by $f(t,x) =tx$. This is continuos and onto. This implies that $S^{1}$ is compact.

Answer 2

Yes. Just consider the map$$\begin{array}{rccc}f\colon&S\times[0,1]&\longrightarrow&S^1\\&(v,t)&\mapsto&tv.\end{array}$$It is continuous and surjective and its domain is compact. Therefore, its range is compact too.

Answer 3

It does imply the sequential compactness. Let $(s_n)\subseteq S^1$ be a some sequence. We first pick a subsequence such that $(\Vert s_{n_k} \Vert )$ converges. Either it converges to $0$ (then we are done as in this case $(s_{n_k})$ converges to $0$) otherwise we consider $y_k:=s_{n_k}/\Vert s_{n_k}\Vert$, which admits a convergent subsequence $(y_{k_l})$ by the compactness of $S$. However, then also $(s_{n_{k_l}})= (\Vert s_{n_{k_l}} \Vert y_{k_l} )$ converges.