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Let $(V,\left\lVert\cdot\right\rVert)$ be a normed vector space whose unit sphere $\left\{v \in V: \left\lVert v \right\rVert = 1\right\}$ is sequentially compact. Show that any closed ball $\left\{v \in V: \left\lVert v \right\rVert \leq R\right\}$ must be sequentially compact. Show that $V$ is complete.

My only idea was to note that if $(v_n)$ is a sequence in the closed ball, then $\left(\frac{v_n}{\left\lVert v_n \right\rVert}\right)$ is a sequence in the unit sphere, so has a convergent subsequence, but that doesn't really seem to help.

Prasiortle
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1 Answers1

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You have started the proof correctly. Now consider two cases:

Case 1) $0$ is a limit point of $(\|v_n\|)$. In this case show that $(v_n)$ has a subsequence converging to $0$.

Case 2. There is a subsequence of $(\|v_n\|)$ converging to some positive number $r$. In this case use $v_n=\frac {v_n} {\|v_n\|} \|v_n\|$ to get a subsequence of $(v_n)$ which converges.

Kavi Rama Murthy
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    You may also argue that the map $(\lambda,x)\in [0,1]\times S^1 \mapsto \lambda x \in B_1(0)$ is continuous and surjective. – Ruy Nov 09 '20 at 13:31
  • @Kavi Rama Murthy, could you please show me the epsilon delta of how $v_N $converges based on the fact that $\frac{v_N}{| v_N | } $converges? – Martin Geller Mar 30 '22 at 10:28
  • @MartinGeller Say $w_n = v_n/\lVert v_n\rVert$. Then the sequential compactness of the unit sphere gives a convergent subsequence $w_{n_r}$. Now $\lVert v_{n_r}\rVert$ certainly lies in $[0,R]$, so by Bolzano-Weierstrass, there's a convergent subsequence $\lVert v_{n_{r_s}}\rVert$. Since $w_{n_r}$ converges, of course $w_{n_{r_s}}$ also does, so $\lVert v_{n_{r_s}}\rVert \cdot w_{n_{r_s}} = v_{n_{r_s}}$ converges (product of convergent sequences converges). Thus $v_{n_{r_s}}$ is the desired convergent subsequence of $v_n$. – Prasiortle Sep 21 '22 at 06:15