Just wanted to follow up on this question: Sequential compactness of unit sphere implies sequential compactness of closed balls
I'm doing the same question, and have finished the first proof. How would I now, as a consequence, show that V is complete? The definition of complete is that every Cauchy sequence converges. Thanks.
If $v_n$ is a Cauchy sequence, $v_n$ is bounded. To see this suppose $\varepsilon = 1$ is given. Then there is $N\in \mathbb{N}$ such that $||v_n- v_m||< 1$ for $n, m\ge N. $ In particular, $$||v_n|| = ||v_n - v_N + v_N||\le ||v_N|| + 1$$ for $n\ge N$. Since there are only finitely many $k< N$ the sequence as a whole is bounded.
So $(v_n)$ is contained in some closed ball, which (due to your assumptoins about sequential compactness) implies it has a convergent subsequence $v_{n_k}$ with $$\lim_k v_{n_k} =: v$$ say. Since $v_n$ is Cauchy, it is now not difficult to see that it is actually the whole sequence $(v_n)$ which converges to $v$ - this I leave as a final exercise to you.
