# Proof theorem 3.17 Rudin's functional analysis

Consider the two theorems (3.15 and 3.16) of Rudin's functional analysis:

Theorem 3.15: If $$V$$ is a neighborhood of $$0$$ in a topological vector space $$X$$ and if $$K = \left\{\Lambda \in X^* : |\Lambda x| \leq 1 \; \text{for every} \; x \in V \right\}$$ then $$K$$ is weak*-compact

Theorem 3.16: If $$X$$ is a separable topological vector space, if $$K \subset X^*$$, and if $$K$$ is weak*-compact then $$K$$ is metrizable, in the weak*-topology.

How are such theorems used to prove the following

If $$V$$ is a neighborhood of $$0$$ in a separable topological vector space $$X$$, and if $$\left\{\Lambda_n \right\}$$ is a sequence in $$X^*$$ such that $$|\Lambda_n x | \leq 1 \;\;\; (x \in V, n = 1,2,3,\ldots)$$ then there's a subsequence $$\left\{ \Lambda_{n_i} \right\}$$ and there's a $$\Lambda \in X^*$$ such that $$\Lambda x = \lim_{i \to \infty} \Lambda_{n_i} x \;\;\; (x \in X)$$ In other words, the polar of $$V$$ is sequentially compact in the weak*-topology

I understand that $$\Lambda_n$$ belong to $$K$$, as defined in the theorem 3.15, also by theorem 3.16 such set is metrizable, but how do I get the conclusion?

Since $$K$$ is metrizable and $$w^*$$-compact, it is sequentially $$w^*$$-compact. This means that for any sequence $$\{\Lambda_n\}_n \subset K$$ we can extract a subsequence such that $$\Lambda_{n_k} \to \Lambda \quad \text{in weak^* sense}$$ to some $$\Lambda\in K$$. Now, convergence in the weak$$^*$$ sense simply means that $$\Lambda x =\lim_{k\to\infty}\Lambda_{n_k}x \quad \text{for all x\in X}.$$
• "Since $K$ is metrizable and w∗-compact, it is sequentially $w$∗-compact", why is that? – user8469759 Jan 16 '19 at 10:43
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