Proof theorem 3.17 Rudin's functional analysis

Question

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?

Answer 1

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 $$ \Lambda x =\lim_{k\to\infty}\Lambda_{n_k}x \quad \text{for all $x\in X$}. $$