Rudin's functional analysis theorem 3.12

Question

Suppose $E$ is a convex subset of a locally convex space $X$. Then the weak closure $\overline{E}_w$ of $E$ is equal to its original closure $\overline{E}$.

The proof starts as follows

$\overline{E}_w$ is weakly closed, hence originally closed, so that $\overline{E} \subset \overline{E}_w$.

I don't get that inclusion, could anyone expound it please?

The proof also continues for the opposite inclusion, but I don't get that either

To obtain the poosite inclusion, choose $x_0 \in X, x_0 \notin \overline{E}$. Part (b) of the separation theorem 3.4 shows that there exists $\Lambda \in X^*$ and $\gamma \in \mathbb{R}$ such that, for every $x \in \overline{E}$ $$ Re \; \Lambda x_0 < \gamma < Re \; \Lambda x $$ The set $\left\{ x : Re \; \Lambda x < \gamma \right\}$ is therefore a weak neighborhood of $x_0$ that does not intersect $E$, thus $x_0 \notin \overline{E}_w$. This proves $\overline{E}_w \subset \overline{E}$

Answer 1

First, the weak topology is weaker than the original topology (it has fewer open and closed sets). This means that every weakly closed set is also a closed set.

Second, the closure of $E$ is the smallest closed set that contains $E$. Since the weak closure of $E$ contains $E$ and is closed, it must include the closure of $E$.

Answer 2

Second part: $\{x: \Re \Lambda x <\gamma\}$ is the inverse image under $\Lambda$ of $\{z\in \mathbb C: \Re z <\gamma\}$ which is open in $\mathbb C$. Since each continuous linear functional is continuous for the weak topology this inverse image is an open set in the weak topology. This open set contains $x_0$ but it contains no point of $E$ (because $\Re \Lambda x >\gamma$ for all $x \in E$). This implies that it cannot belong to the weak closure of $E$. Hence $x_o\notin \overline {E}$ implies $x_0$ does not belong to weak closure of $E$. In other words, weak closure of $E$ is a subset of $\overline {E}$.