Let $(x_d), (y_d)$ be nets such that $x_d \to a$ and $y_d \in \overline{\operatorname{conv} \{x_e \mid e \ge d\}}$. Then $y_d \to a$

Question

In solving Ex 3.13.1 in Brezis's book of Functional Analysis. I come across below claims.

Let $E$ be a locally convex t.v.s. and $(x_d)_{d\in D}$ a net in $E$ such that $x_d \to a\in E$. Let $$ X_d := \operatorname{conv} \{x_e \mid e \ge d\} \quad \forall d \in D. $$

1. Let $(y_d)$ be a net such that $y_d \in X_d$. Then $y_d \to a$.
2. Let $(y_d)$ be a net such that $y_d \in \overline{X_d}$. Then $y_d \to a$.

While I managed to prove claim 1., I'm unable to prove claim 2..

Proof of 1.: Let $U$ be a neighborhood (nbh) of $a$. WLOG, we assume $U$ is convex. There is $d$ such that $x_e \in U$ for all $e\ge d$. This means $X_e \subseteq U$ and thus $y_e \in U$ for all $e \ge d$. Hence $y_d \to a$.

Is claim 2. actually true? If not, could it be true if I impose that $E$ is also Hausdorff?

Answer 1

Let $U$ be an open neighbourhood of $a$. By local convexity (and regularity (we don't need Hausdorff, all uniform spaces are completely regular and regular, whether separated ($T_2$) or nor) there is an open convex neighbourhood $W$ of $a$ so that $\overline{W} \subseteq U$.

As $x_d \to a$ we get that for some $d_0$ we have that for all $d \ge d_0$: $X_d \subseteq \overline{W}$ (as in Kavi Rama's proof, using that $W$ is convex) and so also $(y_d \in) \overline{X_d} \subseteq \overline{W} \subseteq U$, and so $y_d \to a$, as witnessed by $d_0$ for the $U$ etc.

Answer 2

For any neighborhood $V$ of $a$ there is a convex open set $U$ containing $a$ and contained in $V$. The net $(x_d)$ is eventually in $U$ and this implies that $X_d \subset U$ for some $d$. It follows that $y_d' \in U \subset V$ for all $d' \geq d$ so $(y_d) \to a$.