If $S$ is convex then $\text{cl}(S)$ and $\text{int}(S)$ are convex too.

Question

Definition

Let be $X$ a topological vector space. A subset $S$ of $X$ is said convex if the affine combination $$ A:=\{z\in X: z=(1-t)x+y, t\in[0,1]\} $$ is contained in $S$ for any $x, y\in S$.

Statement

If $S$ is convex then $\text{cl}(S)$ and $\text{int}(S)$ are convex too.

So unfortunately I don't be able to prove the last statement: I found the statement here but there are not explanations about it. So could someone help me, please?

Answer 1

Let $x,y$ be in in $\text{cl}(S)$, and $u \in [0,1]$. We want to show that $ux+(1-u)y \in \text{cl}(S)$ as well.

So let $x_i, y_i, i \in I$ be nets with common domain $I$ such that $x_i \to x$ and $y_i \to y$ as nets, and such that $\forall i: x_i \in S, y_i \in S$, by standard theory on nets.

Then as in TVS all linear operations are continuous we have that $\lim_i (ux_i + (1-u)y_i) = ux + (1-u)y$ and so the right hand is a limit of a net from $S$ (as $S$ is convex, we know that all $ux_i + (1-u)y_i \in S$) and so in $\text{cl}(S)$.

This shows the closure is indeed convex. And why nets are a handy tool sometimes.

As to interior, let $U=\operatorname{int}(S)$. Fix $t \in (0,1)$. Then $tU+ (1-t)U \subseteq S$ by convexity of $S$ and $tU$ and $(1-t)U$ are open in any TVS (multiplication by a non-zero scalar is a homeomorphism) and so is their sum $tU + (1-t)U$ (the sum of an open set $A$ and a set $B$ is just a union of translates of $A$ by all members of $B$ etc. standard fact about TVS's). All open subsets of $S$ are a subset of the interior of $S$, which is $U$, so $tU + (1-t)U \subseteq U$ and as $t$ is arbitrary, $U$ is convex.

Answer 2

Henno Brandsma has given a perfect answer. Here I shall give an alternative proof that $\operatorname{cl}(S)$ is convex (which does not use nets).

The map $\phi : X \times X \times [0,1] \to X, \phi(x,y,t) = (1-t)x + ty$, is continuous. Let $x, y \in \operatorname{cl}(S)$. We have to show that $\phi(x,y,t) \in \operatorname{cl}(S)$, that is, each open neighborhood $U$ of $\phi(x,y,t)$ contains a point of $S$. By continuity we find open neigborhoods $V_x$ of $x$ in $X$, $V_y$ of $y$ in $X$ and $W$ of $t$ in $[0,1]$ such that $\phi(V_x \times V_y \times W) \subset U$. There exist $x' \in V_x \cap S$ and $y' \in V_y \cap S$. But then $\phi(x',y',t) \in U \cap S$ because $S$ is convex.