The Closure of a Convex Set is Closed in a LCTVS

# The Closure of a Convex Set is Closed in a LCTVS

Proposition 1: Let $X$ be a locally convex topological vector space. If $K \subseteq X$ is convex then $\bar{K}$ is also convex. |

**Proof:**Let $x, y \in \bar{K}$. We want to show that $tx + (1 - t)y \in \bar{K}$ for every $t \in [0, 1]$. For a fixed $t \in [0, 1]$ and let $k = tx + (1 - t)y$. Hence we want to show that for every open neighbourhood $N$ of $k$ we have that $K \cap N \neq \emptyset$.

- So let $N$ be an open neighbourhood of $k$. Since $X$ is a locally convex topological vector space we have that $0$ has a local base consisting only of convex sets. Since $N$ is open, we can find a convex open neighbourhood $O$ of $0$ such that:

\begin{align} \quad k + O \subseteq N \end{align}

- To show that $N \cap K \neq \emptyset$ it suffices to show that $(k + O) \cap K \neq \emptyset$. We will find an element that is in $k + O$ and $K$.

- Consider $x + O$ which is an open neighbourhood of $x$. Also consider $y + O$ which is an open neighbourhood of $y$. Note that $(x + O) \cap K \neq \emptyset$ and $(y + O) \cap K \neq \emptyset$.

- So take $k_1 \in (x + O) \cap K$ and take $k_2 \in (y + O) \cap K$. Note that since $k_1, k_2 \in K$ we have that $tk_1 + (1 - t)k_2 \in K$ since $K$ is assumed to be convex. Moreover

\begin{align} \quad tk_1 + (1 - t)k_2 \in [tx + (1 - t)y] + O = k + O \end{align}

- Therefore $tk_1 + (1 - t)k_2 \in K$ and $tk_1 + (1-t)k_2 \in k + O$. So $(k + O) \cap K \neq \emptyset$ which shows that $N \cap K \neq \emptyset$ and thus $\bar{K}$ is convex. $\blacksquare$