The Closure of a Convex Set in a TVS

Proposition 1: Let $E$ be a topological space. If $A \subseteq E$ is convex then $\overline{A}$ is convex.

• Proof: Let $A \subseteq E$ be convex. Then, for all $x, y \in A$ and for all $\lambda, \mu \geq 0$ with $\lambda + \mu = 1$, we have that $\lambda x + \mu y \in A$.

• Let $a, b \in \overline{A}$, $\lambda, \mu \geq 0$ with $\lambda + \mu = 1$, and let $U$ be a neighbourhood of the origin. Since $E$ is a topological vector space, there exists a balanced neighbourhood of the origin, $V$, such that $V + V \subseteq U$ (see the proposition on the Bases of Neighbourhoods for a Point in a Topological Vector page). Then $a + V$ is a neighbourhood of $a$ and $b + V$ is a neighbourhood of $b$, so since $a, b \in \overline{A}$ we have that $A \cap (a + V) \neq \emptyset$ and $A \cap (b + V) \neq \emptyset$.

• So take $x \in A \cap (a + V)$ and $y \in A \cap (b + V)$. Then:

(1)

\begin{align} \quad \lambda x \in \lambda (A \cap (a + V)) = (\lambda A) \cap (\lambda a + V) \quad \mathrm{and} \quad \mu y \in \mu (A \cap (b + V)) = (\mu A) \cap (\mu b + \mu V) \end{align}

• Therefore:

(2)

\begin{align} \quad \lambda x + \mu y \in (\lambda A + \mu A) \cap (\lambda a + \mu b + \lambda V + \mu V) \end{align}

• Since $V$ is balanced and since $\lambda \leq \lambda + \mu = 1$ we have that $\lambda V \subseteq V$. Similarly, $\mu V \subseteq V$. Also, $\lambda A + \mu A \subseteq A$, since if $z \in \lambda A + \mu A$ then $z = \lambda a' + \mu b'$ for some $a', b' \in A$, so that by the convexity of $A$, $z \in A$. Thus:

(3)

\begin{align} \quad \lambda x + \mu y &\in A \cap (\lambda a + \mu b + V + V) \\ & \in A \cap (\lambda a + \mu b + U) \end{align}

• Since every neighbourhood of $\lambda a + \mu b$ is of the form $\lambda a + \mu b + U$ (since $E$ is a topological vector space and since $U$ is a neighbourhood of the origin), and since $A \cap (\lambda a + \mu b + U) \neq \emptyset$ from the above inclusion, we conclude that $\lambda a + \mu b \in \overline{A}$. Thus $\overline{A}$ is convex. $\blacksquare$