Topology : Convex set

Question

$A$ is a convex set on a Topological vector space E. How to prove that : $$\overline{\mathring{A}}\supseteq\overline{A}$$ and $$\mathring{\overline{A}}\subseteq\mathring{A}$$

we suppose that $\mathring{A}\neq \emptyset$.

Answer 1

If $x \in \overline{A}$ and $y \in A^\circ$ then $tx+(1−t)y \in A^\circ$ for all $t \in [0,1)$ (this is what convexity buys).

Choosing $t=1-{1 \over n}$ (thanks Daniel) gives a sequence of points in $A^\circ$ converging to $x$, hence $x \in \overline{A^\circ}$ and so $\overline{A} \subset \overline{A^\circ}$.

For the second, we proceed in a similar way:

Suppose $x \in \overline{A}^\circ$, then $x \in \overline{A}$. Choose $y \in A^\circ$ and then, as above, $tx+(1−t)y \in A^\circ$ for all $t \in [0,1)$. Since $x \in \overline{A}^\circ$, here is some $t^* >1$ such that $x^* = t^*x+(1−t^*)y \in \overline{A}$. Hence we can write $x = s x^* + (1-s)y$ for some $s \in [0,1)$ and so $x \in A^\circ$. Hence $\overline{A}^\circ \subset A^\circ$.

Since $A \subset \overline{A}$, we have $A^\circ \subset \overline{A}^\circ$ so it follows that $A^\circ = \overline{A}^\circ$ in this case.

Addendum: To see why the $t^*$ exists, let $p(t) = tx + (1-t)y$ and note that $p$ is continuous and $p(1) =x \in \overline{A}^\circ$. Hence $p^{-1} (\overline{A}^\circ)$ is open and contains $1$, hence there is some $\delta >0$ such that $(1-\delta, 1+\delta) \subset p^{-1} (\overline{A}^\circ)$. Pick $t^* = 1+ {1\over 2} \delta$, then $p(t^*) \in \overline{A}^\circ$.