Topological properties of convex sets I was trying to prove some 'facts' regarding convex sets which seem natural, but I seem to stuck, so I was wondering whether tbey are indeed true, which I'll describe below.
Let $X$ be a topological vector space and let $K_1,K_2\subseteq X$ be closed convex subsets. I think the following are true:

*

*If $\text{int}(K_1)\neq \emptyset$, then $\overline{ \text{int}(K_1) }=K_1$.


*If $\partial K_1=\partial K_2$, then $\text{ext}(K_1)=\text{ext}(K_2)$.
I'm assuming there are counter-examples to these facts, but I can't think of any.
 A: The first is true. The inclusion $\overline{\operatorname{int} K_1} \subset K_1$ holds for all closed $K_1$, convex or not. To see the other inclusion, fix $a \in \operatorname{int} K_1$. For every $b \in K_1$,
$$\gamma_b \colon t \mapsto a + t(b-a),\quad t \in [0,1]$$
is a path in $K_1$ connecting $a$ and $b$, and $\gamma_b(t) \in \operatorname{int} K_1$ for $t \in [0,1)$ since
$$(1-t)\cdot (\operatorname{int} K_1) + tb$$
is an open neighbourhood of $\gamma_b(t)$ contained in $K_1$. Then $b = \lim_{t \to 1} \gamma_b(t)$ shows $b \in \overline{\operatorname{int} K_1}$.
The second doesn't hold. Trivial counterxample: $K_1 = \varnothing$ and $K_2 = X$.
Slightly less trivial counterexample: Let $H$ be a closed hyperplane(1) in $X$ (if that doesn't exist, the trivial counterexample might be the only one in $X$). We can take $K_1$ to be one of the closed half-spaces determined by $H$ and $K_2$ to be either $H$, or the other closed half-space determined by $H$. Then $\partial K_1 = H = \partial K_2$, but the exteriors are different ($\operatorname{int} K_1$ is contained in $\operatorname{ext} K_2$ and not empty).
I don't know if there are other counterexamples. I couldn't think of any, but neither do I see a proof that there are no others.

(1) Viewing $X$ as a real vector space, even if it also is a complex vector space.
