TITLE: Closure of interior of closed convex set
QUESTION [5 upvotes]: Consider a closed convex set with non empty interior in a topological vector space (a vector space endowed with a topology that makes sum and scalar multiplication continuous). Show that the closure of its interior is the original set itself.
I have already proved the case for normed spaces (if $x$ lies in the interior and $z$ is any other point, there is a “cone”, so to speak, whose base is a ball centres at $x$ and whose corner is $z$). But the proof doesn't translate (I am using triangle inequality in the normed case which I don't see how to translate).
Any help is appreciated.

REPLY [3 votes]: Let $X$ be a topological vector space. For a subset of $A\subset X$, 
let $\mathrm{cl}(A), \mathrm{int}(A)$ and $\mathrm{conv}(A)$ denote the 
closure, interior and convex hull of $A$, respectively. The claim you are proving may be slightly 
generalized as follows: If $K$ is a convex set of $X$ and if $\mathrm{int}(K)\neq \varnothing$, 
then 
$$
\mathrm{cl}(K)=\mathrm{cl}(\mathrm{int}(K)). 
$$
To show this, the following lemma is useful. If $U$ be an open neighborhood of $x_0\in X$ and $x\notin U$, then the point $\alpha x_0+(1-\alpha )x$ is an interior point of $\mathrm{conv}(U\cup \{x\})$ for every $\alpha \in (0,1]$ (this can be proved without difficulty) 
We turn to the proof of the last display. We only need to show that 
$\mathrm{cl}(K)\subset \mathrm{cl}(\mathrm{int}(K))$ and first consider the case that $x\in K$. Assume that there exists an open neighborhood $U$ of $x$ such that $U\cap \mathrm{int}(K)=\varnothing$. Choose a point $x_0\in \mathrm{int}K$. By 
the lemma above, $\alpha x_0+(1-\alpha )x\ (0<\alpha \leqslant 1)$ are interior points of $\mathrm{conv}(\mathrm{int}(K)\cup \{x\})$ and a fortiori of $K$. However, by continuity of scalar multiplication, $\alpha x_0+(1-\alpha )x\in U$ for all sufficiently small $\alpha$, a contradiction. Thus, for all open neighborhoods $U$ of $x$, we have $U\cap \mathrm{int}(K)\neq \varnothing$. 
If $x\in \mathrm{cl}(K)$, then for every open neighborhood $U$ of $x$, there exists a point $x^{\prime}$ in $U\cap K\neq \varnothing$. It follows that $U\cap \mathrm{int}(K)\neq \varnothing$. This implies that $x\in \mathrm{cl}(\mathrm{int}(K))$. 
If $\mathrm{int}(K)\neq \varnothing$, we can also prove that
$$
\mathrm{int}(K)=\mathrm{int}(\mathrm{cl}(K)). 
$$