Closure of interior of closed convex set

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.

• what is the proof for normed spaces that won't translate (well, you do indicate using a cone)? For TVS, you may assume vector $0$ is in the interior of the set. Take any point $p$ in your closed convex set $C$, such that $p$ is not in the interior. Consider the line through $0$ and $p$, all points between $0$ and $p$ should belong to the interior of $C$, and $p$ should belong to the closure of this open line segment. I do not remember all relevant definitions, but this is how I would have started, to see if it would work. – Mirko May 19 '17 at 0:35

If $c \in \overline{C}$, $p \in C^\circ$ and $t \in [0,1)$ then the point $p(t) = p + t (c-p)$ is in $C^\circ$.

Since $p \in C^\circ$, there is some open neighbourhood $U$ of $0$ such that $U+\{p\} \subset C$.

Now I claim that if $x \in (1-t)U+\{p(t)\}$, then $x \in C$. In particular, $p(t) \in C^\circ$.

Let $y = {1 \over 1-t} (x-tc)$ and note that $x = (1-t)y + t c$.

Then $y-p = {1 \over 1-t}(x-tc+tp - p) = {1 \over 1-t}(x-p(t)) \in {1 \over 1-t} (1-t)U = U$.

Hence $y \in C$ and so $x \in C$.

• This seems to be the cone and adequate radii (for the $p(t)$) I was looking for but somehow I used triangle inequality for the normable case and I assumed I needed some kind of neighbourhood such that $E + E \subset U$. Anyways, thanks. – Will M. May 19 '17 at 1:46
• The corresponding property holds for the relative interior/closure as well. – copper.hat May 19 '17 at 2:05

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 $$\mathrm{int}(K)=\mathrm{int}(\mathrm{cl}(K)).$$