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.
Best Answer
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 $x \in C$.
Correction: The above is incomplete. It does not use the fact that $c \in \overline{C}$ anywhere.
Suppose $U$ is a convex neighbourhood of $0$ such that $U+\{p\} \subset C$. Note that $c \in C+\epsilon U$ for any $\epsilon>0$. Pick some $t \in [0,1)$ and choose $\epsilon>0$ such that $\epsilon {1+t \over 1-t} \le 1$.
Now suppose $u \in U$ (so that $\epsilon u \in \epsilon U$), then \begin{eqnarray} p(t) + \epsilon u &=& (1-t)p+t c + \epsilon u \\ &\in& (1-t)\{p\} + t (C+\epsilon U) + \epsilon U \\ &\in& (1-t) (\{p\} + \epsilon {1+t \over 1-t} U) + t C \\ &\subset & (1-t)C + t C \\ &=& C \end{eqnarray} In particular, $p(t) \in p(t)+\epsilon U$, so $p(t) \in C^\circ$.
It follows immediately that $c$ is in the closure of the interior.