Closure of interior of closed convex set

Question

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.

Answer 1

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.

Answer 2

Caution: The following proof only works for Hausdorff topological vector space which is different from OP.

For a convex set $C$ in a topological vector space $X$, for any $\lambda \in [0,1)$

$$\lambda \overline{C} + (1-\lambda)C^{\circ} \subseteq C^{\circ}$$

The claim follows from the fact that, $\displaystyle \lambda \overline{C} + (1-\lambda)C^{\circ} = \bigcup_{x \in \overline{C}} \lambda x + (1-\lambda) C^{\circ}$ is a union of open sets, hence open. It suffices to show that, $\displaystyle \lambda \overline{C} + (1-\lambda)C^{\circ} \subset C$.

Since, for any $x \in C^{\circ}$, $C^{\circ} - x$ is an open neighbourhood of $0$ in X. Hence, $$\lambda \overline{C} = \overline{\lambda C} = \bigcap_{0 \in V \underset{\text{open}}{\subset} X} (\lambda C + V) \subseteq \lambda C + (1-\lambda)(C^{\circ} - x)\\ \implies \lambda \overline{C} + (1-\lambda)C^{\circ} = \bigcup_{x \in C^{\circ}} \lambda \overline{C} + (1-\lambda)x \subseteq \lambda C + (1-\lambda)C^{\circ} \subset C$$

Now, letting $\lambda \to 1^-$, we have $\displaystyle \overline{C} \subseteq \overline{C^{\circ}}$ (the other inclusion is trivial).

Answer 3

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