# 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. Commented May 19, 2017 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 $$x \in C$$.

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.

• 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. Commented May 19, 2017 at 1:46
• The corresponding property holds for the relative interior/closure as well. Commented May 19, 2017 at 2:05
• @copper.hat Thank you! The only modification I would make would be to choose $U$ balanced, to ensure that $\varepsilon \frac{1+t}{1-t}U \subset U$.
– Math
Commented Jun 17, 2022 at 10:49
• @VictorHugo the neighbourhood needs to be star shaped with respect to the origin, so a convex neighbourhood will suffice. Commented Jun 17, 2022 at 13:04
• My proof is not as general as hoped, it assumes the tvs is locally convex. Commented Nov 19, 2022 at 22:37

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

• If $U \cap K \neq \phi$ how does that imply that the intersection with int(K) is also non-empty ? Commented Mar 13, 2023 at 8:51

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).