# Given $A$ a convex set of $\mathbb{R}^n$ such that $int(A) \neq \emptyset$, prove that $int(cl(A)) = int(A)$.

I was given this problem and I'm struggling with the $$int(cl(A)) \subseteq int(A)$$ contention. The other contention is trivial and is true regardless of whether $$A$$ is convex (that is that the line segment between any two points $$A$$ is completely contained in A, or rather $$\forall x, y \in A, 0 \leq \lambda \leq 1$$ its $$\lambda x + (1-\lambda)y \in A$$). Furthermore, I was able to come up with an example where the equality is not true if $$A$$ is not convex (take $$\mathbb{R}^n - \{(0,0)\}$$ ). However I really don't know what to do after taking $$x \in int(cl(A))$$ and a ball of radius $$\epsilon > 0$$ such that $$B_\epsilon(x)\subseteq cl(A)$$. I don't know how to incorporate the convexity of $$A$$ into the proof, I've tried assuming there's a point in $$B_\epsilon(x)$$ that is not in $$A$$ to arrive at a contradiction, but there's just something I'm missing.

From Wikipedia, I present one of the versions of the Hahn-Banach Separation Theorem:

Let $$X$$ be a real locally convex topological vector space and let $$A$$ and $$B$$ be non-empty convex subsets. If $$\operatorname{Int} A \neq \emptyset$$ and $$B\cap \operatorname {Int} A=\emptyset$$ then there exists a continuous $$\lambda \in X^*$$ such that $$\sup \lambda(A) \leq \inf \lambda (B)$$ and $$\lambda (a)<\inf \lambda (B)$$ for all $$a\in \operatorname {Int} A$$ (such a $$\lambda$$ is necessarily non-zero).

Fix $$a$$ in the boundary of $$A$$, and let $$B = \lbrace a \rbrace$$. Then $$B$$ is convex, and $$B\cap \operatorname {Int} A=\emptyset$$. Apply the theorem to obtain some such $$\lambda$$. Choose an $$x$$ in the underlying space such that $$\lambda(x) = 1$$ (remember, $$\lambda \neq 0$$), and note that, for $$\varepsilon > 0$$, $$\lambda(a + \varepsilon x) = \lambda(a) + \varepsilon > \lambda(a) = \inf \lambda(B) \ge \sup\lambda(A) = \sup \lambda(\overline{A}).$$ Therefore, $$a + \varepsilon x \notin \overline{A}$$. Note that, if $$a \in \operatorname{Int} (\overline{A})$$, this could not be true for sufficiently small $$\varepsilon$$. Hence, we have proven the contrapositive of what you want proven.

A definitive source is Rockafellar's "Convex Analysis". The result is Theorem 6.3 in said text.

(Well, Rockafellar deals with the relative closure & interior, but it amounts to the same thing if we restrict ourselves to $$\operatorname{aff} A$$.)

Since $$A \subset \overline{A}$$ we have $$A^\circ \subset \overline{A}^\circ$$.

For the other direction, the key result here is if $$y \in \overline{A}$$ and $$x \in A^\circ$$, then for $$\lambda \in [0,1)$$, we have $$(1-\lambda )x+ \lambda y \in A^\circ$$ (Rockafellar, Theorem 6.1). I will give a proof of this below.

Suppose $$z \in \overline{A}^\circ$$ and $$x \in A^\circ$$. Then for some $$\mu>1$$ sufficiently close to one we have $$y=(1-\mu)x+\mu z \in \overline{A}^\circ \subset \overline{A}$$ and $$z = {1 \over \mu} y + (1- {1 \over \mu})x$$ so from above, $$z \in A^\circ$$, hence we have the desired result.

Proof of Theorem 6.1: Rockafellar has a succinct proof, here is one with sequences.

Suppose $$y \in \overline{A}$$ and $$x \in A^\circ$$. There is some sequence $$y_k \to y$$ with $$y_k \in A$$.

Let $$K = \cup_{\lambda \in [0,1)} B((1-\lambda)x + \lambda y, (1-\lambda) \epsilon)$$, it is clear that this is open since it is the union of open sets. Furthermore, it is clear that $$y \in \overline{K}$$. It remains to show that $$K \subset A$$ (and so $$K \subset A^\circ$$).

Pick $$z \in K$$, then $$z = (1-\lambda)x+\lambda y + (1-\lambda) \delta$$, where $$\lambda \in (0,1)$$ and $$\delta \in B(0, \epsilon)$$.

If we let $$\delta_k = { z - \lambda y_k \over 1 - \lambda} -x$$, we see that $$\delta_k \to \delta$$, so for sufficiently large $$k$$, $$\delta_k \in B(0,\epsilon)$$ and since $$z = (1-\lambda)(x+\delta_k)+\lambda y_k$$, we see that $$z \in A$$ and hence $$z \in A^\circ$$.

• This is precisely what I was looking for, thanks a lot! – DeltaAccel Mar 12 at 20:24