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. 
 A: 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: 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 $z \in A^\circ$.
