Why does a convex set have the same interior points as its closure?

Let $C$ be a convex subset of $\mathbb{R}^n$. I've been trying for hours to prove that $\dot{\overline{C}}=\dot{C}$. Somehow my intuition completely fails me. I found a proof in a textbook, but just got stuck on another statement the author considered obvious. Could someone please give a proof that uses little more than elementary linear algebra, topology, and the definition of a convex set?

Edit: The proof mentioned above is from Blackwell and Girshick: Let $y\in\dot{\overline{C}}$ and $T$ be a ball around $y$ contained in $\overline{C}$. Then $C\cap T$ has an inner point, as otherwise $C\cap T$ would be contained in a hyperplane and $\overline{C\cap T}=\overline{T}$ would be contained in the same hyperplane. The problematic statement is "as otherwise $C\cap T$ would be contained in a hyperplane".

Another thing: I would be interested in a proof that doesn't use the theorem about separating a convex set from a point by a hyperplane, as I came across this problem in a proof of that very theorem (in the appendix of Stochastic Finance by Föllmer and Schied). To be more precise, it occurs in the case of the point in question being in the boundary of $C$, when it is tacitly assumed, that it is also in the boundary of $\overline{C}$. I know this isn't strictly necessary, as I could use another proof, e.g. the one referred to by Mike, but now I'm curious.

Solution 1:

Interior points of a convex set $K$ in $\mathbf{R}^n$ are the points in what could be called the "convex interior" of the set: they are inside (the part with positive barycentric coordinates of) at least one nondegenerate n-dimensional simplex with vertices in $K$.

Given a point in the interior of the convex set $\overline{C}$, surround it by a nondegenerate $n$-dimensional simplex with vertices in $\overline{C}$. Because $C$ is dense in its closure, we can perturb the simplex very slightly into one with vertices in $C$, and this simplex will continue to contain the given interior point. (This is because, for example, the barycentric coordinates are continuous functions of the simplex vertices, as long as the simplex does not degenerate, so a small perturbation will keep the coordinates positive and the point strictly inside the simplex).

Solution 2:

One direction is straightforward, so I'll only give the proof in the other direction.

Let $x \in \dot{\bar{C}}$. Then there exists an open ball $B_x$ containing $x$ such that $B_x \subset \bar{C}$.

Suppose $x \not\in \dot{C}$. But $x \in \bar{C}$, which means that $x$ is on the boundary of $C$. Therefore, $B_x$ contains at least one point $z$ not in $C$. Since $C$ is convex, there is a hyperplane that separates $z$ and $C$. But $z \in \dot{\bar{C}}$. Thus there exists another open ball $B_z$ containing $z$ such that $B_z \subset \bar{C}$. Now, intersect $B_z$ with the open half-space (from the separating hyperplane) that does not contain $C$ to obtain set $D$. We have $D \subset \bar{C}$ but $D \cap C = \emptyset$. Thus $D$ is a subset of the boundary of $C$. However, $D$ is open because it is the intersection of two open sets. $D$ thus has the property that it is an open set containing boundary points of $C$ yet fails to contain any points of $C$. Such a set cannot exist, and we have our contradiction.

This proof does bring in convexity via the separating hyperplane theorem - the version with a convex set and a point not in the set - rather than via the definition of a convex set. I'm not sure how to prove this without the separating hyperplane theorem (which is one of the fundamental properties of a convex set). If you want a proof of the separating hyperplane theorem it can be found on page 47 of Mangasarian. I'm also assuming that you're using the usual topology on $\mathbb{R}^n$.