# 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.

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Could you reproduce that proof and explain which is the statement you don't understand? – a.r. Oct 21 '10 at 4:03

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).

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Thank you! As I understand it, one needs convexity only at the very end to argue that the perturbed simplex is contained in $C$. – Stefan Walter Oct 25 '10 at 14:09
Yes, that's true. But the characterization "interior point = convex-interior point" requires the whole, solid (n-dimensional) interior of the simplex to also be contained in the set, and works only for convex K. In this problem the chain of logic (and inclusions of spaces) is interior(closure of C) --> convex-interior(closure of C) --> convex-interior(C) --> solid-convex-interior(C) --> interior(C) so we do need the solid part but only later in the sequence. – T.. Oct 25 '10 at 16:44

I understand the problematic statement to mean that if a convex set $X$ has empty interior, then it is contained in a hyperplane. Let me know if I've misunderstood your problem.

Here's a sketch of a proof by contrapositive. If $X$ is not contained in a hyperplane, then you can find $n+1$ affinely independent points in $X$. Their convex hull is a simplex contained in $X$, and a nondegenerate simplex has nonempty interior. (I'm pretty sure that last statement is true, but I'd be hard pressed to prove it.)

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Every nondegenerate simplex is affinely isomorfic to the standard simplex. Show that the latter has a non-empty interior! – Mariano Suárez-Alvarez Oct 21 '10 at 21:28
@Mariano: Thanks, you've made it look obvious! :) – Rahul Oct 21 '10 at 22:35

This is from a set of class notes.

You need a lemma (interior and closure operators are $\mathrm{int},\mathrm{cl}$, resp.):

Lemma. If $C$ a convex subset of a topological space, and if $x\in \mathrm{int} C$ and $y\in \mathrm{cl} C$, then $[x,y)\subset \mathrm{int} C$.

The half-open segment $[x,y)$ is a right-open convex combination. If $\mathrm{int}C$ is empty, then the above lemma is true. Try to prove why this is true when the interior is not empty.

Proposition. $\mathrm{int} C = \mathrm{int}\;\mathrm{cl}\;C$.

Let $y\in \mathrm{int\; cl\;} C$; position it in an open ball, noting that the in the interior of the closure of $C$ is open, and hence $\exists r>0$ such that $B_r(y)\subset \mathrm{cl\;} C$. Pick a $y'\in \mathrm{int}\;C$. Then there exists an $\epsilon>0$, such that $y'+(1+\epsilon)(y-y') = y+\epsilon(y-y')\in B_r(y)\subset \mathrm{cl}\;C$. At the same time, $y$ belongs to the segment $[y',y+\epsilon(y-y'))$, so that, by the previous lemma, $y$ belongs to $\mathrm{int\;}C$. The reverse inclusion is straightforward, since $C\subseteq \mathrm{cl\;}C$, and thus obtaining the result.

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If $\operatorname{int}(C)$ is empty, how to complete the proof? – Xiang Yu Oct 13 '15 at 10:57

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 $\mathbb{R}^n$.

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Thank you, but as explained in my edit I came across this problem in a proof of the separating hyperplane theorem. As far as I can see, Mangasarian doesn't use that statement, but his proof is quite involved (for me), as it depends on a theorem called "Gordan's" theorem, which again follows from another theorem I've never heard of, and so on. – Stefan Walter Oct 21 '10 at 9:33