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Prove that in $\mathbb R^n$:

If a set $X$ is convex and its complement $X^c$ is convex and non-empty, then the boundary is a hyperplane.

It seems really obvious intuitively. However proving this involves at least one page of work for me. Are there any theorem that I can cite/refer to?

dodo
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    About the validity of the claim, in infinite dimension there is the issue that you may have a discontinuous functional $\phi$ which determines two dense complementary convex sets $\phi^{-1}[0,\infty)$ and $\phi^{-1}(-\infty,0)$. At that point, the boundary of either won't be a hyperplane, but the whole space. –  Dec 19 '19 at 02:49
  • @Gae.S. The typical setting for this type of problem is $\mathbb R^n$, so I don't think the infinite dimensional case is too worrisome. – Math1000 Dec 19 '19 at 03:18
  • You'll need that neither $X$ nor $X^C$ are empty. – Olivier Roche Dec 19 '19 at 06:23
  • @OlivierRoche I thought a null set is a hyperplane? – dodo Dec 19 '19 at 06:24
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    a hyperplane in $\mathbb{R}^n$ has dimension $n-1$. – Olivier Roche Dec 19 '19 at 06:25
  • @OlivierRoche Oh, ok. I appreciate it. Question edited by the way. – dodo Dec 19 '19 at 06:26
  • @user293203 In infinite dimensions, can we claim that $X$ and $X^c$ are half space instead? – dodo Dec 13 '22 at 14:05

2 Answers2

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I don't know any theorem for that, but I have a relatively simple argument:

If $X$ is convex, then it can be written as the intersection of a family $\mathcal F$ of half-spaces (i.e. the part of the space on one side of a hyperplane). Similarly, $X^c$ can also be written as the intersection of another family $\mathcal G$ of half-spaces.

Now I claim that for any element $F\in \mathcal F$ and any element $G\in \mathcal G$, the corresponding hyperplanes must be parallel. This is because the union $F\cup G$ contains $X\cup X^c$, which is the whole space.

Consequently, assuming $X$ is neither empty nor the whole space, all cutting-hyperplanes in the family $\mathcal F$ are parallel, and similarly for $\mathcal G$. This implies that $X$ is a half-space.

I somehow ignored the difference between "open" and "closed" half-spaces, but this is unimportant, and we can of course change the condition to the closures of $X$ and $X^c$ being convex.

WhatsUp
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  • Of course this answer is correct, though I thought there must exist a simple proof for a simple statement like this. – dodo Dec 13 '22 at 14:03
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This is just a separation argument: the sets $X,X^c$ are both convex and non-empty, in addition $X \cap X^c=\emptyset$. So there is a separating hyperplane, i.e., there exists $a\ne0$ such that $$ a^Tx \ge a^T y \quad \forall x\in X, y\in X^c. $$ Define $j:=\inf_{x\in X} a^Tx$. Then the hyperplace $H:=\{ z:\ a^Tz = j\}$ separates $X$ and $X^c$.

Let $z$ be a boundary point of $X$ then there are sequences $x_k \to z$ in $X$ and $y_k\to z$ in $X^c$. Then $a^Tx_k \ge j \ge a^Ty_k$ and passing to the limit implies $z\in H$.

daw
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