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Consider $\Bbb R^n$. We say two convex sets are distinct if their intersection are only boundary points of $C_1$ and $C_2$. Prove zero vector is a boundary point of $C_2 - C_1 = \{x_2 - x_1: x_2\in C_2, x_1 \in C_1\}$ of two "distinct" non-disjoint convex sets $C_1$,$C_2$.

My attempt: if $0$ is not an interior point of $C$, then $B_d(0) \subseteq C$ for some $d>0$. Let $\|x\|<d$, then $\exists x_1,x_2,x_1',x_2'$ s.t. $x_2=x_1+x,x_2'=x_1'-x$, implying $0.5x_1+0.5x_1'=0.5x_2+0.5x_2'\in C_1\bigcap C_2$. However, I am not able to show $0.5x_1+0.5x_1'$ or $0.5x_2+0.5x_2'$ are not boundary points of $C_1,C_2$

Tony
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  • I don't get what you want to prove... Do you want two convex sets $C_1$ and $C_2$ that are "distinct" and such that $0 \in \partial(C_2-C_1)$? – fonfonx May 26 '17 at 03:10
  • Seems false unless the sets are also barely touching. I.e. the intersection of the boundaries is nonempty for the initial sets. Consider the set containing the number 1 and the set containing the number 2 in the real numbers. The set difference is a nonzero singleton point. – Nick Alger May 26 '17 at 03:11
  • Thanks. I missed the condition that they are non-disjoint. I have fixed this in the question. – Tony May 26 '17 at 03:15

2 Answers2

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Let $C_1$ and $C_2$ be as above ("distinct", non-disjoint, convex sets in $\mathbb{R}^n$).

Since they are "distinct" and non-disjoint, the intersection is nonempty and consists entirely of boundary points. So there is some $p\in\partial C_1\cap \partial C_2$. This means $p\in C_1$ and $p\in C_2$ so $0=p-p\in C_2-C_1$.

Now suppose for contradiction that $0$ is not a boundary point of $C_2-C_1$. Then there exists an $\epsilon>0$ such that $B_\epsilon(0)\subseteq C_2-C_1$. Let $s:C_1\times C_2\to \mathbb{R}^n$ be the subtraction map given by $s(v,w)=w-v$. This map is continuous so

$$s^{-1}(B_\epsilon(0))\subseteq C_1\times C_2$$

is an open neighborhood of $\{p\}\times\{p\}$. Since it is open, there must exists a $\delta>0$ such that

$$B_\delta(p)\times B_\delta(p)\subseteq s^{-1}(B_\epsilon(0))\subseteq C_1\times C_2$$

Thus $p$ is an interior point of $C_i$ which is a contradiction.

Thus $0$ must be a boundary point of $C_2-C_1$.

Christian Bueno
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    The open neighbourhood is open relative to $C_1 \times C_2$, but not necessarily open in $\mathbb{R}^n \times \mathbb{R}^n $. – copper.hat May 26 '17 at 05:02
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It is not true.

Consider the following subsets of the plane: $C_1 = [-1,1] \times \{0\}, C_1 = \{0\} \times [-1,1]$. Then $C_1 \cap C_2 = \{(0,0)\}$, $C_1 - C_2 = [-1,1]^2$. Hence $0 \in (C_1-C_2)^\circ$.

copper.hat
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