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A question similar to previous one with a slightly different condition. Consider $\Bbb R^n$. This previous proposition is shown not true by copper.hat

Given a convex set $C$ in $\Bbb R^n$, its affine hull is $A(C) = \{ax+by:x,y\in C, a,b \in \Bbb R, a+b=1\}$. A point $x\in C$ is said to be its interior point if there exists $d>0$ s.t. $B_d(x)\bigcap A(C) \subseteq C$; otherwise $x$ is a relative boundary point. $C$ is relatively open if every point $C$ is a relative interior point.

We say two convex sets are "distinct" if their intersection are only relative boundary points of $C_1$ and $C_2$. The problem is:

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 $C_1$,$C_2$.


Attempt: By Christian Bueno's method, I recap the following until I got stuck.

Let $C_1$ and $C_2$ be as above ("distinct", non-disjoint, convex sets in $\mathbb{R}^n$). Clearly $0 \in C_1 - C_2$

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

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

is open relative to $C_1 \times C_2$. Here "open relative" is a concept from real analysis. I believe this concept should be consistent with the idea of "relative open" w.r.t. affine hull above. I have trouble to go from here to the desired result.

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Since they are distinct we have $ri(C_1) \cap ri(C_2) = \emptyset$ Thus,

$$ 0 \notin ri(C_2)-ri(C_1)=ri(C_2-C_1)$$

(see here for above identity)

Therefore $0$ is not only a boundary point, but also a relative boundary point of the set $C_2 - C_1$.

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  • $\begingroup$ This is a great answer I am looking for! $\endgroup$ – Tony May 27 '17 at 15:55

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