If sets $A$,$B$ in Euclidean space are closed, convex sets, they have the same boundary and their interiors intersection is non-empty, can we say $A=B$? Any suggestions and comments are welcome!
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Convex sets are always path-connected and in particular connected. – k.stm Jul 15 '20 at 12:51
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1When you say Euclidean space, do you mean $\mathbb{R}^d$ for some finite $d$? – Keen-ameteur Jul 15 '20 at 12:54
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3First, drop the word connected, the answer is yes try to prove your statement by contradiction – Bernstein Jul 15 '20 at 12:54
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3Didn't you asked this here already? – drhab Jul 15 '20 at 12:56
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I have asked a weaker condition case already. When I say Euclidean space, I mean Rd for some finite d. – longjian li Jul 15 '20 at 12:59
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@longjianli The questions seem indiscernible right now. – Keen-ameteur Jul 15 '20 at 13:02
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1@drhab in the previous question he originally didn't require the sets to be convex, so the answer was "no" – Cronus Jul 15 '20 at 13:02
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This answer is about "$A,B $are closed set". But I want to show the case that "A,B are closed, convex sets" – longjian li Jul 15 '20 at 13:03
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3@Keen-ameteur I think he just edited his previous question after it was answered, and the answer there is an answer to a different question – Cronus Jul 15 '20 at 13:03
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@Berci This alone doesn’t suffice here now, does it? The premise about the interiors is only that the interiors of $A$ and $B$ meet nontrivially, not that they are the same. Surely it suffices now to show that these are the same, but this is nontrivial, – k.stm Jul 15 '20 at 13:05
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I just edited his previous question after it was answered, and the answer there is an answer to a different question . – longjian li Jul 15 '20 at 13:06
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1@longjianlii it would probably be a good idea to revert back to the original formulation over there in order to avoid confusion – Cronus Jul 15 '20 at 13:07
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Thank you @Cronus – longjian li Jul 15 '20 at 13:09
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@drhab Same question two times. – Unknown Jul 15 '20 at 15:22
2 Answers
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It is enough to show $A^\circ \subset B^\circ$, since the opposite inclusion follows by symmetry. Note that $A^\circ$ is convex (easy exercise) and therefore connected. Also, $A^\circ$ does not intersect $\partial A = \partial B$ and therefore $A^\circ$ is covered by the disjoint open sets $B^\circ$, $(B^c)^\circ$. Since $A^\circ$ has nonempty intersection with $B^\circ$, we must have $A^\circ \subset B^\circ$.
Previous version (a little longer but maybe easier to visualize):
Let $a \in A$; we will try to show $a \in B$. If $a \in \partial A = \partial B$ we are done because $B$ is closed and thus $\partial B \subset B$. So assume $a \in A^\circ$ (the interior of $A$).
By assumption there exists a point $x \in A^\circ \cap B^\circ$. Consider the line segment $L$ joining $x$ to $a$. If we take small open neighborhoods $U_a, U_x$ around $a,x$ that are contained in $A$, then their convex hull is open, contained in $A$, and contains $L$. So every point of $L$ is an interior point of $A$. Now if $a$ were not in $B$, then since $L$ is connected, it would intersect $\partial B = \partial A$, a contradiction.
Nate Eldredge
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I think the answer is yes, going by the complicated tool of Krein-Milman theorem. Assume for now that they are both bounded, then they are compact and by the Krein-Milman theorem it is enough to see that $\text{ext}(A)=\text{ext}(B)$. This is indeed the case since an extreme point of a convex set in in the boundary, see.
You can then define $A_n$ and $B_n$ by $A_n:=A\cap \{x\in \mathbb{R}^d: \; \Vert x\Vert\leq n \}$ and $B_n:=B\cap \{x\in \mathbb{R}^d: \; \Vert x\Vert\leq n \}$.
Since $A=\cup A_n$ and $B=\cup B_n$, it is enough to see that $A_n=B_n$. $A_n$ and $B_n$ are bounded, closed convex sets, so the first case applies, and $A_n=B_n$. You just need to verify that $\text{ext}(A_n)=\text{ext}(B_n)$.
Keen-ameteur
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Sorry, I am not familiar with ext. Can you show how to prove ext($A_n$)=ext($B_n$)? Thank you very much! – longjian li Jul 15 '20 at 13:21
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@longjianli Well that's a little complicated for me to write, but I'll try to write an edit explaining it. – Keen-ameteur Jul 15 '20 at 13:33
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I think there is significant work to be done in the last part, showing that $A_n, B_n$ have the same extreme points. This is going to have to use the assumption that the interiors of $A,B$ intersect, which hasn't been used yet and which is essential, but I don't see how to use it. – Nate Eldredge Jul 15 '20 at 14:23
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An extreme point is in the boundary, but not vice versa, so I'm not sure I understand your proof for the compact case... – Cronus Jul 15 '20 at 14:38