If sets $A$,$B$ are closed, convex sets, they have the same boundary and their interiors intersection is non-empty, does $A=B$?

Question

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!

Answer 1

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 $\text{ext}(A_n)=\text{ext}(B_n)$.

Answer 2

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 $A^\circ \subset B^\circ$.

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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.