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I believe my ideas are correct, but I have a feeling I am expressing them incorrectly in the proof.
Any insight, remarks, or corrections are immensely appreciated.
Thank you.

Note: $\text{cl}(X)$, $\text{int}(X)$, and $\text{bdry}(X$) represent the closure, interior, and boundary of $X$, respectively.


Let $C$ be a nonempty subset of $\mathbb{R}^n$ such that $\text{bdry}\, C$ is convex.


Part (a): Is it necessary that $\text{int}\,C=\emptyset$?

A counterexample is provided.

Let $x = (x_i)_{1 \, \leq \, i \, \leq \, n} \in \mathbb{R}^n$ and $C = \Big\{x \in \mathbb{R}^n : x_1 > 0\Big\} = \mathbb{R}^n_{x_1>0} \subset \mathbb{R}^n$.

$C$ splits the vector space $\mathbb{R}^n$ in half excluding the boundary hyperplane along the $x_1$-axis.

The boundary of $C$ can be obtained as $\text{bdry}(C)=\text{cl}(C)\cap\text{cl}\big(C^\text{C}\big)$.

The complement of $C$ evaluates to $C^\text{C} = \Big\{x \in \mathbb{R}^n : x_1 \leq 0\Big\} = \mathbb{R}^n_{x_1\leq0}$.

The closures $\text{cl}(C) = \mathbb{R}^n_{x_1\geq0}$ and $\text{cl}(C^\text{C}) = \mathbb{R}^n_{x_1\leq0}$ are both clearly convex since any segment formed by their respective elements is their subset (each closure is simply one half of the vector space $\mathbb{R}^n$ including the boundary hyperplane along the $x_1$-axis).

The intersection of convex sets is convex, thus $\text{bdry}(C)$ is also convex, as required.

The interior of $C$ evaluates to $\text{int}(C) = \text{cl}\big(C^\text{C}\big)^\text{C} = \mathbb{R}^n_{x_1>0} = C \neq \emptyset$.

Thus, if $\text{bdry}(C)$ is convex then $\text{int}(C)$ need not be empty.


Part (b): Repeat but with $C$ being bounded.

Prove equivalently the contrapositive: if $\text{int}(C)\neq \emptyset$ then $\text{bdry}(C)$ is not convex.

Let $x,y \in \text{bdry}(C)$ and since $\text{int}(C)\neq \emptyset$ then $\exists z \in \text{int}(C)$.

Since $\text{int}(C)\neq\emptyset$ and $C$ is bounded, then it is possible to choose $x$, $y$, and $z$ such that $z = (x+y)/2 \in \text{int}(C)$. Therefore, $\big[ \lambda x + (1-\lambda)y\big]_{\lambda \, = \, 1/2} \notin \text{bdry}(C)$.

Therefore, $\text{bdry}(C)$ is not convex.

Thus, for a nonempty bounded set $C$, if $\text{bdry}(C)$ is convex then $\text{int}(C) = \emptyset$.

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    $\begingroup$ for part $b$, it is not clear to me how boundedness is used; you seemingly hide all difficulty in "then it is possible to choose" $\endgroup$
    – LinAlg
    Sep 13, 2018 at 16:31
  • $\begingroup$ If you draw the case for $\mathbb{R}^2$ or even imagine it for $\mathbb{R}^3$ it is evident. I agree about the huge leap you are speaking of but here is my reasoning: (1) "$\text{int}(C)\neq\emptyset$" lets me put elements in $\text{int}(C)$. (2) "$C$ is bounded" lets me choose points on the boundary that have a midpoint in the interior, there is literally no way to 'convexify' the boundary without flattening it into a segment. (3) When $C$ is unbounded you can just have a convex (or affine) boundary on one side and let the rest extend indefinitely → convex boundary and non-empty interior. $\endgroup$
    – ex.nihil
    Sep 13, 2018 at 17:01
  • $\begingroup$ Essentially what I am trying to say with boundedness is: (1) If $C$ was unbounded you only need a convex boundary on one side and the rest is convex simply by extending indefinitely, allowing a nonempty interior. (2) If $C$ is bounded, then graphically "there is no escape", the interior must flatten in order to 'convexify' the boundary of the set. $\endgroup$
    – ex.nihil
    Sep 13, 2018 at 17:02

1 Answer 1

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I confirm the the proof is correct but Part (b) needs elaboration as below:


Part (b): Repeat but with $C$ being bounded.

Prove equivalently the contrapositive: if $\text{int}(C)\neq \emptyset$ then $\text{bdry}(C)$ is not convex.

Since $\text{int}(C)\neq \emptyset$ then $\exists z \in \text{int}(C)$.

Let $L$ be any line passing through $z$.

Since $C$ is bounded, then $L$ will intersect $\text{bdry}(C)$ at least twice.
Indeed, if not then $L$ will extend infinitely in one or two directions without intersecting $\text{bdry}(C)$ making $C$ unbounded in those directions.

Let $x,y\in L \cap \text{bdry}(C)$.

Therefore, $z=\lambda x + (1-\lambda)y$ for all $\lambda\in\;]0,1[$.

Since $z \notin \text{bdry}(C)$, then $\text{bdry}(C)$ is not convex.

Thus, for a nonempty bounded set $C$, if $\text{bdry}(C)$ is convex then $\text{int}(C) = \emptyset$.

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