0
$\begingroup$

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 $\text{int}(C) = \emptyset$.

$\endgroup$
  • 1
    $\begingroup$ 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 '18 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 '18 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 '18 at 17:02
0
$\begingroup$

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 $\text{int}(C) = \emptyset$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.