I am self-studying topology using Viro et al.'s Elementary Topology Problem Textbook. The question above occurs early in the text, and though I have a solution, it feels a little ugly to me, and I would appreciate knowing if it can be made more elegant.
Here's what I have:
Fr $A$ is convex if Fr $A$ is contained in an $(n-1)$-dimensional affine subspace of $\mathbb{R}^n$. If $A$ is bounded and not contained in any $(n-1)$-dimensional subspace of $\mathbb{R}^n$, then Fr $A$ is not convex.
For the first claim, let Fr $A$ be contained in an $(n-1)$-dimensional affine subspace of $\mathbb{R}^n$, and let $x, y \in$ Fr $A$, $z \in [x,y]$. Since Fr $A$ = Cl $A \setminus$ Int $A$, $x,y \in $ Cl $A$. We already know that if $A$ is convex, Cl $A$ is convex as well. So $z \in$ Cl $A$. Now let $U$ be a neighborhood of $z$. Then there exists $\epsilon > 0$ such that $B_\epsilon(z) \subseteq U$, where $B_\epsilon(z)$ is the open ball of radius $\epsilon$ centered at $z$. Since $B_\epsilon(z)$ is an $n$-dimensional set while Fr $A$ is contained in an $(n-1)$-dimensional subspace, we cannot have $B_\epsilon(z) \subseteq Fr A$. Thus $U \not \subseteq$ Fr $A$, which means $z \notin$ Int $A$. So $z \in$ Fr $A = $ Cl $A \setminus$ Int $A$, which means that Fr $A$ is convex.
For the second claim, assume that $A$ is bounded and not contained in an $(n-1)$-dimensional affine subspace of $\mathbb{R}^n$. Then $A$ contains an open ball; let $B_\epsilon(z)$ be such a ball, where $\epsilon > 0$ and $z$ is the ball's center. Since $z \in B_\epsilon(z)$ and $B_\epsilon(z)$ is open, $z \in$ Int $A$.
Since $A$ is bounded, Cl $A$ is bounded, and certainly $z \in$ Cl $A$. Let $u \in \mathbb{R}^n$. Then $\{x \in \mathbb{R}^n|x = z + \lambda u; \lambda \in \mathbb{R}\}$ is a line passing through (z). Since Cl $A$ is bounded, we can define $\lambda_x = \sup\{\lambda > 0|z + \lambda u \in A\}$. By the definition of $\sup$, for any $\delta > 0$ we can find $\epsilon_0$ with $0 \leq \epsilon_0 < \delta$ such that $z + (\lambda - \epsilon_0)u \in A$ and $\epsilon_1$ with $0 \leq \epsilon_1 < \delta$ such that $z + (\lambda + \epsilon_1)u \notin A$. So $z + \lambda_xu \in$ Cl $A$ and $z + \lambda_xu \notin$ Int $A$. Thus $z + \lambda_x u \in$ Fr $A$. Let $x = z + \lambda_x u$. Similarly, define $\lambda_y = \sup\{\lambda < 0|z + \lambda u \in A\}$. Setting $y = z + \lambda_yu$, we have $y \in$ Fr $A$ as well. Now, $z \in [x,y]$, but since $z \in$ Int $A$, $z \notin$ Fr $A$. So Fr $A$ is not convex.
As I say, I do not believe my reasoning is wrong, so far as it goes; I just worry that I haven't covered all the bases. It would feel better to me if, for example, the boundedness requirement didn't put in an appearance, or if I could show that Fr $A$ is convex if and only if Fr $A$ is contained in an $(n-1)$-dimensional subspace of $\mathbb{R}^n$, but somehow something so general eludes me. Any thoughts or insight would be appreciated.
