Must every bounded nonempty set in $\Bbb R^n$ have a non empty boundary? I can't think of a counter example for this statement.
If any of you can find one or show that it is true, I would be appreciated.
Thanks.
 A: Suppose $U$ is a bounded non-empty set with empty boundary.
Then for every point $x\in U,$ there is a neighborhood of $x$ entirely contained in $U,$ and for every point $x\not\in U,$ there is a neighborhood of $x$ entirely contained in the complement of $U.$
So both $U$ and its complement are open and non-empty, contradicting the fact that the space is connected.
A: Suppose a nonempty bounded set $X$ has an empty boundary. Denote an $\epsilon$-ball of $x$ by $B(x,\epsilon)$ Then for every $x\in \Bbb R^n$, there is some $\epsilon>0$ such that $B(x,\epsilon)$ either contains only points in $X$ or only points in $\Bbb R^n\setminus X$. Of course, $x$ itself is in $B(x,\epsilon)$, so $B(x,\epsilon)$ must contain only points in $X$. In other words, every points in $X$ is an interior point, so $X$ is open.
Now do the same thing with $\Bbb R^n\setminus X$. For every $y\in \Bbb R^n\setminus X$, we have some $\epsilon>0$ such that $B(y,\epsilon)$ must contain only points in $\Bbb R^n\setminus X$. Thus $\Bbb R^n\setminus X$ is also open.
But now we have two non-empty, disjoint, open sets whose union is $\Bbb R^n$. This is impossible since there is no separation of $\Bbb R^n$.
