Find all of the subsets $M \subset \Bbb R^n$ such that $ \exists r>0 \; \forall x \in M: K_r(x) \subset M$.
(Where $K_r(x) = \{a \in \Bbb R^n : ||a-x||<r\}$)
My attempt:
After thinking about this for a while, the only two sets I could come up with were the empty set and $\Bbb R^n$. I think these are the only sets that have this property. My intuition is that such an $M$ would have to be unbounded in every direction or empty.
I want to prove it.
My idea is to show that the only sets that can have such property would have to be open and closed, hence only $\emptyset$ and $\Bbb R^n$.
It is clear that such a subset $M$ is open, just pick $\epsilon = r$. It is not clear to me that such a set $M$ is closed.
We have
$$m \in M^c \iff \forall r >0 \; \exists x \in M \; K_r(x) \not \subset M $$
How can we conclude that $M^c$ is open? Is that even the case?
It's clear that such an $M$ is open (all points are interior points with the same $r>0$ even). If $M$ is not closed, we can find $p \in \overline{M}\setminus M$. Then $p \in \overline{M}$ allows us to find some $m \in M$ with $d(x,p) < \frac{r}{2}$, and then $p \notin M$, yet $p \in K_r(m)$ contradicting $K_r(m) \subseteq M$.
So $M$ is also closed. Connectedness tells us that $M=\emptyset$ or $M=\mathbb{R}^n$. This argument works in any connected metric space.