From Munkre Topology page 246. Let U be an open set in a metrizable space X. The set $S_\epsilon(U)$ is obtained by "shrinking" U a distance of $\epsilon$.
$$S(U)=\{x|B(x,\epsilon)\subset U\}$$
Munkre states that this set is closed, which I am able to prove. One can prove $U-S(U)$ is open by picking any point $y\in U-S(U)$ and prove by contradiction that there must exist a small ball around y such that $B(y,\delta)\subset U-S(U)$.
My question is that if one can claim a similar statement: let U be an open set in a metrizable space X. Then the set
$$C(U)=\{x|\bar{B}(x,\epsilon)\subset U\}$$
is open ($\bar{B}(x,\epsilon)$ denotes the closure of the ball).
I try to prove the statement using the same approach as above but it did not go through. I was wondering if this statement is true or if someone can come up with a counterexample.
Best Answer
Consider $X = \Bbb R \setminus \{-1,1\}$ and $U = (-1,1)$ and $\varepsilon = 1$.
Then, $C(U) = \{0\}$ is not open.