Let $f$ be continuous on a compact subset $X$ of a metric space. If we put $A_h=\{x\in X:f(x)
Edited: Theo already showed that there are counterexamples. Is it true then that $A_h = B_h^\circ$?
Let $f$ be continuous on a compact subset $X$ of a metric space. If we put $A_h=\{x\in X:f(x)
Edited: Theo already showed that there are counterexamples. Is it true then that $A_h = B_h^\circ$?
On Ilya's request, I'm posting my second example as an answer.
Let $X = [-N,N]$ and $f(x) = \begin{cases}x &\text{if }x \leq 0,\\0 &\text{if }x \gt 0.\end{cases}$ For $h = 0$ we have $A_{0} = [-N,0)$, $B_0 = [-N,N]$. Thus, neither $\overline{A_h} = B_h$ nor $A_h = (B_h)^{\circ}$ in general.