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Let $f$ be continuous on a compact subset $X$ of a metric space. If we put $A_h=\{x\in X:f(x) and $B_h=\{x\in X:f(x)\leq h\}$ - when is it true that $B_h = \overline{A_h}$? Is it true if and only if $A_h$ is not empty?

Edited: Theo already showed that there are counterexamples. Is it true then that $A_h = B_h^\circ$?

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    @Theo: thanks. Would you provide it in the form of the answer with just the $2$nd example? I'll admit.2011-07-07

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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.