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Let $f$ be continuous on a compact subset $X$ of a metric space. If we put $A_h=\{x\in X:f(x)<h\}$ 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 $A_h = B_h^\circ$?

SBF
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    No. Consider $f(x) = x$ for $x \leq 0$ and $f(x) = \sin{x}$ for $x \geq 0$ (on a large interval $[-N,N]$). Then $A_{-1}$ is $[-N,-1)$, while $B_{-1} = A_{-1} \cup {0 \lt x \leq N : x = 2k\pi + 3\pi/2}$. – t.b. Jul 07 '11 at 10:40
  • @Theo: you're right, so there are connections with local minimums? – SBF Jul 07 '11 at 10:43
  • Well, I wouldn't quite put it this way (there is no reason for the points in $B_{h} \smallsetminus \overline{A_h}$ to be local minima. But of course, no local local minimum in $X \smallsetminus \overline{A_h}$ can be reached from $A_h$. – t.b. Jul 07 '11 at 10:55
  • @Theo: Could you provide an example when $B_h\setminus\overline{A_h}$ does not contain a local minimum? – SBF Jul 07 '11 at 11:00
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    $f(x) = x$ for $x \leq 0$, $f(x) = 0$ for $x \geq 0$ and $h = 0$. – t.b. Jul 07 '11 at 11:03
  • @Theo: Thanks, I should think more about it. I edited my question. – SBF Jul 07 '11 at 11:53
  • Concerning your follow up: no, as the second example shows. – t.b. Jul 07 '11 at 11:55
  • @Theo: thanks. Would you provide it in the form of the answer with just the 2nd example? I'll admit. – SBF Jul 07 '11 at 12:03

1 Answers1

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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.

Srivatsan
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t.b.
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