Let $f: D \rightarrow \mathbb{R}$ be a continuous function, and let $S \subset D \subset \mathbb{R}^n$.

I want to check whether $f$ on $S$ fulfills the premises of the extreme value theorem. Is it true that the following two statements are equivalent because “compactness is a topological property?”

  • $S$ is closed and bounded in $D$.
  • $S$ is closed and bounded in $\mathbb{R}^n$.

In other words, can I decide which of these two statements to check?

An example: Let $f: x \rightarrow \ln(1 - x^2)$, so the domain $D$ of $f$ is $(-1, 1)$. Let $S$ be $[0, 1)$. Then $S$ is closed in $D$ but not bounded, and $S$ is bounded in $\mathbb{R}$ but not closed. Either way I conclude that the premises of the extreme value theorem are not fulfilled.

Any flaws with this?


For $S\subset D\subset X$ with $D$ equipped with the subspace topology, the set $S$ is compact in $D$ iff $S$ is compact in $X$ (see Q1 and Q2). Then by Heine-Borel theorem you need to check whether $S$ is a closed and bounded subset of $\mathbb{R}^n$ (and not of $D$).

  • $\begingroup$ Thank you. So, $S$ is closed and bounded in $D \Leftrightarrow S$ is closed and bounded in $\mathbb{R}^n$ does not hold in general? Could you provide a counterexample? $\endgroup$ – bbrot Jul 4 '18 at 7:55
  • $\begingroup$ It is your example. $S=[0,1)$ is closed in $D=(-1,1)$, but not in $\mathbb{R}$... $\endgroup$ – d.k.o. Jul 4 '18 at 8:43
  • $\begingroup$ Okay, but then again $S$ is not bounded in $D$ or is it? $\endgroup$ – bbrot Jul 4 '18 at 10:07
  • $\begingroup$ $S$ is contained in $B(0,1)$. en.wikipedia.org/wiki/Bounded_set#Metric_space $\endgroup$ – d.k.o. Jul 4 '18 at 14:45
  • $\begingroup$ Thank you. German Wikipedia says that the ball needs to be closed, but I guess that’s a matter of definition $\endgroup$ – bbrot Jul 5 '18 at 8:59

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