S is compact iff every continuous $ f: S \to \mathbb{R} $ has a maximum

compactnesscontinuitygeneral-topology

Given a set $ S\subseteq \mathbb{R} $ I have to prove that $S$ is compact iff every continuous function $f:S\to \mathbb{R}$ has a maximum.

$\Rightarrow$) $S$ is compact $\implies f(S)$ compact $\iff f(S)$ closed and bounded. Every closed and bounded subset of $\mathbb{R}$ has maximum and minimum.

I have few problems with the other part of the proof. In order to show that S is compact I tried to use the following theorem:

Let $f$ be a function $f:X \to Y$. If $Y$ is compact and $f^{-1}(y)$
is compact $\forall y \in Y$, then $X$ is also compact.

but I fail to prove that $X=f^{-1}(\{y\})$ is compact.
I know that it's closed because f is continuous, so the preimage of the closed set $\{y\}$ is closed, but how to prove that it's bounded?

Best Answer

If $S$ is not compact, then it is unbounded or it is not closed. So:

  • if $S$ is unbounded, take $f(x)=\lvert x\rvert$;
  • if $S$ is not closed, let $a\in\overline S\setminus S$ and take $f(x)=\frac1{\lvert x-a\rvert}$.

In both cases, $f$ is unbounded above and therefore it has no maximum.

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