Extreme Value Theorem

See Compactness Is Preserved By Continuity which this corollary comes from.

If $f$ is continuous on a closed interval then $f$ attains an absolute maximum and an absolute minimum in that interval.

Proof

Let $a < b \in R$ . Take $f : [a, b] \to R$ to be continuous. By the Heine-Borel Theorem (Compact Equivalences) then our $[a, b]$ is compact. Therefore, by Compactness Is Preserved By Continuity then $f ([a, b])$ is compact. Compact gives that $f ([a, b])$ must be bounded, so then we can define $m := inf f ([a, b])$ and $M := sup f ([a, b])$ both exist.

$f ([a, b])$ are both closed (by being compact), so then our $m, M \in f ([a, b])$ . So then $\exists x_{1}, x_{2} \in [a, b]$ where $f (x_{1}) = m$ and $f (x_{2}) = M$ as desired (since these points $x_{1}, x_{2}$ ) are our minimum/maximum points.

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