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Can we prove the extreme value theorem by merely showing that a maximum exists (if $f$ is continuous and defined on a closed, bounded interval in $\mathbb{R}$) because then we'd apply this "half" of the theorem on $-f$ defined by $-f(x)$ which too is continuous on a closed bounded interval and since it has a maximum, $f$ must have a minimum?

The proof in my book deals with both cases separately (first showing maximum, then minimum), so I'm thinking something might be fishy about an argument like above?

Idelae
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1 Answers1

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No, that would be a valid, and shorter proof.

This is actually the method used on the wikipedia article (http://en.wikipedia.org/wiki/Extreme_value_theorem)

"We look at the proof for the upper bound and the maximum of f. By applying these results to the function –f, the existence of the lower bound and the result for the minimum of f follows."

WMycroft
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