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Showing range is countable

I was going through my old notebooks and found this. I thought it was good enough to share.

Problem: Let $f(x)$ be a function $\mathbb R \rightarrow \mathbb R$, with the only restriction on it being that it has a local extremum at each point. Prove it has countable range.

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    Bravo! Great poser!2012-12-20
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    So, was this problem brought to your attention by a cute undergraduate?2012-12-20
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    @WillJagy Unfortunately not :(.2012-12-20
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    This is similar to the question [Showing range is countable](http://math.stackexchange.com/questions/45035/showing-range-is-countable). I actually voted to close as duplicate, but perhaps hastily as here it is a little more general, assuming extremum rather than local minimum. [Shai Covo's answer](http://math.stackexchange.com/a/45047/) there is on this generalization.2012-12-20
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    @Jonas: one of the answers in the question you linked answers the case of arbitrary extrema (and gives references, too).2012-12-20
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    It is literally the same proof. You can do the proof for local minimums from the link, then do the proof for maximums (which is the same). Then notice the range is a subset of the union of two countable sets. Further, if this function is continuous, it is constant.2012-12-20

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