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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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    It is literally the same proof. You ca$n$ do the proof for loc$a$l minimums from the link, then do the proof for m$a$ximums (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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