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## Homework Statement

Let ##f:\Bbb{R} \to \Bbb{R}## be a function such that ##f## has a local minimum for all ##x \in \Bbb{R}## (This means that for each ##x \in \Bbb{R}## there is an ##\epsilon \gt 0## where if ##\vert x-t\vert \lt \epsilon## then ##f(x) \leq f(t)##.). Then the image of ##f## is countable.

## Homework Equations

N/A

## The Attempt at a Solution

Let ##f:\Bbb{R} \to \Bbb{R}## be a function such that ##f## has a local minimum for all ##x \in \Bbb{R}##. Consider the arbitrary interval ##(r- \epsilon, r+ \epsilon)##. Since if ##q \in \Bbb{Q}## is in ##(r - \epsilon, r+ \epsilon)##, then ##(r - \delta, r + \delta) \subset (r - \epsilon, r + \epsilon)## such that ##\vert q-r\vert=\delta##. Hence we can define a mapping ##g:f(\Bbb{R}) \to \Bbb{Q}## as ##g(f(r))=\delta##. Note that we can assign a unique correspondence between elements in ##f(\Bbb{R})## and ##\Bbb{Q}## due to the density property; which states ##\forall x,y\in\Bbb{R} \exists q' \in \Bbb{Q}, x \lt q' \lt y##. Since ##\Bbb{Q}## is countable then so is ##f(\Bbb{R})##.