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Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuous function such that $f(\mathbb{R}$) is countable. Show that $f$ is constant.

Some ideas: proving the contrapositive. To suppose that $f$ is not constant and to use the intermediate value theorem to prove that $f(\mathbb{R})$ is not countable.

Any help?

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marked as duplicate by Clement C., user296602, carmichael561, Xander Henderson, JonMark Perry Sep 12 '17 at 4:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The image of a connected space under a continuous map is connected. The only connected subsets of $\mathbb{R}$ are one-point sets and intervals, but intervals are uncountable. QED.

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If $f$ is continuous on $\mathbb{R}$, then it satisfies the intermediate value property on $\mathbb{R}$. That is, if $a < b$ and $y$ is between $f(a)$ and $f(b)$, then there exists some $c \in (a,b)$ such that $f(c)$ is between $f(a)$ and $f(b)$.

If $f$ takes more than one value, say $f(a) \ne f(b)$ (in fact, assume without loss of generality that $f(a) < f(b)$), then for any value $y\in (f(a),f(b))$, there must be some $c\in (a,b)$ such that $f(c) = y$. But the interval $(f(a),f(b))$ is uncountable, thus if a continuous function takes more than one value, then it's range must be uncountable.

From this it follows that if a continuous function takes only countably many values, then it must be constant.

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If $x,y\in f(\Bbb R)$ and $x<y$, then $[x,y]\subset f(\Bbb R)$.

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