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A continuous function $f:\mathbb R \to \mathbb Q$ must be constant.

My Attemps: Suppose it is not. Then there must be $x,y\in \mathbb R $ such that $f(x) \neq f(y)$. Since $f$ is continuous it achieves all values between $f(x)$ and $f(y)$.

Later on, I couldn't move. Could you help me? Thanks in advance.

Fuat Ray
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  • $f(x)$ and $f(y)$ are distinct rational numbers. What can be said about the values between them? – Integrand Sep 06 '23 at 20:48
  • You are almost there. Hint: use the density of $\mathbb R\setminus\mathbb Q$ in $\mathbb R$. – Joe Sep 06 '23 at 20:50
  • Thank you very much for the hint. There is a irrational number between two rational number call it $s$. As I stated beforee, $f$ achieves all values between $f(x)$ and $f(y)$ however $s$ is irrational. Impossible. – Fuat Ray Sep 06 '23 at 20:56
  • The continuous image of a connected set is connected. What are the connected subsets of $\mathbb{Q}$? – MPW Sep 06 '23 at 20:59

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