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Let $\Bbb Q$ be the set of rationals and $f:\Bbb Q\to \Bbb R$ be a continuous function. Then $f$ is bounded on some interval? If not, what happen if in addition $f$ satisfies $f(xy)=f(x)f(y)$ for all $x, y\in \Bbb Q$?

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Let $x\in \mathbb Q$. Then as $f$ is continuous, there is $\delta>0$ such that

$$|f(y) - f(x)|<1$$

whenever $|y-x| < \delta $ and $y\in \mathbb Q$. Then

$$|f(y)| < |f(x)| +1$$

for all $y \in [x-\delta/2, x + \delta/2] \cap \mathbb Q$.

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  • $\begingroup$ It's perfect! thank you so much. $\endgroup$ – Chung. J Mar 21 '14 at 9:15
  • $\begingroup$ And the reasoning is good for any continuous $f:X\longrightarrow\Bbb R$, $X$ metric space. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 21 '14 at 10:29

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