Let $\phi : \mathbb{R} \to \mathbb{R}$ a continuous function. Show that for all $r \in \mathbb{Q}$, if $r$ is such that $\phi(r)=r$, than $\phi$ is the classic affine function.
This is a personnel question, I am not able to answer. Any helps?
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If $\phi$ is continuous, then as $\Bbb{Q}$ is dense in $\Bbb{R}$, $\phi$ is the identity function on $\Bbb{R}$. – Jan 22 '16 at 17:49
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2This question makes no sense as stated. Do you mean $\phi(r) = r$ for all $r \in \mathbb Q$? – Robert Israel Jan 22 '16 at 17:50
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I have modified the question – Jan 22 '16 at 17:51
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2This question (and its answers) might be what you are looking for. – Pierre-Guy Plamondon Jan 22 '16 at 17:53
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For any $x\in \mathbb{R}$, we can take a sequence of rational numbers $\{a_n\}$ which converge to $x$ because the rationals are dense in the reals. By (sequential) continuity, $$ \phi(x)=\lim_{n\to\infty} \phi(a_n)=\lim_{n\to\infty} a_n=x $$
TomGrubb
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