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Is there a function $f\colon\mathbb R \to \mathbb R$, such that its limits at rational points approach infinity?

Eman Yalpsid
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Taa
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  • The answer I posted, which you accepted, was wrong. I assumed that $f(x)\gt n$ in a neighborhood $U_{n,q}$ of $q,$ which is nonsense; that inequality only has to hold in a deleted neighborhood $U_{n,q}\setminus{q}.$ I have edited my answer; hope it's right now. – bof Mar 17 '17 at 12:03

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Suppose $\lim_{x\to q}f(x)=\infty$ for each $q\in\mathbb Q.$ Given $q\in\mathbb Q$ and $n\in\mathbb N,$ there is a neighborhood $U_{n,q}$ of $q$ such that $f(x)\gt n$ for all $x\in U_{n,q}\setminus\{q\}.$ For each $n\in\mathbb N,$ the set $U_n=\bigcup_{q\in\mathbb Q}U_{n,q}$ is a dense open set, and $f(x)\gt n$ for every irrational $x\in U_n.$ By the Baire category theorem, the set $$\left(\bigcap_{n\in\mathbb N}U_n\right)\setminus\mathbb Q=\left(\bigcap_{n\in\mathbb N}U_n\right)\cap\left(\bigcap_{q\in\mathbb Q}(\mathbb R\setminus\{q\})\right)$$ is nonempty, i.e., there is an irrational number $x\in\bigcap_{n\in\mathbb N}U_n.$ Then $f(x)\gt n$ for all $n\in\mathbb N,$ which is absurd.

bof
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