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Is there any function $f:\mathbb R\rightarrow \mathbb R$ that has an infinite limit at all points?

There exist functions that are not bounded at any neighborhood at all points. For example $$f(x)=\begin{cases} n &\text{ for }x=\frac{m}{n}, \text{ for integers } m , n \text{ with }(m , n)=1 , n\neq0 \\ 0 & \text{otherwise.}\end{cases}$$ is a function with the above property. As you see this $f(x)$ is a function that has no limit at any points. It is natural to ask about functions that has an infinite limit at all points.

Arctic Char
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ali reza
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    Do you mean at any arbitrary point or at all points? – user2628206 Aug 12 '21 at 16:05
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    You might want to consider that all rationals sufficiently near $m/n$, but not equal to $m/n$ have arbitrarily large denominators. I would agree that you have a function that is not bounded in any neighborhood of $m/n.$ However, every neighborhood of $m/n$ also contains an irrational, where your function is $0.$ How do you reconcile that with the requirements for the existence of a limit? – Chris Leary Aug 12 '21 at 16:26

1 Answers1

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There is no such function.

Proof: For all $N \in \mathbb N$ and for each $y\in \mathbb R$, since $f(x) \to \infty$ as $x\to y$, there is $\delta_y >0$ so that $$|f(x)|\ge M,$$ for all $x\in \mathbb R$ and $0<|y-x|<\delta_y$. Then

$$\{ ( y-\delta_y, y+\delta_y) : y\in \mathbb R\}$$

forms an open covering of $\mathbb R$, and thus has a countable subcover, which we called

$$\{ ( y_n-\delta_{y_n}, y_n+\delta_{y_n}) : n\in \mathbb N\}.$$ As a result, the set
$$\tag{1} \{ y\in \mathbb R : |f(y)|<N\}$$ is countable since it's inside $\{y_1, y_2, \cdots, y_n, \cdots\}$.

But this is impossible, since

$$ \bigcup_{N\in \mathbb N} \{ y\in \mathbb R : |f(y)|<N\} = \mathbb R$$

and $\mathbb R$ is uncountable.

Remark: This question is very similar, in essence, to this question, and the technique is similar.

Remark Another way to show that the set in (1) is countable is by contradiction: if it is uncountable, then it has a limit point $x\in \mathbb R$. By definition of a limit point, there are $y_n \to x$ so that $|f(y_n)|<N$. But that contradicts to the assumption that $f(y)\to \infty$ as $y\to x$.

Arctic Char
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