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I was wondering whether there exists a function $f:\Bbb R\to\Bbb R$ that satisfies: $$\text{For all } y\in\Bbb R: \lim_{x\to y} f(x)=\infty.$$

Intuitively it seems to me like this is impossible. But I don't see how to prove it.

By definition we would have $$\forall y \in \Bbb R: \forall r \in \Bbb R_+: \exists \delta > 0: \forall x \in (y-\delta, y+\delta)\setminus\{y\}: f(x)>r,$$ and not I don't know how to proceed.

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Such a function does not exist, because $\mathbb{R}$ is uncountable and complete.

Let $A_n = \{x \in [0,1]: |f(x)| \leq n\}$. If $A_n$ was infinite, we could extract a strictly monotone subsequence $x_k$ converging to some $x \in [0,1]$ by compactness. Since $|f(x_k)| \leq n$ we have $\lim_{x_k\rightarrow x}f(x_k) \neq \infty$, which violates the assumption. Thus, $A_n$ is has to be finite, but then $[0,1] = \bigcup_{n=1}^\infty A_n$ can only contain countably many points, which is a contradiction.

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