# Function that converges to $\infty$ at every point

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.

• Wouldn't $\{x:f(x)\le n\}$ have to be a nowhere dense set? Doesn't Baire have something to say about that? – bof Jun 10 at 9:40
• – Minus One-Twelfth Jun 10 at 9:41
• Blumberg's Theorem shows that there is no such function. See encyclopediaofmath.org/index.php/Blumberg_theorem – Kabo Murphy Jun 10 at 10:08
• Note there are pathological functions which satisfy this if $\lim$ is replaced with $\limsup$. – MathematicsStudent1122 Jun 10 at 14:21

## 1 Answer

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.