How to show that the set is at most countable?

Question

$f(x)$ is a real-valued function on $\mathbb{R}$. How to show that the set $E=\{x\in\mathbb{R}: \lim_{y\to x}f(y)=+\infty\}$ is at most countable?

Answer 1

$a\in E$ means that for every $M\in\Bbb N$, we can pick $u,v\in\Bbb Q$ with $u<a<v$ and $f(x)>M$ for all $x\in(u,v)\setminus\{a\}$. Write $(u,v)=\phi(a,M)$. We may assume wlog. that $\phi(a,M+1)\Subset \phi(a,M)$.

Start with $E_0:=E$ and assume $E_n$ is uncountable. As $\phi(a,n)$ can take only countably many values, there exists $(u_n,v_n)$ such that $\phi(a,n)=(u_n,v_n)$ for uncountably many $a\in E_n$. Let $E_{n+1}=\{\,a\in E_n\mid \phi(a,n)=(u_n,v_n)\,\}$ and recurese.

This gives us a nested sequence of uncountable sets $E_n$, accompanied by a sequence of nested intervals $(u_1,v_1)\Supset (u_2,v_2)\Supset\ldots$. Pick $\alpha\in\bigcap(a_n,v_n)$ and let $M=\lceil f(\alpha)\rceil$. Then for $a\in E_M$, we have $f(x)>M$ for all $x\in (u_M,v_M)\setminus\{a\}$. As $\alpha\in(u_M,v_M)$, this means $\alpha=a$, contradicting $|E_M|>1$.