# Must the set of explosion points be countable?

Let $$f:(0,1) \to {\mathbb R}$$ be a map, which is not assumed to be continuous or even measurable. I say that $$x_0\in (0,1)$$ is an explosion point if $$\lim_{x\to x_0,x\neq x_0}|f(x)|=\infty$$.

My question : Must the set of explosion points be at most countable ?

My thoughts : So far, I cannot even show that not every point is an explosion point. If $$[0,1]$$ is replaced by $$[0,1]\cap {\mathbb Q}$$, it is easy to construct a pathological example where every point is an explosion point : let $$(q_k)_{k\geq 1}$$ be an enumeration of $$[0,1]\cap\mathbb Q$$, and define

$$f(q_n)=\max_{k\leq n}\frac{1}{|q_n-q_k|}$$

Then for every $$k$$, $$f$$ satisfies $$|f(x)|\geq \frac{1}{|x-q_k|}$$ for all but finitely many $$x$$, so $$q_k$$ is an explosion point.

Let $$A$$ be the set of explosion points. I claim that $$A$$ is countable. It will suffice to show that, for each $$n\in\mathbb N$$, $$\{x\in A:|f(x)|\le n\}$$ is countable.

Consider a fixed $$n\in\mathbb N$$. Each point of $$A$$ is covered by an open interval $$I$$ such that $$|f(x)|\le n$$ holds for at most one point in $$I$$. The set $$A$$ is covered by countably many of those intervals.

• I don't quite follow the reasoning in the very last sentence. Put $A_n = \{x\in A:|f(x)|\le n\}$. For each $a\in A_n$, we have an interval $I_a$ as you described, with $I_a \cap A_n = \lbrace a \rbrace$ and $A_n \subseteq \bigcup_{a\in A_n} I_a$. How do we extract a countable subcover ? Mar 22, 2020 at 10:45
• For a set $A\subseteq\mathbb R$, every open cover of $A$ has a countable subcover. For instance, we could have chosen intervals with rational endpoints, and then it's obvious, because there are only countably many intervals all told.
– bof
Mar 22, 2020 at 10:49
• en.wikipedia.org/wiki/…
– bof
Mar 22, 2020 at 10:53
• Got it. Thanks for the explanation Mar 22, 2020 at 10:54

I have yet no idea wether the set of explosion points can be uncountable. But the following is true:

Let $$A \subset \mathbb{R}$$ be non-empty and closed. Let $$f: A \rightarrow \mathbb{R}$$ be any map, then the set of explosion points is not equal to $$A$$.

Proof.
Assume the statement is false, i.e., that the set of explosion points is equal to $$A$$. Define for each $$n \in \mathbb{N}$$, $$A_n = \{ x \in A \mid |f(x)| > n \}.$$ We claim that each $$A_n$$ is open and dense in $$A$$.
A_n is Open: Take $$a \in A_n$$, because $$\lim_{x \rightarrow a} |f(x)| = \infty$$, there exists $$\delta >0$$ such that for all $$x \in A$$ with $$0<|x-a|<\delta \Rightarrow |f(x)| >n$$. So that $$B(a,\delta) \subset A$$.
A_n is dense: Let $$a \in A$$ and $$\delta >0$$, because $$\lim_{x \rightarrow a} |f(x)| = \infty$$, there exists $$\epsilon >0$$ such that for all $$x \in A$$ with $$0<|x-a|<\epsilon \Rightarrow |f(x)| >n$$. In particular $$a+\frac12 \min \{\epsilon, \delta\} \in A_n$$.

By Baire's category theorem $$\cap_{n=0}^\infty A_n$$ is still dense in $$A$$. However $$\cap A_n$$ is empty and hence cannot be dense in $$A$$.

• Nice ! This also implies that there is a non-explosion point in every nonempty closed subset of $A$ ; in particular, the set of non-explosion points is dense in $A$. Mar 22, 2020 at 10:11
• You are right! I had the impression that something more general could be said. Mar 22, 2020 at 10:15