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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.

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2 Answers 2

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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.

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  • $\begingroup$ 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 ? $\endgroup$ Mar 22, 2020 at 10:45
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    $\begingroup$ 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. $\endgroup$
    – bof
    Mar 22, 2020 at 10:49
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – bof
    Mar 22, 2020 at 10:53
  • $\begingroup$ Got it. Thanks for the explanation $\endgroup$ Mar 22, 2020 at 10:54
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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 $A$.

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    $\begingroup$ 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 $A$. $\endgroup$ Mar 22, 2020 at 10:11
  • $\begingroup$ You are right! I had the impression that something more general could be said. $\endgroup$
    – abcdef
    Mar 22, 2020 at 10:15

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