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Problem: Let $f(x)$ be a real valued functions defined on $\mathbb{R}$, prove that the point set $$E = \{x\in \mathbb{R}: \lim_{y\rightarrow x} f(y) = +\infty\}$$ is a finite or a countable set.

Proof: Let $g(x) = \arctan f(x), x \in \mathbb{R}$. Then the point set $E$ can be written as $$E = \{x\in \mathbb{R}: \lim_{y\rightarrow x} g(y) = \frac{\pi}{2}\}$$ Therefore $E$ is a finite or a countable set.

The above proof is from a textbook of real analysis. How can we see that the set $E$ in this proof is finite or countable?

ScienceAge
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    That's no proof at all! Labelling the two expressions in the obvious way, showing that $E_2$ is countable is no easier than showing that $E_1$ is countable. (Unless they've already proved the corresponding result for $f:\Bbb R\to(a,b)$...) – David C. Ullrich Jan 03 '19 at 14:14

1 Answers1

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Right now, I have no idea why we can see immediately that $ E=\{ x\in \mathbb{R} \;|\;\lim_{y\to x} g(y) = \frac{\pi}{2}\} $ is at most countable. But I've found the following argument that shows the set $E$ is at most countable.

Proof: Assume $x \in E$. If $f(x)< n$, then since $\lim_{y\to x}f(y) =\infty$, we can find a $\delta>0$ such that $$ (x-\delta,x)\cup (x,x+\delta) \subset f^{-1}((n,\infty)). $$ Now consider the open set $$U_n=\text{int}f^{-1}((n,\infty)) = \bigcup_{k=1}^\infty (\alpha_k, \beta_k)$$ where $\{(\alpha_k, \beta_k)\}$ is a disjoint family of open intervals. The above argument shows that $x$ is one of the end points $\alpha_k$ or $\beta_k$ of $U_n$ for some $n\ge 1$. Since there are at most countably many such end points, it follows that $E$ is countable. $\blacksquare$

After watching the above argument, I felt it is not trivial and it made me think that it is probably author's confusion, or there may be some results that the author did not mention.

Myunghyun Song
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  • Thanks, but why can we write $U_n=\text{int}f^{-1}((n,\infty))$ as a disjoint union of open intervals? This is what I am not clear about. – ScienceAge Jan 05 '19 at 07:49
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    Oh, I was not so clear about that. Actually, it is just writing $U_n$ as a disjoint union of its components, and most topology textbooks perhaps would contain this proposition. If you are interested in the proof of it, you can see e.g. this earlier post https://math.stackexchange.com/questions/318299/any-open-subset-of-bbb-r-is-a-at-most-countable-union-of-disjoint-open-interv. – Myunghyun Song Jan 05 '19 at 07:57
  • Thanks. I wonder whether you think that if $(x-\delta,x)\cup (x,x+\delta) \subset f^{-1}((n,\infty))$, then $(x-\delta,x)\cup (x,x+\delta) \subset \text{int} f^{-1}((n,\infty))$? – ScienceAge Jan 06 '19 at 03:24
  • Yes, since the left-hand side is open in $\mathbb{R}$, its interior is itself. Now we can take $\text{int}$ operator both sides to see that it is true. – Myunghyun Song Jan 06 '19 at 03:48