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Let $f : \mathbb{R}\to \mathbb{R}$ be a monotone function and consider the set $S$ of all points $a\in \mathbb{R}$ such that

$$\lim_{x\to a^-}f(x)\neq \lim_{x\to a^+}f(x).$$

I want to show that this set is countable, but I'm not finding any way to do it. I thought one way would be to show that this set is discrete, so that it is countable. For that to happen, I would need to show that there is no convergent sequence $(x_n)$ of points of $S$ all different than $a$ whose limit is $a$.

My idea then is to suppose there is such a sequence $(x_n)$. Then $x_n\in S$ and $\lim x_n = a$ with $x_n\neq a$. In that case we have $\lim_{x\to x_n^-}f(x)\neq \lim_{x\to x_n^+}f(x)$.

Intuition says this should contradict $f$ being monotone, but I'm not finding a way to prove it.

So how can I finish this proof? Is my strategy correct? If so how do I finish it? If not, how should I prove this result?

Gold
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1 Answers1

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HINT: Let $X=\{a: \lim_{x\rightarrow a^-}f(x)\not=\lim_{x\rightarrow a^+}f(x)\}$, and assume $f$ is nondecreasing.

For $a\in X$, let $I_a=(\lim_{x\rightarrow a^-}f(x), \lim_{x\rightarrow a^+}f(x))$ be the "gap" between the left and right limits.

What can you say about $I_a$ and $I_b$ for $a, b$ distinct elements of $X$?

Noah Schweber
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  • The idea is that $\forall x \in R , \forall n\in Z^+ [\infty>f(x)-f(x-n)\geq \sum_{a\in X\cap [x-n,x]} I_a \geq 0] $ and that $I_a>0$ for any $a\in X$. – DanielWainfleet Oct 20 '15 at 04:10
  • @user254665 Bwah? I can't figure out what that means. Also, $I_a$ is an interval, not a number, so I'm not sure what "$I_a>0$" means. Regardless, it's needlessly complicated - there's a much simpler approach . . . – Noah Schweber Oct 20 '15 at 06:18
  • Sorry.I should have written $|I_a|$ and also $x\in Z$ and fixed $n=1$. If $X $ is uncountable then $X\cap[x-1,x] $ls uncountable for some $x\in Z$. But then ,for some $q\in Q^+$ we have ${a\in X\cap [x-1,x] : |I_a|>q}$ uncountable. – DanielWainfleet Oct 20 '15 at 17:18
  • @user254665 I see. This is still needlessly complicated - since $I_a$ is a nonempty interval for $a\in X$, $I_a$ contains a rational $q_a$. Distinct $I_a$s are disjoint. So there's an injection from $X$ into $\mathbb{Q}$. – Noah Schweber Oct 20 '15 at 18:53
  • works anyway. i always find a more complicated proof. – DanielWainfleet Oct 21 '15 at 00:07