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I would like to prove that the number of simple jump discontinuities of any function is countable.

Can someone point me some material where the proof is or explain the proof here?

Thanks.

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The point is that any jump discontinuity has a neighborhood with no other jump discontinuity, and that the real line is Lindelof. @xavierm02: there are only countably many integers. –  ronno Dec 22 '12 at 8:55

2 Answers 2

up vote 3 down vote accepted

Let $f:(a,b)\to \mathbb{R}$ and $$A=\left\{x\in (a,b):f\text{ has a jump discontinuity at $x$}\right\}$$ Now $$A=A^{+}\cup A^{-}$$ where $$A^{+}=\left\{x\in (a,b):\lim_{y\to x^+}f(y)>\lim_{y\to x^-}f(y)\right\}$$ and $$A^{-}=\left\{x\in (a,b):\lim_{y\to x^+}f(y)<\lim_{y\to x^-}f(y)\right\}$$ I will show $A^{+}$ is countable and leave the rest to you. Fix $x\in A^{+}$ and then $\exists q\in \mathbb{Q}$ so that $$\lim_{y\to x^+}f(y)>q>\lim_{y\to x^-}f(y)$$ (why???). This means that $\exists \delta>0$ so that $$x-\delta<y<x<z<x+\delta\implies f(z)>q>f(y)$$ and so (why?) $\exists n\in \mathbb{N}$ so that $$x-\frac1n<y<x<z<x+\frac1n\implies f(z)>q>f(y)$$ If we let $$A_{q,n}=\left\{x\in (a,b):x-\frac1n<y<x<z<x+\frac1n\implies f(z)>q>f(y)\right\}$$ ($q\in \mathbb{Q}$,$n\in \mathbb{N}$) then by our previous discussion $$A^{+}\subseteq\bigcup_{q\in \mathbb{Q}}\bigcup_{n\in \mathbb{N}}A_{q,n}$$ Therefore the problem moves to proving that $A_{q,n}$ is countable. This follows from the fact $A_{q,n}$ is isolated (show this!).

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Wow! This is an amazing solution. I really like it. Very elegant! –  elaRosca Dec 22 '12 at 9:20
    
The proof for the other case follows in similar manner, just with the sign changed in the RHS of the implication, where we define Aq,n. q exists by density of Q in R. –  elaRosca Dec 22 '12 at 9:29
    
@elaRosca Indeed. How would you answer the first "why"? –  Nameless Dec 22 '12 at 9:30
    
Q is dense in R. –  elaRosca Dec 22 '12 at 9:31
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I wrote the code in latex but it does not seem to recognize it here. –  elaRosca Dec 22 '12 at 9:56

any jump discontinuity has a neighborhood with no other jump discontinuity, Associate to each such neighbourhood a rational number inside that.so there is a bijection between a subset of rationals and jump discontinuity.

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this shows no. of jump discontinuities can be at most countable –  Koushik Dec 22 '12 at 9:08
    
This is very clear and short, gives a very intuitive answer. Thanks –  elaRosca Dec 22 '12 at 9:13
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Notice that you can choose a rational in an interval with a canonical way; thus, the result doesn't depend on the axiom of choice. –  Seirios Dec 22 '12 at 9:16
    
Can you explain this a bit, please? I am not sure I understand what you mean –  elaRosca Dec 22 '12 at 9:22
    
In general, to choose an element in a set you need the axiom of choice, but to pick out a rational from $(a,b)$ you can define $q_0 = \min \{ n >0 : \frac{1}{n} < b-a \}$ and $p_0= \min \{ n \geq 0 : \frac{p_0}{q_0} \in (a,b) \}$ if $b>0$ or $p_0= \max \{ n \leq 0 : \frac{p_0}{q_0} \in (a,b) \}$ otherwise, and $p_0/q_0$ works. Here, you only need the archimedean property. –  Seirios Dec 22 '12 at 9:48

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