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Given the following:

If $f : (a,b)\rightarrow \mathbb{R}$ is monotone, then $f$ has at most countably many points of discontinuity in (a,b), all of which are jump discontinuities.

We must deduce that a monotone $f: \mathbb{R} \rightarrow \mathbb{R}$ has points of continuity in every open interval.

My work:

W.L.O.G. assume $f$ is increasing. Because $f$ is monotone (in this case increasing), I know that the only discontinuities that can exist are jump discontinuities (meaning that the left hand limit is strictly less than the right hand limit). We also know that these limits exist. The intervals of discontinuity are also disjoint. Given an interval $(a,b)$, a point $c \in (a,b)$ can be discontinuous if it meets the criteria mentioned above; that is $f(c-) < f(c+)$. There are points in an interval $(x,c)$ and $(c,y)$ which are continuous (for the limits to exist); where $x,y \in (a,b)$

I'm sure I'm missing a lot of fine details, and may even have the wrong idea on how to approach it. Any critique and clarification would be greatly appreciated.

zodross
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  • The set of rationals? – zodross Aug 25 '18 at 02:23
  • So like.. pick a point rational point in for every discontinuity? Then since Q is countable then so are the discontinuities. So does this imply that there must be points of continuity in every open interval, because there are countably many discontinuities? I thought countable meant it could be enumerated to N, and not finite. – zodross Aug 25 '18 at 02:30
  • @zodross See also How to show that a set of discontinuous points of an increasing function is at most countable. – dxiv Aug 25 '18 at 02:34
  • Sorry, my bad. I have misread your question. What I actually commented is the proof of the fact given at first in your post, not your question. Sorry for wasting your time. – xbh Aug 25 '18 at 02:47
  • Do not worry about it, it actually helped me to understand that since the discontinuities are countable and any sub interval of the Real number line is uncountable, then there must be points in an open interval (a,b) that are not discontinuous. You're guidance was not a waste of time, but a reinstatement of meaning of the properties of countable sets. thank you – zodross Aug 25 '18 at 03:04

1 Answers1

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HINT: You know that $f$ has at most countably many discontinuities in $(a,b)$. Is $(a,b)$ countable?

ervx
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  • (a,b) is not countable, correct? That would imply that the Real line is countable? – zodross Aug 25 '18 at 02:24
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    $(a,b)$ is uncountable, so there must be a point of continuity in $(a,b)$ since the set of discountinuities in $(a,b)$ is countable. – ervx Aug 25 '18 at 02:25
  • Oh, I see. So that must mean that there is an uncountable amount of continuities? I am not sure what I can do with this information if it is correct. – zodross Aug 25 '18 at 02:25
  • You are assuming that you have only countably many discontinuities. So you have a contradiction. – ervx Aug 25 '18 at 02:27
  • I see what you are saying. That was what I was trying to get at with the whole inclusion of points x and y. Those are points of continuity. – zodross Aug 25 '18 at 02:27
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    The fact that $(a,b)$ is uncountable, while the set of discontinuities of $f$ is countable is all you need for the proof. Obviously a countable set and an uncountable set cannot be equal to one another. – ervx Aug 25 '18 at 02:30
  • Oh my god, so simple. Thank you sir. That clears it all up. – zodross Aug 25 '18 at 02:33
  • You are welcome. P.S. There are women on this site too. – ervx Aug 25 '18 at 03:16
  • My sincerest apologies, madam. – zodross Aug 25 '18 at 06:13