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I am trying to understand why the set of discontinuities of an increasing function $f: \mathbb R \to \mathbb R$ must be finite or countable. I showed that such function can only have jump discontinuities. It is not clear to me why the discontinuities can't be uncountable and it is also not clear to me for given $f$ how to find a bijection between discontinuities of $f$ and a subset of $\mathbb Q$. Please can somebody explain me why it should be true?

mathematics2x2life
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blue
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1 Answers1

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At each jump, draw an open interval on the $y$-axis which fits in between the bottom of the jump and the top of the jump.

After you are done, you have a string of disjoint intervals on the $y$-axis (because the function is increasing.)

Now choose a rational number in each of these intervals. This gives you a bijection between the intervals and a subset of $\Bbb Q$.

The main idea is that there is a limit to how big a disjoint collection of open sets in $\Bbb R$ can be. The generalized idea is that of the Souslin number of a topological space. In a nutshell, it's "the largest cardinality of a set of disjoint open sets."

You can see that the same argument works with $\Bbb R$ replaced with any separable linearly ordered topological space.

rschwieb
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  • I may be wrong, but I can't see why "increasing" prohibits an accumulation point. For example, consider jumps at $\left{\frac{1}{n}\right}$ and I don't believe your bijection works at $0$. – cderwin Jan 24 '14 at 14:06
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    @dirtyderwin I am reading discontinuity here as "jump discontinuity" and I take it to mean a point $a$ where $\lim_{x\to a^-}f(x)<\lim_{x\to a^+}f(x)$. Between these two distinct numbers on the $y$-axis, you can fit an interval. The only way "increasing" is involved here is that it lets me conclude that the right limit is greater than the left, and that the intervals are disjoint. – rschwieb Jan 24 '14 at 14:11
  • @dirtyderwin But yeah, if you read the question as "any discontinuity" including removable discontinuities, then the theorem just isn't true :) You can pluck out uncountably many points from a continuous curve and have a counterexample. – rschwieb Jan 24 '14 at 14:15
  • yea that's what I thought.. functions can be really nasty – cderwin Jan 24 '14 at 14:16
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    @rschwieb The function has to be defined in $\mathbb{R}$ so how do you pluck out points from its curve? – JiK Jan 24 '14 at 14:16
  • @JiK Good point, the question is already phrased to avoid removable discontinuities :) I'm accustomed to thinking about functions of real variables being from subsets of $\Bbb R$ to $\Bbb R$. – rschwieb Jan 24 '14 at 14:17
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    @rschwieb To avoid any doubts about the existence of the (one-sided) limits, one can note $\sup f|{(-\infty,a)}\le \inf f|{(a,\infty)}$ with equality iff $f$ is continuous at $a$. Then continue as before, i.e. the intervals $(\sup f|{(-\infty,a)}\le \inf f|{(a,\infty)})$ are pairwise disjoint and the nonempty ones contain a rational. – Hagen von Eitzen Jan 24 '14 at 14:18
  • @HagenvonEitzen That's also a good point: thank you for contributing it! – rschwieb Jan 24 '14 at 14:23
  • For it to work you need to first prove that there are at most countably many jump discontinuities, no?(if you don't know about Souslin number) – blue Jan 25 '14 at 08:20
  • @blue I'm not sure what you mean. This is a proof that there are at most countably many jumps. You don't need to talk about the Souslin number of a space, I just mentioned that it's the idea that generalizes the concept here. – rschwieb Jan 25 '14 at 13:39
  • @blue So you see that you can build the intervals, right? And it doesn't matter how many jumps there are, you can build these intervals. And finally, you can see they don't overlap, right? – rschwieb Jan 25 '14 at 13:58
  • But if there are $\mathbb R \setminus \mathbb Q$ many jumps then what happen? – blue Jan 27 '14 at 09:07
  • @blue If you look at what I'm doing, you will see why that is not possible. I'm going to assume for a moment that you believe this set of intervals exists. The next step is this: pick a rational in each interval. This creates a mapping from the intervals $\to\Bbb Q$. Can you see why this mapping is injective? (You need to use the fact the intervals don't overlap.) – rschwieb Jan 27 '14 at 12:44
  • @Masacroso You have evidently have misunderstood something. There is no contradiction at all with the discontinuities occurring on a dense subset as they do in that example. In the example you linked, the jumps occur exactly at the rationals: a countable set. The definition of "jump discontinuity" implies such intervals exist. – rschwieb Jun 28 '16 at 00:08
  • @Masacroso It could be, for example, that you're confusing the denseness of the points where discontinuity occur (on the $x$-axis) with their images which are not dense in the interval they occupy. – rschwieb Jun 28 '16 at 00:15