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I came across this argument but I'm having some trouble with it. $f$ is from $\mathbb{R} \to \mathbb{R}$, $f(a\pm 0)$ is the right and left handed limits of $f$ at $a$ respectively.

Assume that $f$ is monotone increasing on an interval $I$. If $f$ is not continuous at a $c \in I$, then $f(c-0)\lt f(c+0)$. Let $r(c)$ be a rational number for which $f(c-0)\lt r(c) \lt f(c+0)$. If $c_1 \lt c_2$, then by the monotonicity of $f$, $f(c_1 + 0) \le f(c_2 - 0)$. Thus if $f$ has both $c_1$ and $c_2$ as points of discontinuity, then $r(c_1) \lt r(c_2)$. This means we have crated a one-to-one correspondence between the points of discontinuity and a subset of the rational numbers. Since the set of rational numbers is countable, $f$ can only have a countable number of discontinuities

However, I cant see what's wrong with saying, that since the irrationals are everywhere dense, you choose an irrational number $s$ such that $f(c-0)\lt s(c) \lt f(c+0)$, and conclude that you have a one-to-one correspondence of the discontinuities and a subset of the irrational numbers, concluding that there can be an uncountable number of discontinuities, which makes me think I don't fully understand the argument.

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    $\begingroup$ Finding an injective map from a set $A$ into $2^\Bbb{Q}$ doesn't prove that $A$ is uncountable; it just proves that $|A| \le |2^\Bbb{Q}|$, which is the cardinality of the continuum. But, $|A|$ could be finite, or countably infinite, or possibly something in between (depending on your opinion of the continuum hypothesis). e.g. the inclusion map of $\{1\}$ into $\Bbb{Q}$ is injective, but $\{1\}$ is not uncountable. $\endgroup$ Jul 25, 2022 at 0:14
  • $\begingroup$ @TheoBendit Am I wrong in thinking a set with the cardinality of the continuum is uncountable? Also I'm not trying to say that the subset of irrationals corresponding to the discontinuities is uncountable, but just mirroring the first argument, saying that it can be. Is this valid do you think? $\endgroup$
    – emillike
    Jul 25, 2022 at 0:18
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    $\begingroup$ No, you are not wrong in thinking the cardinality of the continuum is uncountable. But, the assertion that $|A| \le |\Bbb{R}| \implies A \text{ is uncountable}$ is not true. Saying the set may be uncountable requires something stronger: you need a concrete example to show it's possible. Don't forget: if $A$ is countable, then $|A| \le |\Bbb{R}|$ will still be true! $\endgroup$ Jul 25, 2022 at 0:21
  • $\begingroup$ @TheoBendit Right I agree. So I guess you would say that the first argument works because every subset of $\mathbb{Q}$ is countable, whereas if you make a bijection from the set of discontinuities to $\mathbb{R}\smallsetminus\mathbb{Q}$, since a subset may be countable or uncountable, you can't conclude anything? $\endgroup$
    – emillike
    Jul 25, 2022 at 0:25
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    $\begingroup$ Yes, but replace "bijection" with "injection". If you can find a bijection with $\Bbb{Q} \setminus \Bbb{Q}$ (or any other uncountable set), then you know the cardinalities are the same, and so the set is uncountable. Both arguments construct an injection, so you know that the set of discontinuities is at most $|\Bbb{Q}|$ or $|2^\Bbb{Q}|$, depending on the argument used (the former argument is stronger). $\endgroup$ Jul 25, 2022 at 0:33

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This is a standard case of arriving, at some time during solving a problem, at a situation where you can make many different choices to proceed. You could chose a rational number from that interval, an irrational number, a transcendental number or many other things.

However, for the goal you want achieve by solving the problem, some choices are better than others.

To give an example, you leave a meeting at 1PM and need to be at the train station at 2PM, to catch your train. Depending on how far away the train station is, how much luggage you have, making a walk on foot may be perfectly reasonable, so you don't waste your money on a taxi (cab). Or taking public transport is feasable. Or maybe your only option is to really take a taxi and promise the driver a bonus if they reach it before 1:55.

All of those options may be available, but for your specific purpose, many could work, or maybe only one or even none.

It's the same with your proof. The goal of the proof is to arrive at an "as low as we can get" upper bound on the number of discontinuities of such functions. Since each discontiuity is a real number, knowing there are "at most as many discuntinuities as there are real numbers" is the starting point.

If you know a bit about irrational and transcendental numbers, you know they could also have been picked from in the proof. However, the result those choices would give is just not any better than what we knew beforehand. Of course to verify that you need to know that the cardinality of the irrational and transcendental numbers is the same as the cardinality of the real numbers.

That explains why choosing the rational numbers instead of the other mentioned is better for this specific purpose, because their cardinality is actually less than those of the real numbers.

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A subset of the set of all irrational numbers may be countable, so a set which is in bijection with such a set can be countable.

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  • $\begingroup$ Right, I'm not trying to say that it is uncountable, just by the same logic of the first argument that it could be. Which I don't think is true. So I think I'm missing something $\endgroup$
    – emillike
    Jul 25, 2022 at 0:20
  • $\begingroup$ @emillike The set of natural numbers is in bijection with a subset of $\mathbb R$. Does that make it uncountable? $\endgroup$ Jul 25, 2022 at 4:59
  • $\begingroup$ No. However nowhere in my post did I argue that the set of discontinuities was actually uncountable. $\endgroup$
    – emillike
    Jul 25, 2022 at 18:40

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