Clarification on why there can only be a countable number of jump discontinuities

Question

I have been trying to find a proof I can follow (I am not familiar with metric spaces, for instance) for the fact that any given function only has a countable number of jump discontinuities.

I have been looking at the first answer from this post, however I can't understand why the set $$A_{q,n}=\left\{x\in (a,b):x-\frac1n<y<x<z<x+\frac1n\implies f(z)>q>f(y)\right\}$$ must be countable.

One of the post's comments suggested that, if a set is uncountable, then it must contain an interval, and sketched a proof starting from there. However, I don't think that this statement is true; a proper counterexample would be the Cantor set.

I hope that I am not breaking any rules by referencing another post on the site. Looking in the post's comments, I saw that there were some other people who had the same question as me, but didn't receive any answers, so I hope that someone can shed some light on the issue. Thanks in advance!

Answer 1

There is now a very nice, short answer explaining the proof.

This is a follow-up, a suggestion for further study, a push for a different perspective, a plea for greater understanding. Also an excuse to promote a very nice survey article by a friend (see [1]).

Instead of thinking of this as a theorem for understanding of jump discontinuities, there is a broader point of view. Indeed if $f:(a,b)\to \mathbb R$ is an arbitrary function the collection of all points $x\in (a,b)$ at which both $f(x-)=\lim_{t\to x-}f(t)$ and $f(x+)=\lim_{t\to x+}f(t)$ exist and yet are different is a countable set.

Students usually see this in the context of monotone functions. But you can see from this proof that it is true for a completely arbitrary function.

Now think about it this way: this is not a statement about jump discontinuities it is a statement about asymmetry! Something is happening on the right that is different from what is happening on the left.

One of the oldest asymmetry theorems is due to Beppo Levi. He proved that a function can have only a countable number of points where the right-hand and the left-hand derivatives both exist but are different.

Probably inspired by the Beppo Levi theorem, the Young family (William Henry Young and Grace Chisolm Young) in the early 20th century in the UK followed a program investigating asymmetry problems.

Here is one of theirs. [William published it but Grace contributed, of course, as she did all of their efforts. He needed the credit; she raised the children.]

Definition. Let $f:(a,b)\to \mathbb R$ be an arbitrary function. For each $x\in (a,b)$ let $L^+(f,x)$ be the set of all right limit numbers of $f$ at $x$, i.e., all numbers $r$ (including $\pm\infty)$ such that $f(t_n)\to r$ as $t_n \searrow x$ for some decreasing sequence $\{t_n\}$. Similarly $L^-(f,x)$ be the set of all left limit numbers of $f$ at $x$.

You can see this a generalization of the situation of a jump discontinuity. At such a point the sets $L^+(f,x)$ and $L^-(f,x)$ just contain one value, different on the right from the left.

Here is the theorem that Young called his Rome theorem, because he presented it a meeting in Rome. He was justly proud of this since it is probably the first genuine theorem that applies to a completely arbitrary function--no assumptions about continuity or monotonicity, etc.

Theorem. [Youngs' Rome Theorem (1906)] Let $f:(a,b)\to \mathbb R$ be an arbitrary function. Then the set $$ \{x\in (a,b): L^+(f,x) \not= L^-(f,x)\}$$ is countable.

As a special and very weak corollary it follows that an arbitrary function can have no more than countably many jump discontinuities.

So think asymmetry. No jumping!

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REFERENCES:

[1] Scientific Issues, Jan Długosz University in Częstochowa Mathematics XXIII (2018) 11–24DOI http://dx.doi.org/10.16926/m.2018.23.01 ASYMMETRY IN REAL FUNCTIONS THEORY, JACEK MAREK JĘDRZEJEWSKI

Answer 2

If $f$ is less than $q$ on the interval $(x-\frac{1}{n},x)$, and $f$ is greater than $q$ on $(x,x+\frac{1}{n})$, then clearly there cannot be an element of $A_{q,n}$ on either interval.

Therefore, for any $x_1,x_2\in A_{q,n}$, we see that $x_1 \neq x_2 \implies |x_1-x_2|\geq\frac{1}{n}$.

So, for $k\in\mathbb{Z}$, every interval of the form $(\frac{k-1}{2n},\frac{k+1}{2n})$ has at most one element in $A_{q,n}$.

It easily follows that $A_{q,n}$ is countable.

Answer 3

We will explicitly construct a mapping between a countable set and the jump discontinuities of a function. To being we know that the left and right hand limit exist there left-continuous and right-continuous neighborhoods near each jump discontinuity. Because the rational numbers are dense in the reals and we can make these neighborhoods arbitrarily small we can cover the discontinuity with an interval of the form $(q-r,q+r)$. There is an obvious bijection between intervals of that form and the element $(q,r) \in \mathbb{Q}^2$. Since the latter set is countable, and every jump discontinuity can be covered by such an interval there can only be countably many such discontinuities.