The Popcorn Function

Here is a very crazy function in math called the Popcorn function (or Thomae's Function) which has the property that it is discontinuous at every rational number and continuous at every irrational number. So it kind of has jumps and no jumps at the same time. Enjoy this spooky mathematical adventure!

[samratyadav]

04:33 today I came to know the meaning of wtf for mathematician is 'what to find' 😅😅

[stlemur]

I love this function because it's perverse. It's a great one to pull out for when students need a counterexample.

[Kdd]

To be honest I was having popcorn and watching yt when your notification came. . . Such a coincidence

[thedoublehelix]

Amazing idea to use the fact that a rational sequence converging to an irrational number must eventually have as complicated of a denominator as you want!!!!

[georgesanxionnat]

The reference to Stromae though <3 U Peyam

[noahtaul]

All these discontinuities are “removable”, meaning that limits exist at all points, the function just isn’t equal to the limit at the rationals. Fun theorem: it’s impossible to have a limit at every point but an uncountable number of removable discontinuities. So this is the worst we can do and still have limits everywhere.

[aishayo]

This is so beautiful! Thank you very much, Dr Peyam, for such a detailed explanation. I finally got it!

[honghongnt]

The function is cool and all but it's quite amazing to know that you can limit the size of the denominator of the rationals between any 2 irrationals to any number that you want!

[Qermaq]

I love how "want to find" is written WTF. :D

[tomkerruish]

John Conway called this function "Stars Over Babylon".

[ВасилийТёркин-кх]

So basically it discontinuous in a countable set of points

[tim-cca]

That’s easy to understand！
But why we can restrict the delta, to exclude an infinite set of form p/n.

[kanewilliams]

How on earth do you make a video every single day?? Incredible!

[iabervon]

I thought it would be easiest to show that the limit is 0 everywhere. If x_0 is rational, choose delta to be 1/(N!) such that 1/N<epsilon. The only rational number in that interval with denominator less than N is x_0 itself, so the function is closer to 0 than epsilon for all x such that 0<|x-x_0|<delta. If x_0 is irrational, approximate it with a p/q such that q>N and (p/q-1/(N!), p/q+1/(N!)) contains x_0, then choose delta such that that interval also contains (x_0-delta, x_0+delta). Then f(p/q) = 1/q, and everywhere else in the interval, f(x) < 1/q. Then the function is continuous where it is equal to 0.

[worldnotworld]

This freaks me out. How can this be? I too happened to be eating popcorn while watching this and I almost never eat popcorn.

[kenharari]

Thank you Dr Peyam, I've read a lot of explanations to this problem and yours is definitely the best one.

[colin]

delta epsilon proof for discontinuity on rationals. Let x0 = p/q in Q, x0 in [0, 1]. Then f(x0) = 1/q. Let epsilon < 1/q. Let delta > 0. By the density of irrational numbers in R, B_delta(x0) contains an irrational number y0. We have that |x0-y0| < delta, but |f(x0) - f(y0)| = |1/q - 0| = 1/q > epsilon, so f is discontinuous for rational numbers

[FT]

wow, this is quite beautiful. I think of it as a bunch of frogs that hop 1/2, 1/3, 1/4, ..., 1/N starting from 0, and they all hop over this very tiny range (x_0 - delta, x_0 + delta).

[JPiMaths]

Stromae + an insanely cool function; what more could you ask for? Amazing stuff!!

[cparks]

For the selection of $\delta$, you only need to show that the set $\mathbb{Q}_N = \{p/q \mid 0 \le q \leq N \}$ is closed where $p$ and $q$ are integers. (Since it's a finite union of the closed sets $\mathbb{N}/n$, this is easy.) You then take $delta = d(x, \mathbb{Q}_N)$ where $d$ is the distance function. Since $\mathbb{Q}_N$ is closed and does not contain $x$, we know $\delta$ is not zero.