Dirichlet Function: The King of Calculus Counter-Examples

Reference:
demonstrations.wolfram.com/Th...
mathworld.wolfram.com/Dirichl...
math.fel.cvut.cz/mt/txtb/4/tx...
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0:00 Five Anti-intuition Calculus Questions
0:20 Derive the Dirichlet Function
1:35 Dirichlet Function Formal Introduction
2:07 Can A Function Be Defined Everywhere but Discontinuous Everywhere
3:37 Can The Sum of Two Discontinuous Function Be Continuous?
4:09 Can The Product of Two Discontinuous Function Be Continuous?
4:25 Can A Function Be Continuous Only at Given Points?
5:54 Can A Periodic Function Have No Minimal Period?
7:10 The Outro
7:29 Bonus Clip

[@antonemberbroque4452]

Now I am become counter example, destroyer of proofs.

[@Winium]

Hidden destroyer of the freshmen who got pranked by their seniors.

[@andreasxfjd4141]

It is not a proof, if it is rebuttable

[@koifish528]

throughout discrete math and real analysis, i alone am the counter example

[@danielevilone]

Moreover, Dirichlet's function is integrable by Lebesgue but not by Riemann.

[@ron-math]

Yep. I wanted to include this point in the video as well.

[@fakezpred]

Funny how a huge troublemaker to the riemann integral follows immediately from definition of the Lebesgue integral actually

[@lperezherrera1608]

​@@fakezpredI mean the proof for it not being Riemann-integrable is pretty trivial too so I wouldn't call it a huge trouble maker

[@fakezpred]

@@lperezherrera1608 by troublemaker I mean being nonintegrable with the riemann definition, not the proof.

[@rwiturajgoswami5001]

Because the set of rationals has a measure equal to zero. So Dirichlet's function is the zero function almost everywhere.

[@douglasstrother6584]

My Calculus Professor (Tony Tromba, UC Santa Cruz 1981) dropped the Dirichlet Function on us at the end of a Friday lecture to give something to discuss at Happy Hour.

[@albertemc2stein290]

Small remark to 4:38. f(x) = 1/((x-1)(x-e)) is actually continuous on its whole domain. Since it is not defined in either 1 or e it cannot be discontinuous at those points. Only if you defined it to be some real number in those points, it would be discontinuous. This is because in the definition of continuity, only points from the domain are considered for the condition |x - x0| < delta => |f(x) - f(x0)| < epsilon

[@ron-math]

Yes. Great catch!

[@set.theory]

I had a smile on my face this whole video :) I hope more people come across this channel! Can't wait for the next upload

[@Senshidayo]

Yes this is a really cool video, and I’m also a fan of Micaiah!

[@rosettaroberts8053]

Another example of a periodic function with no smallest period is f(x) = 0

[@comedyfriendsenglish]

Also another example for two discontinuous functions with a continous sum is any discontinous function f and -f

[@naufalfadhlurrahman5092]

​@@comedyfriendsenglish what if the function is discontinuous because it's not defined at some points? Then the sum will still be discontinuous.

[@comedyfriendsenglish]

No. Because the zero function is continuous no matter on what subset of say R you define it. In fact any map between topological spaces that maps all values to a single point is automatically continuous. You can not say a functions continuity is violated on a point on which the function isn't even defined

[@moritzalshuth7239]

This function is a smart bomb on "Let epsilon > 0"

[@stevenfallinge7149]

Next, the Cantor set indicator function. Discontinuous at an uncountable set of points but still riemann integrable.

[@MikeT10101]

Excellent video! Thank you.

[@estebanvasquez-giraldo5770]

You did pass a lot of joy, thanks!

[@dnhatanh]

Amazing! Both the original construction and your explaination.

[@ron-math]

Cheers!

[@rssl5500]

Amazing ! Well done

[@ron-math]

Thank you! Cheers!

[@Rory626]

This funxtion absolutely carried me through undergrad

[@chixenlegjo]

I would like to mention the “periodic but no smallest period” has the counterexample f(x)=0. Here’s another related counterintuitive fact: All linear functions are the sum of two periodic functions.

[@farfa2937]

I don't think you consider 0 as a period. Otherwise all functions are periodic.

[@tomtomspa]

@@farfa2937no, but 0 has no smallest period

[@andy02q]

f(0) is not period because there's no p for which f(0)≠f(p+0).

[@omp199]

1. That's an example, not a counterexample. 2. How?

[@Troloze]

f(x) = 0 then f(x) = f(x + k) = 0 where k is the interval, and therefore f(x) = 0 is periodic

[@Just_Ava]

My calculus lecturer gave us the task to proof whether or not the dirchelet function is a regulated function. So its kinda funny to see this video a week later :)

[@klausolekristiansen2960]

Conway's base 13 function is discontinous everywhere.

[@ron-math]

On my to-do list :)

[@_P_a_o_l_o_]

Do the Devil's Staircase function now! Also, I recommend the book 'Counterexamples in Analysis'

[@SamarthPatil-my5mh]

How about Weirstrass Function which is continuous but non differentiable everywhere

[@ron-math]

on the to-do list.

[@alonsoviton8278]

Great video, but in 5:49 you say that a similar method could be used to create a function that is continuos only on the integers. I remember reading that due to Baire's cathegory theorem cannot be continuos on infinite sets with zero measure like the rationals, is that the case? If so, why does it fail on the rationals but not in the integers even thought the latter also have 0 measure?

[@ron-math]

Good point! Rationals are dense while integers are not.

[@jensraab2902]

I have heard the name Dirichlet before but I've never had a close look at the Dirichlet function. It's truly a crazy function. But what I found most fascinating is that it's not just one of those artificial functions where you define the different parts but that you get this weird beast from what looks like this ordinary double limit! Cool video! PS: Did you make a mistake at 5:36? You say that the same argument applies for the "e case". But is this correct since e is not a rational number? Wouldn't the limit be 0 when x goes to e because when x=e the lower definition (for x not element of Q) applies?

[@ron-math]

Hi! I was referring to the artificial function defined as you see on the screen 5:36, not the original Dirichlet function. You are right that when x=e then f(e) is indeed 0. And by definition of the limit, the limit of f(x) when x- > e is also 0.

[@3141minecraft]

2:59 by the way, you forgot to mention 1 thing: e is not only irrational, e is trancendental. That means e², e³, e⁴ etc. cannot be rational numbers. (By the way, this is not true for all irrational numbers. For example: sqrt(2) is irrstional, bit sqrt(2)²=2, which is rational.)

[@danielc.martin1574]

Fun, but funnier when you try to do something like the following: Write the sine function buuuut at x=69 is 69 aaaand can only be written as a combination of elementary functions and limits (no use of the brackets)

[@PixelSergey]

wait, what? you said that lim_{k->inf}(k!x) is an integer iff x is rational. But for x=1/2, lim_{k->inf}(k! * 1/2) = infinity, which isn't an integer. Is there some more precise way of phrasing this? (I'm guessing "eventually all terms become integers" or something)

[@markopanev3317]

The x is inside the cosine function hence it's limited by -1 and 1 and in the limit even if the cosine is 0.99999999 as j tends to infinity it goes to 0, so only from the rationals can we get 1 for the expression

[@lox7182]

1:17 I get what you're trying to say with lim k to infinity (k factorial * x) is in Z, but that limit (unless x is 0) would actually be positive or negative infinity. I think it would be better to say something along the lines of (for all k bigger than a certain natural number N (k!x) is an integer)

[@ron-math]

Right. Good point.

[@lox7182]

0:23, for number 4 can't you just have the constant function c?

[@ron-math]

As mathematicians will say, it is trivial 🤣

[@emilyscloset2648]

A Weierstrass function is also an answer to most of those questions

[@ron-math]

On the todo list.

[@parthhooda3713]

Can the greatest integer function be the answer to 2nd question ❓⁉️

[@ron-math]

Mind be more specific? You mean the ceil() function?

[@Setiny]

Now find a function that is discontinuous at any rational number but continuous otherwise

[@mr_1wr572]

He be like Gojo fr

[@_dd-n9zl]

Herkes için altyazı lütfen

[@ron-math]

Hi bro. Click the `cc` button on the lower right corner, select `auto-translate` and find the language you are using. Most of the time, it is good enough.

[@andy02q]

But... There are infinitely more irrational numbers than there are rational numbers, I'm sure I learned that some years ago. So shouldn't there be examples for two irrational numbers so close that there's no rational number in between? And if not, why not?

[@ron-math]

Good question. You can construct a rational number between any two irrational numbers. Because I can't type LaTeX in the comment, here is the link: math.stackexchange.com/questions/421580/is-there-a-rational-number-between-any-two-irrationals

[@omp199]

The cardinality of the set of irrational numbers is higher than the cardinality of the set of rational numbers, which is a more precise way of saying that there are more irrational numbers than there are rational numbers, so yes to your first question. But to your second question, no, there can't be two irrational numbers so close that there's no rational number in between them. Given any two distinct irrational numbers a and b, the distance d between them is given by d = |a - b|. Since a and b are distinct, a - b is not zero, so d > 0. However small d is, you can always find a fraction 2^(-n) that is smaller than d for a large enough positive integer n. Given that consecutive integer multiples of 2^(-n) are always a distance 2^(-n) apart, which is smaller than d, it must be true that at least one such integer multiple of 2^(-n) falls within the interval between a and b. And any integer multiple of 2^(-n) is a rational number.

[@ron-math]

Thank you! @@omp199

[@omp199]

@@ron-math You're welcome.

[@user-mf7li2eb1o]

Okay the way you pronounced continuous and other things are just wrong. Except that the video seems great

[@ron-math]

Can you help me identify the list of mispronunciations? Thanks!

[@HJ-gg6ju]

The pronunciation is fine. Your ears function poorly.

[@finthechat2717]

@@ron-math your accent is on the thicker side so words like graph become gruff between 1:45 and 1:55 I feel are the most noticeable, like the word atoms sounds like items, those are just some examples

[@ron-math]

Thank you so much for pointing it out! I will definitely work on it in the future videos.@@finthechat2717

[@BridgeBum]

Shortly after being introduced to this function, my calc 2 teacher posed to the class to give an example of a function continuous at only 1 point. His solution was the Dirichlet-adjacent function f(x)={x if rational, -x iff irrational}. It isnt hard to see that f(0) has a straightforward delta epsilon proof.