The Dirichlet Function is Nowhere Continuous - Advanced Calculus Proof

[@rickyoon1446]

As an 2nd year analysis undergrad student, this is super clear, straightforward, and helpful, thank you! 🙏

[@ved1749]

anyone discovered this masterpiece in 2022 ?

[@XZMath]

also can tie this nicely with Riemann-Darboux integration! Although the Dirichlet function is bounded, it is nowhere continuous, and in order for the Riemann integral to exist, f(x) must also be continuous almost everywhere. Since the upper sums equal 1 and lower sums equal 0 you can't integrate this function, so the condition that f(x) is bounded is insufficient.

[@mariomuysensual]

The Lebesgue integral exists in [0,1], why?

[@amritlohia8240]

@@mariomuysensual This function is Lebesgue integrable because the rationals are countable, so have measure 0. Thus the function is in fact 0 almost everywhere, so its Lebesgue integral is 0.

[@tarun-prakash]

THIS QUSTION WAS THERE IN THE BOOK Problems in Calculus of One Variable - I. A. Maron

[@TheMathSorcerer]

[@oscardavidalarcon2673]

I need a similar proof wtith the Thomae function.

[@douglasstrother6584]

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

[@kcwilliamson50]

very helpful for my analysis class. thanks a lot.

[@TheMathSorcerer]

np:)

[@iamkhang3952]

I have a question why the epsilon is 1 but not a different number

[@theaniviamain7709]

you can pick anything for epsilon but the precise definiton of continuity says every epsilon grater than zero, must be satisfy for at least 1 delta. so picking epsilon as 1 is making this statement much more easier. because if its not satisfys you can say this function is not continuous.

[@newGam3Trailers]

Why are we allowed to pick our X our of the complement of x0? Doesn't that mean, that for all other X it isn't proven?? (eg when you set X0 = rational, so you picked X = irrational and proved it for that. But what's the case, when X0, as well as X are irrational??) Thanks!

[@mariomuysensual]

I have this question

[@ansper1905]

the thing is x can be anything in the domain (x_0-delta,x0+delta). By the assumption that f is continuous somewhere, we get the claim that it's true for all x, then to disprove it, we only need to take only one value of x from the interval that gives a contradiction.

[@indiaThroughDrones]

@@ansper1905 cleared my doubt... thanks mate!!

[@franziscoschmidt]

This would be also really nice using sequences to define continuity.

[@shunpillay]

Awesome video. Thanks.

[@ikhouvandewii2]

Thanks for this proof. How would we proof this function is not Riemann integrable?

[@amritlohia8240]

The upper Riemann sums are 1 and the lower sums are 0.

[@Zinebzineb-o3f]

scuse me sir,, but why do you set epsilon as greater than 1. genuine question

[@yifanzhang6556]

What if x and a are both rational/irrational?

[@CaptchaSamurai]

By definition it can't be. Rational numbers are all numbers which can be expressed as fractions p/q, where p and q are integers (and q≠0). Irrational numbers are those numbers which can't be expressed in fractions p/q. I think we agree, that we can't have both x = p/q, and x ≠p/q :D. Cheers!

[@CaptchaSamurai]

You might also think what about function like f(x) = = 2, when x is irrational = 1, when x is rational = 0, when x is a integer This is ill defined function, thus not a function! Because it outputs two values for every input which is a integer (since integers are both integers and rational numbers).

[@atharvsawant7509]

The choice of epsilon is arbitrary right? So what if we choose it to be greater than 1. How do you prove it in that case?

[@alexandruionita7385]

In both cases basically is it 1 < ε so you can take whichever ε > 1 you'd like.

[@duckymomo7935]

Why is Dirichlet discontinuous everywhere but popcorn function is continuous sometimes?

[@amritlohia8240]

Essentially because in any interval, you can find rationals p/q with arbitrarily large q, and 1/q (the output of the popcorn function at rational inputs) can thus get arbitrarily close to 0 (the output at irrational inputs).

[@Alienman1212]

is there a reason you picked epsilon to be 1, could it have been any other number, i.e. 2 or 3...etc?

[@Vidaljr88]

Can any reasonable epsilon be greater than one? The distance between 0 and 1 is no more than 1. That's why I think these proofs use epsilon to be 1 or less.

[@emilmullerquintanar5055]

You could pick any epsilon smaller than 1, 1 is just when the contradiction starts happening

[@fraliak]

Wow this makes so much sense. :)

[@TheMathSorcerer]

:)

[@anjanajs9705]

Thanks!

[@TheMathSorcerer]

You are welcome!

[@ikersanchez8222]

Is this function continuous at [0,1]nQ ?

[@amritlohia8240]

Well, if you restrict the domain to rationals, then the function is just constant - it's always equal to 1, so of course it's continuous.

[@Sheikxlove123]

Amazing!

[@TheMathSorcerer]

thanks!!!!!!!

[@notadreamer-u2l]

How would u prove that (x^2)f(x) is differentiable at x=0

[@ansper1905]

you need some conditions there, buddy. C.E. : f : x --> 1/x^3

[@mikeking4311]

Nice! But I want to know how to proof that xD(x) is discontinuous at R but not at x = 0. ( D(x)is Dirichlet function )

[@monkeytrollhunter]

Thank you so much! You're great!

[@arijitmajumder2638]

Thanks sir ❤️

[@TheMathSorcerer]

You are welcome!

[@martaschaefferbutzge946]

thanks so mutch

[@TheMathSorcerer]

👍