"On sets of measure zero, always bet on Lebesgue... and his overwhelming integrability."

@reedoken6143

2 ай бұрын

bear witness to the one who left it all behind, the one who is free

@aokiest

2 ай бұрын

hey where did u study lebesgue integrals on i ve seen quite an amount of this but i gotta see more of it

@Sir_Isaac_Newton_

2 ай бұрын

💀💀💀 Fraudmat about to get the airport treatment with this one

@botondbalogh6017

2 ай бұрын

Found the jujutsu kaisen fan

@DerPoto

2 ай бұрын

As the king of calculus, Bernhard Riemann faced the Dirichlet function.

Life has much improved since I stopped worrying about analytic monstrosities and decided to live C_infinity

@_Xeto

2 ай бұрын

Welcome to physics

@whatisrokosbasilisk80

2 ай бұрын

@@_Xeto There's a derivative for that

@Oxygenationatom

2 ай бұрын

What the? I don’t understand the last part

@patrickdevlin8553

2 ай бұрын

C_infinity refers to functions that have infinitely many derivatives. In other words, the commenter means they decided to exclusively work with very smooth functions (a nice place indeed, but not as nice as functions that are equal to their Taylor series)

@jankriz9199

2 ай бұрын

​@@whatisrokosbasilisk80 certainly they are not. Some of the numerical approximating methods which are PRECISELY what physicists do when describing the real world is densely packed with many very nice very smooth functions, say gauss curves for example. Smoothness is nice and convenient, thus it is used.

Interesting to think that these jumps occur at infinite scales too. Whether you zoom in or zoom out, the density of the discontinuities is going to be just as thick!

@shimrrashai-rc8fq

2 ай бұрын

Yeah, you can't really visualize it perfectly ... a line riddled with infinitely tiny holes packed infinitely dense, or two complementary lines where one has holes exactly where the other is filled.

@TheLukeLsd

2 ай бұрын

Irônicamente elas tendem a ter densidade "completa" porque a quantidade de números irracionais é um infinito além do infinito dos números racionais e pode-se dizer que a quantidade de irracionais entre dois racionais é maior que de todos os racionais. E a integral dessa função tende ao valor da constante do número irracional. Mas ao mesmo tempo entre quaisquer dois irracionais definidos haverá pelo menos um racional e entre quaisquer dois racionais definidos haverá pelo menos um irracional. Uma coisa louca lidar com infinidades, infinitos e infinitesimais.

@reedoken6143

2 ай бұрын

that's a pretty good description of the concept of dense sets in calculus and topology

My Calculus Professor dropped this on us at the end of a Friday lecture to give something to discuss at Happy Hour.

To prove "a point" 😂

@LeandroSehnemHeck

2 ай бұрын

OP has proved his point, twice.

A slightly more fun example: consider a function that is equal to 0 for all irrational arguments and 1/q for any rational number p/q, where p and q are coprime and q is positive. It's not hard to show that such a function is discontinuous for all rational values and continuous for irrational ones.

@Grizzly01-vr4pn

2 ай бұрын

Thomae's function

@TheLuckySpades

2 ай бұрын

The modified Dirichlet function is fun It is also neat to show that the Dirichlet function is not Riemann integrable, but the modified one is

@matheusjahnke8643

2 ай бұрын

Nitpick police coming in That function is also continuous for 0;

@pavlosurzhenko4048

2 ай бұрын

@@matheusjahnke8643 The nitpick police is wrong. Zero is a rational number of the form 0/1 (0 and 1 are coprime since gcd(0,1) = 1, and 1 is positive), so the value of my function there is 1/1 = 1. But there are irrational numbers in any neighborhood of zero, and those are sent to zero, so the function is discontinuous at zero too.

@polvoazul

2 ай бұрын

@@pavlosurzhenko4048 so i guess you are... the nitpick fbi? heahaeheha

In the description you wrote "In 1729" and I believe you meant to write "In 1829" since this would be roughly around Fourier's time..

@chloeeeeeeeeeeeeeeeeee

22 күн бұрын

ramanujans typo strikes again....

But what happens to the FT of that function??? Also converges to 1/2 ?

@NuptialFailures

2 ай бұрын

Because the function is nowhere continuous it doesn't meet the criteria to have its Fourier transform taken. The point of this function was to essentially construct a function for which the Fourier transform couldn't be taken

@hansyuan4116

2 ай бұрын

I believe it will be constantly 1. Observe that if f and g are Lebesgue integrable functions and the measure of {x|f(x) != g(x)} is zero, then the Fourier series of g equals the Fourier series of f. The result follows by letting f=1 for all x and g be the dirichlet function.

@NuptialFailures

2 ай бұрын

@@hansyuan4116 this will depend on how we are defining the Fourier Transform. If we use the Riemann Integral, then, as the video is discussing, the lack of piecewise continuity will make it so we can’t take the Fourier Transform. If we use the Lebesgue Integral then, as you suggest, we can extend the functions for which we can apply the Fourier Transform to.

@pierro281279

2 ай бұрын

I was about to comment that your pronunciation of Dirichlet may have been wrong as all teacher I've had pronounced "dirikley". They were all wrong, my life was a lie !

@user-gd9vc3wq2h

2 ай бұрын

​​@@pierro281279He was German with French-speaking ancestors (from what is now the French-speaking part of Belgium). In Germany, his name is usually pronounced "Dirikley" nowadays, but the French way, i.e. "Dirishley", isn't wrong either, imo.

What a smooth explanations of proofs! Great job !

Whatta rebel my man Dirichlet! Great video!🎉

Whoa, yeah it becomes continuous. That raises moe questions… Great video! I hope you’re going to do a follow-up to this.

Lovely as always. 💜

Mindblowing. Great explanation!

I love Will Wood's music, but he's an even better mathematician :)

Very cool!

This just makes me realize that the intuition I had developed to explain why the cardinality of the reals is larger than the cardinality of the rationals is not correct Well that's fun. Just gotta relearn my fundamental understanding of infinity again

@jordanrodrigues1279

2 ай бұрын

Intuition is kinda cursed when thinking about these things. Notice that we're really close to stating the continuum hypothesis, and that can't be proven or disproven.

@serraramayfield9230

2 ай бұрын

@@jordanrodrigues1279Indeed, it was proven to be independent in the 90s

@matheusjahnke8643

2 ай бұрын

@@jordanrodrigues1279 you mean those things are cursed while intuition fails because it isn't;

This might be the greatest math-pun I've ever read

Great video! At 0:30, that infinite sum of trig functions would also be a weierstrass function right? Would love to see you do a video on that topic, a similarly weird function to the one you talked about this time

@MeButOnTheInternet

2 ай бұрын

Fourier series are especially useful for solving PDEs such as the heat equation. such a series would certainly be differentiable whereas the point of the wierstrass function is that it is differentiable nowhere.

Dirac's delta function has entered the chat

@ThePiotrekpecet

2 ай бұрын

Dirac delta is continuous when you define it properly, as a linear functional on a Schwartz space (a distribution) since it is bounded (as an operator)

@ThePiotrekpecet

2 ай бұрын

Also Fourier transform is defined differently for distributions i.e. Fourier transform (Dirac delta(f))=Dirac delta(- Fourier transform(f))

@pneujai

2 ай бұрын

dirac delta has nothing to do with today’s topic

Why does it become continuous if you take only the rational or irrational parts? Wouldn't there be an infinitely dense forest of Removable points of discontinuity if you do that?

@shimrrashai-rc8fq

2 ай бұрын

Continuity means different things depending on the domain. In fact, calculus books that call a function with points excluded from its domain "discontinuous" at non-domain points are kinda misleading because such a thing is technically not a function at all on the "full" reals (rather it's what is called a "partial function"), since the definition of a function requires that each point of the domain be associated with _some_ point of the codomain. If you restrict the domain to, say, only rational points, then what happens is that in effect this domain cannot "see" where what you are calling "discontinuities" are, and thus it "thinks" the function is continuous. (Think about what'd happen if the reals just didn't exist, because you did not define them yet!)

@argon7624

2 ай бұрын

Well imagine if you only have rational numbers, for each x in Q, f(x) = 1, you are continuous on your domain

@hanskywalker1246

2 ай бұрын

​@@shimrrashai-rc8fqbut then 1/x must be continuous too?

@skallos_

2 ай бұрын

If you exclude 0 from the domain, yes, 1/x is continuous. If you do include 0, then no matter what you define 1/0, the function will be discontinuous.

@hanskywalker1246

2 ай бұрын

@@skallos_ but since everything divided by 0 doesn't exist can't you just say that these functions then are contiuous. Or if they are just contiuous on a given interval aren't they just partly contiuous?

Hey this isnt the Normal Album!

@the_person

Ай бұрын

Was searching for a Will Wood (the artist) comment

Love those multilevel narketing campaigns

This is so cool

OK ngl, that last bit killed me ☠

always coming across this function in physics

The c for rationao and d for irrational idea was the first thought I had when I saw the thumbnail. Kinda proud of having basic understanding of maths.

Is this a reupload? I remember seeing this long ago

The most epic pun

The last point really drives home what a monstrosity this function is

There is a definition of phi as the most irrational number, where it’s written as a continuous fraction of ones all the way down. If two takes the place of the last digit in this continuous fraction construction for phi, then those two numbers are not separated by a rational number. I think this runs into some methodological issues. However, you can do this digit substitution with finite values of phi, for which it is possible to find a rational number. It is only problematic to find it when the digits occur at the end. Still, any continuous fraction can be flipped inside out, though you may run into problems with infinity, though an equality sign by definition makes an equation finite if one side has finite value.

I didn't know that there's a rational number between any two irrationals. Because there's also an irrational between any two rationals, does that mean you can't go from one irrational to another without going through a rational, so the number line alternates in some sense between rationals and irrationals?

this is for 2:21 ) also have you heard of the musician will wood? also good stuff

@johnmartorana196

2 ай бұрын

Thank you. We'll need another at 3:57 if you've got one. It was driving me more crazy than it had any right to.

@hihello8601

2 ай бұрын

will wood fan spotted

With the proof you just gave, could you say irrational numbers and rational numbers alternate in a certain way? Or not?

Would be interesting to know about the significance of this. What has this function been used for?

@Rantalaiho74

2 ай бұрын

For proving a point.

This is incredible. I have so many questions. If you can find a rational number between any two irritational numbers, and an irritational between any two rational numbers, doesn't that imply that there are an equal number of rational and irrational numbers, since you cannot have a continuous sequence of either? Then why are there infinitely more irrational numbers than rational numbers? The function is interesting, but the two small proofs used to construct it are way more interesting. They imply a lot of crazy things. Can you make a follow up video covering a section of the number line, and why there must be more irrational numbers? Or does the proof only work for the entire number line, rather than a finite section? If that's the case, infinities are even more confusing.

@lah30303

26 күн бұрын

There are more irrational numbers between any two irrational numbers than there are rational numbers between those two (or any other two) irrational numbers.

Isn't that there are more irrational numbers than rational numbers? Won't this affect the density of the lower line? What would the Fourier transformation for this function look like?

1:13 should this not converge to x = 1 at 0, because the original function is equal to x = 1 at 0?

If "Because FU, that's why" had a function.

Okay, but what's the Fourier transform of this function?

Wait, how is one part or the other, contiuous, at the end of video?

So, can you Fourier transform the Dirichlet function?

how does the function become continuous if restricted to just one of the domains?

how does it become continuous if you restrict the domain just to rational or irrational? won't you have holes?

@unknowntimelord9557

22 күн бұрын

Well yes but actually no. Rationals and reals are both what we call "dense", meaning at any Intervall no matter how small you have an infinitely many numbers from both. So that makes the holes kinda disappear. So you might say if we forget about all reals there would obviously be a "hole" at π to which I would say: well yes but I can find a rational thats as close to it as you like. So basically the holes are infinitely small and everything still works just fine

While I understand each of them in isolation, it makes my brain hurt to try to reconcile that a) between any two rationals there is an irrational and between any two irrationals there is a rational, and b) that rationals are countably infinite and irrationals are uncountably infinite. It feels like A should imply they have the same cardinality, even though i know that it doesn't!

@SioxerNikita

Ай бұрын

You can systematize a mathematical way to represent all rationals, you can't with irrationals.

@lawrencebates8172

Ай бұрын

@@SioxerNikita That's just another way of saying that the rationals are countably infinite and irrationals are uncountably infinite, it doesn't really help with building an understanding or intuition for why.

@bebe8090

22 күн бұрын

@@SioxerNikita Could I not say, list out the irrationals by pairing them up with one of the rational numbers? Say I have irrationals a, b, c... and rationals A, B, C... defined such that a < b < c... and A is the rational between a and b, B is the rational between b and c, so on. Since the rationals are listable, would I not have all the irrationals listed out by pairing a with A, b with B, c with C, ect?

@SioxerNikita

22 күн бұрын

@@bebe8090 You wouldn't be able to, because irrationals are infinitely long, so by the mathematical standards of that, you can't.

@SioxerNikita

22 күн бұрын

@@bebe8090 Essentially, you can't list irrationals like: "I start with 0.00000...1...", because there is one that has one more zero before the 1. While rationals, you can start systematicizing it. Position 0,0 could be 0.0, then 0,1 would be 0.1, and so on. 1,0 could be 1.0, etc. Listable via system. You cannot do this with irrationals. You can't make a "start point".

I guess the merely countably infinite (actually constructivistically "existing") subset of the irrationals (that are computable by a necessarily contably infinite set of all combinatorically possible different algorithms) are sufficient here. So is this compatible with constructivistic math? I'm not sure.

Its almost like the function is in a "super position" of sorts?

Wait!! That last sentence of the video felt like mike drop. Why dies the function become continuous when we restrict the domain to just the rational or just the irrational? Wouldn't this be a line with infinity many holes in it?!

What would be the integral of this function ? (between 0 and 1 for example) Undefined ? 1/2 ? some other ratio ?

@ironicdivinemandatestan4262

7 күн бұрын

The function is non-integrable.

@isaz2425

7 күн бұрын

@@ironicdivinemandatestan4262 Thanks. Yeah, I guess that makes sense.

Hi is the Weierstrass function related to this subject?

I learned about this function in college.

how does restricting it to the rational inputs make it continuous?

@magma90

2 ай бұрын

The function is defined as f(x)=a if x is rational, and f(x)=b if x is irrational, therefore if we restrict it to rational inputs, the function cannot give the value of b, as b is output when x is irrational, therefore as a is the only value of the function when x is rational, and we only can input rational numbers, the function simplifies to f(x)=a, which is continuous.

@pizzarickk333

2 ай бұрын

@@magma90 I understand that as we restrict the domain to the rational inputs, f takes only the value a (which is 1 in pur case). But I don't understand how that makes it continuous. My understanding is that if a function is any close to being continuous on some interval I, it needs to be defined on all real numbers in I. If, for example, I define the function g(x) that equals x² when x is an integer, and 0 when it is not. If I restrict the domain to only integers, my functions would g(x) = x². But it is not continuous since it is defined only for integer values. Am I missing something here?

@pavlosurzhenko4048

2 ай бұрын

@@pizzarickk333you're missing the more general definition of continuity. The most general version of it is that between arbitrary topological spaces. You treat both the domain and the codomain as their own spaces (so for our purposes irrational inputs simply don't exist). If you can't detect any discontinuities, then the function is continuous. For example, if you define a function on rational values to be 0 for x = 0, you can still find that the function is discontinuous at 0, but for constant functions there aren't any discontinuities at all. Now the induced topology on natural numbers is a discrete one, which means that every function with natural numbers as its domain and some other topological space as a codomain is continuous. People generally don't talk about continuous functions on natural numbers because it's not very meaningful - they are _all_ continuous. As for more precise definitions, I'm not going to give you the most general one, since it requires some intuition building first, but if you have two sets X and Y with a well defined concept of distance between two points (let's write it as d(x, y)) then you can say that a function f from X to Y is continuous at a point x iff for any epsilon > 0 there is such a delta > 0, such that for all x' if d(x', x) < delta then d(f(x'), f(x)) < epsilon. Hopefully you can see how it's basically the same definition as in your real analysis class.

@natebowers7024

2 ай бұрын

​@@pizzarickk333 I think some of your confusion may come from the (opaque) definition of continuous functions on weird sets. In your example on the integers, we have a function g:Z->Z with g(x) = x^2. The usual topology on the integers is the discrete topology; that is, ever subset of Z is open. By the definition of topological continuity, it is quite easy to see g(x) is continuous as every function from a discrete topology to an arbitrary topology is continuous (see note below) If you haven't studied topology, sadly, this approach isn't too intuitive. The key idea, however, is that defining a function consists of three things: the domain, the co-domain, and the map. It's normally implied that the domain is the real numbers, but, for domains like Z, definitions that are formulated over the reals no longer make sense. Topology helps us deal with these cases. I hope this helped! N.B. For two topological spaces (X,𝜏x), (Y,𝜏y), we say a function f:X→Y is continuous if for every open V⊆Y, its inverse image is open; that is f^-1(V)={x∈X|f(x)∈Y}∈𝜏x. A direct consequence of this definition is that every function from a discrete topology to an arbitrary topology is continuous. As the discrete topology is simply 𝜏x=P(X) (the power set of X), trivially for any V we have f^-1(V)∈𝜏x as the power set contains all subsets. Thus, the claim holds.

@farfa2937

2 ай бұрын

@@pizzarickk333 Continuous always implies "within the domain". If you evaluate continuity as "it needs to be defined on all real numbers in I" you haven't restricted the domain.

3:55 you forgot the ) after irrational

That is actually so fucking genius

2:02 not differentiable anywhere? Weierstrass moment

Wouldn't splitting the function into two restricted functions not be continuous because there exists both rational and irrational numbers in all real numbers? For example, on the irrational restriction, wouldn't the function be discontinuous at every rational number like 0,1,2??

@hhhhhh0175

2 ай бұрын

talking about continuity where the domain isn't an interval (or even connected) is indeed pretty weird. if we use the normal definitions of continuity for these domains, the restrictions of the dirichlet function are continuous - but so is, for example, the indicator function of x^2 < 2 on the rational numbers, which clearly has jumps

@mirkotorresani9615

2 ай бұрын

The restriction to some subset is a new function, that only sees that subset. And if on that subset the function is constant, then is continuous. We are teaching math at high-school in the same way that mathematicians did before Cauchy: everything must be an interval, and function have always a "natural" domain. The problem is that Cauchy died in 1857

One thing that bothers me a bit. Aren't irrational numbers suppose to be denser than rational ones since they are alef_1? If there is at least one rational number between two irrational and one irrational between two rational it sounds like there is the same density and same amount of numbers which cannot be due to uncountable/countable Infinity

@canaDavid1

2 ай бұрын

There is not just one between them; there is a coutably infinite rationals and uncountably many irrationals. It's like there is always an integer between two numbers with difference > 1, and also a real between them, but there are more reals between.

@alexanderf8451

2 ай бұрын

They do have the same density. Nothing requires sets of different cardinalities to have different densities.

@legendgames128

2 ай бұрын

Even if the irrational numbers are more numerous, I'd be inclined to think that irrationals and rationals have the same *density*

@SioxerNikita

Ай бұрын

​@@canaDavid1You cannot have more irrationals than rationals, if you can always find a rational in between any two irrationals and vice versa. Saying there is now more of one than the other is... Well... Logically wrong. We can say there are more reals than integers, because there is an infinite amount of reals in between any two integers.

I expected this to lead somewhere.

2:21 um... you are missing a closing parenthesis on "(irrational" and then again at 3:59.

Wait! So does the dirichlet function even have a fourier expansion? My instinct is that it can't as it is everywhere discontinuous, and so it can't have a derivative anywhere. It also doesn't satisfy an alternate sufficiency for f-series by not being of bounded variation (Wik, '72).

@adammillar1324

2 ай бұрын

Unless I am mistaken, your instinct is backed up by a theorem! Isn’t it wonderful when that happens. If f is differentiable at a point x=c, it is continuous there. Contrapositively, if f is discontinuous at x=c, it cannot be differentiable there.

@pavlosurzhenko4048

2 ай бұрын

You need Integrability for the Fourier expansion, not differentiability. But yeah, it's not Riemann-integrable. However, the Lebesgue integral of this function on any interval is 0, and it will also be the case after multiplying by a sine or cosine, so the Fourier series will have all zero coefficients and will converge to 0 (which is equal to the Dirichlet function almost everywhere, so it's not actually that bad).

@timseguine2

2 ай бұрын

In fourier theory you can ignore an arbitrary measure zero set and get the same answer. In this context, the rationals are countably infinite and so have measure zero, so we can choose to just ignore the rationals. If we do that, the function is exactly zero everywhere that remains, so the fourier transform is exactly zero everywhere.

@xizar0rg

2 ай бұрын

@@timseguine2 Aren't fourier transforms reversible, though? Seems like this destroys the original function. (asking, not arguing)

@timseguine2

2 ай бұрын

@@xizar0rg Reversible in the function space L1. And the Dirichlet function is equivalent to zero in the L1 norm. As far as the L1 norm is concerned the Dirichlet function is the zero function. One way of interpreting it is that the dirichlet function has zero content at any finite frequency. Which the video hinted at.

Can quantum nonlocality/entanglement be represented by a Dirichlet function?

@kikivoorburg

2 ай бұрын

They’re not really directly related. Quantum entanglement comes down to interconnected probabilities, and extends far past 2 connected states (entire systems of particles can entangle together). Also, which state is which is an arbitrary choice, which the rational/irrational inputs in the dirichlet function aren’t. (Obvious example: there are more irrational numbers than rational ones, by Cantor’s diagonal argument.) The fact we can distinguish these two means it can’t correspond to the indistinguishable entangled states.

Oh my god, I understood that...

But what about Fourier transformation of this function?

@btf_flotsam478

2 ай бұрын

Fourier transforms involve integrals and the rational numbers have measure zero over the real numbers. The Fourier transformation would just be zero.

@farfa2937

2 ай бұрын

afaik the point this was invented to prove is that it cannot have one.

@ThePiotrekpecet

2 ай бұрын

​@@farfa2937it has one if you define Fourier transform in a modern way by Lebesgue integral. Fourier transform of this function is zero since Q has Lebesgue measure 0

......so what was the fourier transform of the dirichlet function? You said he invented it to prove a point but never said how it proves that point

So between any two points in the real numberline, there are infinitely many rational numbers, and infinitely many irrational numbers for each of those rational numbers. Yet between any two irrational number there exists a rational number inbetween them. I guess the pigeon hole theorem doesn’t really hold once we have infinite holes.

Isn’t this basically a fractal?

I feel like with infinite precision the function would be continuous for both rational and irrational numbers

Well, I would argue, that y=Dirichlet(x) where x is from Q is dense, but it is not continuous, since Q, while dense, is only countably infinite, |Q|=|N|, i.e. does not have the cardinality of continuum, while R is uncountably infinite, thus R\Q is also uncountably infinite, |R| > |Q|, |R\Q| > |Q|, and thus y=Dirichlet(x) where x is from R\Q is really continuous subset. In other words, rational line y=1, is infinitely-times sparser than Irrational line y=0. Therefore, on average, the Dirichlet function is zero. Only occasionally, on those infinitely rare rational countable moments in the irrational continuum, is it 1.

@YouTube_username_not_found

Ай бұрын

>> "I would argue, that y=Dirichlet(x) where x is from Q is dense, but it is not continuous, since Q, while dense, is only countably infinite" Continuity does not care about the cardinality , it only cares about the behaviour of the function in the neighbourhood of the non isolated points of the domain and the behaviour at that point. Whenever there is a sequence that converges to a point, we can define the limit of the function at that point. Note that this point need not be in the domain. And since Q is dense in R, then every real is not an isolated point, so we can define the limit of the Dirichlet function at any real number. This limit is determined by the behaviour of the function in the neighbourhood of the non isolated point. Since the restriction of the Dirichlet function on Q (respectively on R\Q) is constant, the behaviour of the function around any point would be constant, so any limit value would be that same constant. Now, we recall the definition of continuity: f is continuous at a ⇔ lim(x→a) f(x) = f(a) . In order for us to talk about the continuity at a point, that point must be both an non isolated and a point in the domain. This is already verified for the Dirichlet function restricted to Q (respectively R\Q) , all of the points of the domain are non isolated. The function is constant, so, ∀a∈Dom(f) , f(a) = c and lim(x→a) f(x) = c , they are both equal! Thus, any constant function is continuous. Reference: Real Analysis 26 | Limits of Functions

@YouTube_username_not_found

Ай бұрын

Just to clarify; the reference is a video on KZread

ok but what happens with the fourier transform of that function?

@ThePiotrekpecet

2 ай бұрын

If we talk about Fourier transform defines by riemman integral it doesn't exists since 1_Q is not riemman integradable so Dirichlet would probably end at that but if you define Fourier transform by Lebesgue integral Fourier series is the 0 function so Fourier series converges everywhere to 1_Q :)

But aren't "just the rationals" countably infinite thus not continuum in a sense

@semicolumnn

Ай бұрын

You can define continuity on discrete sets by generalizing from the real line. In fact, one can define continuity for functions from set to set, like f: R -> Q

Where Fourier?

Anyone remember when KZread was literally just cat videos? Yeah... I dropped out of maths at university right before the change happened.

Infinite decimals does not mean necessarily that the number is irrational, take for example 1/3

Inadvertently there is a proof in here (or at least a basis for one) that the number of rationals and number of irrationals is actually the same, and thus they are the same size of infinity. This would contradict other proofs that show that there are more irrationals than rationals, infinitely many more making the counts different orders of infinity. I wonder if you could find a flaw in one of these proofs using the other, or if they can both be true (which would be weird, but infinities can sometimes just be weird like that)

@semicolumnn

Ай бұрын

Between two irrationals a,b there are countably infinitely many rationals and uncountably infinitely many irrationals. (This immediately follows from the fact that (a,b) is uncountably infinite and that Q is dense in R as demonstrated in the video). While Q is dense in R, it can still be denumerated.

@TheFinagle

Ай бұрын

​@@semicolumnn Ok, but according to the Dirichlet function rationals and irrationals strictly alternate. Because they strictly alternate each rational is followed by one and only one irrational, and each irrational is followed by one and only one rational. Thus according to Dirichlet and the proof started around 2:18 there is an exact 1 to 1 relationship between rationals and irrationals. Somewhere between the different proofs exists a contradiction, but as they are describing different facets of the number line and I don't think they directly contradict each other. My quandary which I am not knowledgeable enough to follow through properly is if they really are in contradiction where only one could be true or if it could be solved an shown that both things are true simultaneously - that there are infinitely many more irrationals than rationals AND an exact 1 to 1 relation between rationals and irrationals (which as I said would be weird, but sometimes infinities just do weird stuff)

@semicolumnn

Ай бұрын

​@@TheFinagle A rational isnt followed by exactly one irrational. In constructing the irrational, Dr. Wood uses (a + r(b-a)) where r is an irrational number between 0 and 1. There are uncountably many irrational numbers between 0 and 1 (Because (0,1) is uncountable and Q is countable) and thus there are uncountably many irrational numbers between two rationals. The other direction is much simpler. We know that there are infinitely many rationals between any two reals, and that the rationals are only countably infinite. Thus, the number of rationals between two irrationals is countably infinite. To recap: Between any two real numbers a and b, there exist uncountably many irrationals and countably many rationals. No paradox.

@TheFinagle

Ай бұрын

@@semicolumnn Then it should be possible to formally write that proof in a way that directly disproves that Dirichlet's function as being strictly discontinuous, ie that at some points you must have more than one irrational in a sequence because there are more irrationals than rationals. Thus forcing a continuity to exist within the function where irrationals happen in sequence. In doing so you would break one of the 2 proofs he used to show that they strictly alternate such that the function can be discontinuous between ALL real numbers.

@semicolumnn

Ай бұрын

@@TheFinagle You can't have any real numbers in a sequence. If you could, their difference would be the smallest positive real number, but no such number exists.

After watching the video I understand the problem but not the solution. How does the formula look like for that rational/irrational function? How doe the Fourier of it look like? is this infinitely zoomable and if so why? Why is it not continuous and why does every other point need to be irrational. Can't I just make up a function where every odd measurement point is 1 and every even measurement point is 2?

@alexanderf8451

2 ай бұрын

Like calculating if a number is irrational? That's not needed here, only the existence and properties the rationals and irrationals. Its not continuous because if you pick any two points on the real line the function will have at least discontinuity (a switch between 0 and 1, in this case) between them (in fact it will always have infinitely many). You can have a function that is 1 on the odds and 2 and the evens, yes, though that doesn't matter here are parity is only defined for integers (whole numbers) and those are sparse in the reals.

@benrex7775

2 ай бұрын

​@@alexanderf8451 I don't think your answer really helped me understand the situation any better. Perhaps let's start with a simple question. Is this a function which you can formulate in the following matter: f(x)=a*x+b

@alexanderf8451

2 ай бұрын

​@@benrex7775 No, the function can't be written the way you describe.

@benrex7775

2 ай бұрын

@@alexanderf8451 How is it written then?

@alexanderf8451

2 ай бұрын

@@benrex7775 He shows you how its written in the video.

People had a lot of free time those days.

Damn, how they become continuous by themselves?

Isn't the number of irrationals greater than the number of rationals (different infinities), yet it appears as if they are exactly alternating (an irrational between two rationals, and a rational between two irrationals). Are there gaps in the continuum? ;-) Insights anyone?

@alexanderf8451

2 ай бұрын

The irrationals are uncountable while the rationals are countable. However that doesn't matter here only their density does and both have the property of being dense in the reals. That means given any rational (or irrational) there is no meaningful "next rational" (or "next irrational"), specifically if you chose any rational (or irrational) and then declare another value to the next one there are infinitely many counterexamples. Thus they don't alternate because to alternate you'd have to be able to pick the next number.

@philipoakley5498

2 ай бұрын

@@alexanderf8451It's how to explain that while you can put the numbers apparently in commuting ascending order (rational, irrational, rational,..) that you{we} don't actually have what you{we} thought you{we} had, possibly because you don't 'sort' the numbers like that to actually count the rational numbers (it may also be co-related to primes and co-primes and finding them).

@kikivoorburg

2 ай бұрын

They’re not “alternating”, as mentioned in the video we can’t call it “oscillating” specifically because of the fact that _there’s actually infinitely many rationals and irrationals between any two (ir)rational numbers._ It’s basically impossible for our minds to conceptualize a function like this, since infinity is weird. There’s no notion of “zooming in far enough” until you see rationals and irrationals as separate, because they just aren’t ever separate. Personally I interpret it as two vaguely ethereal “constant (line) functions” which look solid from a distance but are actually filled with holes. Kinda like matter in the universe, I guess

@semicolumnn

Ай бұрын

@@philipoakley5498 It’s provable that you can’t exhaust the real numbers using a sequence of the form {rational, irrational, rational, …}.

@philipoakley5498

Ай бұрын

@@semicolumnn the question is how you explain it...

What in banach-tarski‘s two dimensional cousin is this?

I'm pretty sure his name's pronounced Deer-ee-cleh rather than deer-ee-shleh, at least that's how Wikipedia and my professor say his name

Still waiting to hear what the Fourier transform is...this was false advertising 😂

This makes intuitive sense to me, but wouldn't this also prove that there is an equal number of rational and irrational numbers?

@surelyred

2 ай бұрын

you can’t, it isn’t really true, look it up

I thought there were more rationals than irrationals

The emphasis in "Dirichlet" is on "e", not on either of "i".

How can it be continuous on just the rationals if there are irrationals in between? How would such a function even be defined?

Does this have any applications for quantum mechanics and super positions?

@alexanderf8451

2 ай бұрын

No.

"Restricting it to just the rationals or just the irrationals, the function becomes continuous"...but it is not defined at every point, correct? So if we say y=1 for x=rational, then we pick an irrational number (x = pi), then the function y becomes undefined. Therefore, the function is not continuous because it has to be defined for all x.

@alexanderf8451

2 ай бұрын

If you restrict it to the rationals then you can't pick an irrational number, it doesn't exist in the domain of the function.

@mirkotorresani9615

2 ай бұрын

But the restriction has a new domain, the one you are restricting on. And if I have a function f from X to Y, I don't need to have an outside world where X lives in, for speaking about continuity. And since our restricted function is constant on all his domain, then it is continuous. Since, in absolute generality, a constant function g:X -> Y brween topological spaces is always continous

1:12 🙋‍♂

4:45 I think Cantor would disagree, for the set rational numbers cannot constitute a continuum. It is discrete!

@ThePiotrekpecet

2 ай бұрын

It doesn't matter, euclidean metric restricted to Q induces a topology on Q and therefore we can talk about continuity

@ThePiotrekpecet

2 ай бұрын

You can even talk about continuity on Z with discrete metric (but then every function is continuous so it's not as interesting). Topology, and therefore the notion of continuity, can be introduced on any nonempty set

@mirkotorresani9615

2 ай бұрын

You know that math has advanced a little since Cantor times?

Hi

Siri help

Does that mean there's an equal number of rational and irrational numbers?

@alexanderf8451

2 ай бұрын

No, density and cardinality are different properties of sets.

>> 4:38 - in fact, restricting the function to just the rationals or just the irrationals gives you a constant function No. It gives you a function with infinite number of holes. Function does not exists in that holes or not defined. That is not a constant function.

@semicolumnn

Ай бұрын

in R, yes. but were restricting to Q, where it is a constant function. the difference is always 0 so itll always beat any epsilon regardless of the delta.

so the function is continuous but not dereivable. ¿Weierstrass?

@edsangha3724

2 ай бұрын

the function is not continuous in R

@alexanderf8451

2 ай бұрын

No, this function is discontinuous everywhere and thus, unsurprisingly, doesn't have an derivative. Weierstrass's function is continuous everywhere but still has no derivative anywhere, which is even stranger.

@Xayuap

2 ай бұрын

what about the length of a weierstrass segment?

I clicked BECAUSE of the clever title

:O

This function is 0 almost everywhere, so it might as well be 0.

@SpinDip42069

2 ай бұрын

Imagine how useless mathematics would be if we applied your logic to everything

@brightblackhole2442

2 ай бұрын

@@SpinDip42069 "real numbers are irrational almost everywhere so they might as well all be irrational"

@lumina_

2 ай бұрын

what.

@cara-seyun

2 ай бұрын

The engineer

@Malthusia

2 ай бұрын

@@cara-seyunreal

I guess you could say Dirichlet's function is pretty much...broken

My Calculus Professor dropped this on us at the end of a Friday lecture to give something to discuss at Happy Hour.

I'm sure if the function is discovered in 1900s it would be called schrödinger function. f(x) is in superposition of both 1 and 0 until x is observed to either be rational or irrational. Upon observation f(x) collapses to either 0 or 1.

@edsangha3724

2 ай бұрын

very cool link to physics here maboi

@philipoakley5498

2 ай бұрын

Love it. !

@semicolumnn

Ай бұрын

Probably not... unless Schrödinger were to come up with it. In reality it would probably be called the indicator function on Q, since the reluctance of mathematicians to name things after people increases with time.

@user-jn9hs5ry7h

24 күн бұрын

No it's stupid. It is like saying f(x) = |x|/x is schrödinger function because f(x) is in superposition of both 1 and -1 until x is observed to either be negative of positive.

@edsangha3724

24 күн бұрын

@@user-jn9hs5ry7h the example you’ve given seems to fit the description of Schrödinger function, your point is?