(0,1) vs. the real numbers

[@XanderGouws]

Could you just use the sigmoid function as a mapping from all real numbers to the interval (0,1)

[@drpeyam]

Xander Gouws That would work too!

[@TheRealSamSpedding]

Very good, would like to see more analysis videos like these in the future.

[@TheRedfire21]

Great! More topology and set theory please this are great

[@jeromesnail]

Great video as usual! The notation (0,1) is really confusing IMHO, it can be mixed with the ordered pair (0,1) if not enough context is given. In France we use the notation ]0,1[ for open intervals and [0,1] for closed ones. I wonder where else those notations are used in the rest of the world :)

[@drpeyam]

I agree, I went to a French school as well :P

[@jeromesnail]

-Also couldn't we have simply studied the function (2/π)*tan(x) ?- Ignore my question it was a brain fart...

[@jeromesnail]

Dr. Peyam's Show ahah I saw the video where you talk in like 10 different languages, you're really awesome!!

[@theoleblanc9761]

Personally, (I am french) and every time I see (1,0) I don't known if it is [0;1] or ]0;1[

[@eleonard0o]

In Peru we use for open intervals and [0,1] for closed intervals. Some of us use the same notation you mentioned :')

[@adumont]

I love all your videos Dr PiM. It reminds me when I studied Math Spe M' at Ginette, Versailles, back in 1993-1995.

[@drpeyam]

Spé Maths 🥰 Good old times

[@adumont]

@@drpeyam thanks I studied that I can enjoy your videos! They are like little pills of Spe Math. Refreshers. I am usually nostalgic of what I remember I managed to learn and actually understand at that time, and now so long after I feel like an idiot, not remembering nor using any of it... So this pills 💊 of Math you give us are very welcome! Thanks. And I like your style.

[@ShubhayanKabir]

So glad you quoted "Eat, Pray, Love". That was really romantic. 😍

[@supertren]

Please Dr: a brief introduction of Topology will be great!.

[@drjoaoventura]

The sound of math, blackboard, and chalk is so relaxing... #ASMR

[@RealEverythingComputers]

Awesome tutorial! helped me out for a math proofs course. Wondering if there is an alternative method to this proof - for starters, is there another function aside from tan(x)?

[@MrRyanroberson1]

by 8:30 it's just like "now that we proved 1=2, we can prove 1=infinity", since 1=2=1+1, and 1=2, 1=4, 1=8, 1=2^(set of all integers), and also 1=2^infinity=infinity

[@9wyn]

Good camera angle. 🙏

[@drpeyam]

I’ll let the cameraman know :)

[@Shadow4707]

I agree.

[@BuddyVQ]

Keep up the great work! I learned a lot!

[@thesakinator7639]

can you not adjust the tan function so that its bounds can be 0,1 and have it approach -inf at 0 and inf at 1? such as tan(pi(x+0.5))

[@Charm98thanks]

Amazing video as always! I just have a couple comments to point out some minor technicalities in this proof. First, when you proved that f was onto, you should also evaluate f(x) after finding x to prove that f(x)=y. You essentially still want to show that it actually equals y (the way it is written, you have proven that if an x exists such that f(x)=y, then it must take the given form; however you must also show that the given x actually satisfies the equation). Moreover, the derivative of tan(x) is equal to 0 at x=0, so the increasing argument (correct me if I'm wrong) does not hold. Perhaps a couple more theorems might have made this result more rigorous. I think this could have been shown with the definition fairly easily as done for f. Regardless, I love your videos and your enthusiasm. This was a great video, and I can't wait to see more of your brilliant, enthusiastic work!

[@drpeyam]

PcRules1 Thanks so much, I really appreciate it! :) And actually when I solved for x, all the steps I did were reversible, so there’s no need to show f(x) = y, it automatically holds in this case! And sec^2 is always positive, so tan is strictly increasing!

[@Charm98thanks]

Right! Thanks for the clarifications. I also for some reason thought sec^2(0) was 0 haha. My mistake! Keep up the excellent work :)

[@TheMauror22]

But what's so special about (0,1)? Because, in that sense its possible to show that an interval (a,b) with a and b contained on the real numbers and different from each other, is isomorphic with R, doesn't it?

[@drpeyam]

You’re absolutely right, there’s nothing special about (0,1), any interval (a,b) would do!

[@TheMauror22]

Thank you Dr. Peyam! I love your videos, I always see them all!!

[@wherestheshroomsyo]

Can we simplify this by first proving that, (0,1) is isomorphic to (-r,r) and then obtain any real number by allowing r to increase to any size? The limit as r -> infinity of (-r,r) approaches the set of real numbers by definition I think.

[@drpeyam]

But then you’d have to prove that for any set if An goes to A (which you have to define rigorously), then the size of An goes to the size of A, which is a nontrivial Statement (and might not even be true)

[@wherestheshroomsyo]

I think I see what you're saying. So, the size of the interval (-r,r) having the same size as the reals is nontrivial even if (-r,r) does approach the reals. It would need to be proven, but not using the theorem from your video or else I'd be trying to prove a theorem using itself. My idea seemed so solid though. proving (0,1) is isomorphic to (-r,r) is straightforward, it uses the same argument as the one from your video with different numbers. The parameterization I used was f(x)=-r+2rx The derivative is 2r, and r is positive so f is always increasing. f is continuous on the closed interval [0,1] so by the intermediate value theorem f attains all values from -r to r. So if r is an arbitrary positive real (0,1) could be mapped to any finite interval centered on 0. Is this formulation not strong enough to prove that (0,1) is isomorphic to the set of real numbers? Thanks Dr. Peyam!

[@drpeyam]

That’s just half of the proof, basically the same thing as me showing that (0,1) has the same size as (-pi/2,pi/2). The nontrivial part is the second part, showing that the latter has the same size as R.

[@drpeyam]

That’s the tricky thing about analysis btw, just because it’s true for every r doesn’t mean it’s true for R. For example, any continuous function has a maximum on [-r,r] for finite r, but this property is not true for R.

[@wherestheshroomsyo]

I see, rigor is annoying sometimes. I thought that f being bijective and lim of f(0) -> -infinity and f(1) -> infinity as r -> infinity was good enough, but it needs something more. Can't wait for |R|=|R^n| video!

[@renzalightning6008]

Can you do a video showing examples or even proofs of when composite functions are bijective given the initial functions are injective and/or surjective? i.e. does an injective function composed with a surjective function end up as an overall bijective function?

[@drpeyam]

It doesn’t, unfortunately, take f(x) = e^x, which is injective, and g(x) = x, which is surjective; then g o f is e^x, which is not bijective!

[@renzalightning6008]

@@angelmendez-rivera351 Why not both? XD That's an interesting one though, does the order of functions matter? I'd imagine it does.

[@yugerten_a]

f(x)=ln(-ln(x)) is a bijection from (0,1) to R

[@drpeyam]

Correct but tan is easier

[@МаксимЮрченков-ы5ь]

(2x-1)/(1-abs(2x-1)) is easier, isn't it?

[@drpeyam]

Both ways are equally nice :) Yours might be a bit harder to prove that it’s 1-1, but it’s doable!

[@kamehamehaDdragon]

Epsilon is indeed like casting a spell!

[@random19911004]

A first I thought of just using logistic function (or the inverse of logistic). Was wondering why you went through changing it to (-pi/2 to pi/2). Logistic regression maps (-inf, inf) to a probability that is in (0,1). I guess they are both good.

[@random19911004]

I have a statistics background, not a pure math background.

[@Sam_on_YouTube]

Haven't watched the video yet, but I'm noticing it is 20 minutes long. Can't you do this in 2 minutes by drawing a semicircle of length 1 and projecting rays from the center through it into the real number line?

[@drpeyam]

Yep, that works too! The video would still be 20 mins long, though, if you want to prove everything rigorously!

[@Sam_on_YouTube]

Dr. Peyam's Show If you want to do it rigourously, 2 minutes is too fast. But I don't think it would take that long. You don't need 2 separate functions that way, for example. You don't even need to know the function. Indeed, there are infinitely many of them depending on how far away you place the real number line. You just need to show that as the ray approaches 0 degrees and pi degrees, it approaches positive and negative infinity. Also, it is continuous and increasing. And since the length of the line is irrelevant, you can do the general case off the bat with no need to first do a specific length like 0 to 1.

[@drpeyam]

Yeah, but actually if you think about it, it’s basically the same function, tan maps the semicircle to the whole real line :)

[@Sam_on_YouTube]

Dr. Peyam's Show Yes. The advantage of your method is that you do get a known function out of it. The advantage of the other method is you don't need to. You can just show that a function exists.

[@emanuellandeholm5657]

Just a nitpick, but don't you also have to prove that the relation R(x, y) "x is isomorphic to y" is transitive, ie. that the composition of two bijections is also a bijection.

[@emanuellandeholm5657]

I googled "bijection of bijections" and I found a nice proof that, alas, I cannot fit into this youtube comment. It was like a two page pdf. I really appreciate that once you took a flight to enjoy a torus tho.

[@RyanLucroy]

Well, first I wanted to ask, why you didn't use f(x) = - cot(π*x) till I found out that it's the same as tan(πx-π/2) which is the function you defined at the end :)

[@rizkyagungshahputra215]

aleph 1 and |ℝ|, which one is bigger?

[@drpeyam]

They’re the same :)

[@yuvalpaz3752]

I proved it using limits(i also showed that this is true for all open sets in the real numbers, excluding (a,a)) Let's construct a sequence a_n=(-n,n) (1)lim n-> infinity a_n=the real numbers (2)We can construct the function f(x)=kx+p, thus |(0,1)|=|(p,k+p)|, k=/=0. because k is arbitrary we have |(a,b)|=|(c,d)| for all a,b,c,d in the real numbers and a,b are not the same and c,d are not the same. (3)By (2) i know that |a_i|=|a_j| for all i,j (4)By (1) and (3) we have |the real numbers|=|lim n-> infinity a_n |=|a_i|. Using (4) and (2) again we have |the real numbers|=|a_i|=|(b,c)| which implies that |the real numbers|=|(b,c)| for all b,c in the real numbers and not the same

[@drpeyam]

That’s a good idea BUT the issue is: how do you know that cardinality is preserved under limits? In other words, how do you know that |lim An| = lim |An| ? Stuff could break down at that point, unfortunately!

[@yuvalpaz3752]

the set {|a_n| ; for all n in the real numbers} is closed set, the sequence |a_n| is by definition inside of the set for all n in the real numbers therefore the limit is also inside of the set(i think this is enough to prove that)

[@drpeyam]

Yuval Paz I’m not 100% sure, but consider for example f(A) = 1 if A is finite, and 0 if A is infinite, then f(An) = 1 but f(A) = 0, where An = (-n,n), and A = R, so you can’t really pass to limits like that even if the set of f(An) is closed

[@kamoroso94]

Hey Dr. Peyam, would f(x)=1/x-1/(1-x) work too?

[@MrRyanroberson1]

Begin: let the number 1 be in the set of natural numbers. Let all sums N1+N2 be in the set of natural numbers, for N1,N2 in natural numbers. Let all differences N1-N2 be in the set of integers. Let all quotients Z1/Z2 be in the set of rational numbers, for Z1,Z2 in integers and Z2 not = 0 Let all point values which divide the range of rational numbers into two sets of numbers respectively all greater or all not greater than R be in the set of reals. Let all sums a+bi be in the set of complex numbers, for a,b in the reals, and where i is the "principal" solution to x^2 +1=0, noting that -i should remain distinct from i.

[@eipiwau]

Please use the terms injective for 1-1 and surjective for onto and bijective for both of them.

[@RJYounglingTricking]

Loved it! Ty brother

[@ericlizalde5362]

Love also this type if videos

[@theoleblanc9761]

Easier: Use inverse tanh (For all x in ]0;1[, argtanh(x)=0.5ln((1+x)/(1-x))

[@drpeyam]

Ummmm, not sure if it’s easier...

[@gregorio8827]

tan(pi × X) or something like that isnt it?

[@gregorio8827]

For the interval [a,b] it could be tan(pi ×((X-a)×b)) ?

[@fadhlurrahmanal-akbar7946]

thx for the explanation Dr Peyam (^.^)

[@Hepad_]

Can we say that (0,1) = {floor(x)/x, x€R}?

[@drpeyam]

I think that should be [0,1] because if x = 1/2, then 0 is in your set and if x = 1, then 1 is in your set.

[@Hepad_]

Then what does (0,1) represent? I thought (0,1) and [0,1] meant the same thing

[@willnewman9783]

Hepad (0,1) does not include 0 and 1, while [0,1] does

[@Hepad_]

Oh ok, I've always seen ]0,1[ for 0 and 1 excluded. Different countries, different ways.

[@ekadria-bo4962]

Proof that Cardinality of all dimension is same as (0,1)

[@AvinashtheIyerHaHaLOL]

Enrian Sholeh I want that as well

[@drpeyam]

By that do you mean that the cardinality of Rn is the same as R?

[@AvinashtheIyerHaHaLOL]

Dr. Peyam's Show Yes

[@ekadria-bo4962]

yes, i think will realy cool. i have read some pdf on cantor's proof but i don't got it well...

[@Jonathan_Jamps]

Dr.Payam's...you're awesome and looks like Rajesh of The Big Ben Theory hahaha...

[@吳京-f9i]

Thanks. Saved my ass.

[@youtubercocuq]

parlamışsın peyam

[@StevenPhD4]

Please use a microphone next time!!!