How Big are All Infinities Combined? (Cantor's Paradox) | Infinite Series

[@jorants]

At 10:44 you have not yet proven that the cardinality has to be greater than aleph_0. All you have shown is that it does not behave as the ordinal omega, it might behave as omega+1 in which case its cardinality is still aleph_0. Good video by the way :-)

[@pbsinfiniteseries]

Good point. That was sloppy logic on my part. Dammit! Ironically enough, in the episode 2 weeks from now, I actually make a very similar point to the one you just made... and yet somehow, I let a bonehead thing like that creep into this video. Ugh! Anyway, that's why the audience is so useful here -- thanks for pointing out the logical lapse. I'm upvoting this so that (I hope) other people see it.

[@yuvalpaz3752]

Well, as we don't actually have a set here it is okay(cheat the system)

[@Macieks300]

yeah, I thought it was not right

[@moshesinclair2620]

@@yuvalpaz3752 היי גם אתה רואה את זה

[@4thDimensionalCube]

Mahlo cardinal:M Weakly compact ordinal:W Grahams number:g(64) TREE:TREE(3)

[@YatriTrivedi]

This is probably one of those Infinite Series videos I just can't quite grasp my first few times through - without the follow up. To be fair, I also had a hard time with VSauce's Counting Past Infinity one, too. Which is to say, now I'm having a lot of fun!

[@sulferx6370]

I understand stuff like this and I’m only 11, 5th grade is too easy for me!

[@Rakadis]

*Sigh.* I am going to need more than the beer in my hand to figure this out.

[@JM-us3fr]

That's why mathematicians drink coffee

[@nickellis1553]

Ellobats it's an empty set so empty set so 0 cardinality.

[@jeffirwin7862]

I drank a beer. One beer leads to another. By induction, the whole case was finished.

[@dlevi67]

As per Erdős's definition: "a mathematician is a machine to turn coffee into theorems". Note he never said whether the machine halts...

[@davidwright8432]

A beer in the hand is worth two in the keg!

[@bobfish7699]

It's turtles all the way down..

[@dlevi67]

Ah, but since the turtles are not rational animals, should we assume they are dense and real instead?

[@atomic_soup_juice]

I found 1.4142135623730950488… of a turtle. Where do I go now?

[@lightlysalted7790]

Is that a JOHN Greene reference?

[@Kalumbatsch]

all the way up

[@qabasak]

Brief history of time?

[@MoonDystopia]

So, does the set of all possible infinite sets contain itself? I'm going for a walk..

[@johnkentzel-griffin2855]

Go home, Bertrand. You're drunk!

[@royjify]

first prove that it's a set at all, then we can talk...

[@gregoryfenn1462]

It's not clear that that exists, much like "the largest natural number" does not exist. After all, given such a set S ("of all possible infinite sets"), one could form a set which consists of the union of S and {S}, which would be a new infinite set, meaning that S was NOT the set of all possible infinite sets in the first place.

[@goderator2002]

the set of all possible infinite sets *is* itself. it doesn't need to 'contain' it. what gabe doesn't understand here about unioning sets together, is that unioning a set with it's powerset, does not necessarily create anything new. the power set of A already contains all the members of A, so the union of A with its powerset is exactly the same set as just it's powerset. unioning a set of chain power sets (for example A, P(A), PP(A), etc), all together, is exactly the same size of whatever the largest powerset is, in the chain. if that chain is infinitely big, i don't think this changes anything. i'm not really sure there is a paradox here, just a misunderstanding of how set union functions.

[@lidiaalfaro9354]

No set can have itself as an element

[@antoinebrgt]

There is a slight inaccuracy concerning the order of quantifiers at 8:05 "There is at least one subset of A, namely B, that no function that maps elements to subsets ever outputs . " This sentence as it is, is incorrect, because B depends on the function. Better say "for any function, there is a set B such that ...".

[@pbsinfiniteseries]

That's what I meant in that whole section. In other words, I didn't mean "there is at least one subset of A..." in the mathematical sense of first-order quantifiers; rather, I meant the example here with one A, one B, and one G to be representative, with a tacit "Now generalize to all G", ergo my diction when I say "But we didn't specify what G was" yadda yadda. It's funny, even though I'm now doing a math channel, it doesn't always occur to me that a substantial fraction of the viewership reads things as mathematicians tend to read them. ;) This actually gives me an interesting idea for an episode. There are some fun open-ended questions to which mathematicians/computer scientists tend to give one class of answer while physicists/engineers on the other give different answers, and the nature of that split seems to be feature nontrivially in difficulties that early undergrad STEM students who are of one bent vs the other face in their introductory calc and linear algebra courses. Some very interesting stuff at the interface of math and math pedagogy. Maybe we could attach this to a poll and get a rough snapshot of the Infinite Series audience like this...

[@worldfacade]

I think Scientia Egregia put things too gently. I believe the statements you made will be misleading or confusing to ordinary viewers who try to follow your reasoning carefully (as well as being technically incorrect). It's as if someone said: Pick any integer g. there is an integer strictly greater than g, call it b. Now, since > is antisymetric g is not greater than b. But there's nothing special about g, the same holds for any other integer. So "there is at least one integer, namely b, such that no integer is greater than it." (Exactly parallel to your sentence. Do physicists read that sentence in an innocuous sense?)

[@abdelarmstr5173]

How do you know it exists ? I mean how can you prove it's not the empty set ?

[@totietje]

Abdel Armstr Then every output of G would contain its input, and therefore never be empty - so then the empty set would never be the output of G.

[@1991dmj]

even if it is an empty set, the proof stays correct, because empty set is also an element of P(A).

[@Thee_Sinner]

My guess is that they’re infinitely big.

[@eufalesio1146]

agreed

[@MrBrew4321]

My guess is that you are a troll... but I will bite today. What they? The infinite sets!? Of course THEY are infinite... Lol. The problem here is essentially that some infinites have different sizes and when you ask how many sizes of infinity there are you get a paradox.

[@MrInside20]

And infinitely empty.

[@ilxstatus]

But which infinity?

[@eufalesio1146]

is it infinitely infinite

[@mitchelljacky1617]

It is so good to finally have popular mathematics channels which cater to the more advanced layperson! Thanks guys!

[@elliott614]

Just found this channel and it's like ... FINALLY not being talked down to

[@schlaier]

All I keep hearing is "Olive Knot"

[@dlevi67]

Lucky you. I keep hearing "all of naught".

[@josephcohen734]

"The cardinality of the totality of these cardinalities" Such poetry

[@gibsonman507]

This guy finally helped me understand spacetime. Glad he popped back up

[@Simon-ps3oj]

Well, the number of finite numbers is too big to be finite, so I think that the amount of infinite numbers is too big to be infinite.

[@rmsgrey]

Meaning it's finite? ("infinite" literally means "not finite")

[@JM-us3fr]

lol Interesting reasoning

[@MikeRosoftJH]

You are on the right track. The set of all possible infinite sizes is indeed too big to be infinite. So it isn't infinite; it simply ... isn't. Formally, the class of all cardinal numbers is not a set; it is a proper class. This means that statement "x is a cardinal number" can be defined with a formula, but a set of all cardinal numbers can be proven not to exist; the assumption of its existence would yield a paradox.

[@bobdole69420]

Using the Axiom of choice you can define a cardinal so large that it cannot be reached using any amount of operations on infinite cardinals. These are called Inaccessible cardinals (the same way infinity is inaccessible using any amount of operations on finite numbers). They are the first level of Large Cardinals above infinite. The contraction of U=P(U) is the largest possible cardinal and is written 0=1 as is it so large it is by definition a contradiction. websupport1.citytech.cuny.edu/faculty/vgitman/images/diagram.jpg

[@phoenixstone4208]

The 0=1 marking at the top of that diagram, then, implies the existence of so many things that finite numbers are no longer distinguishable in size? (Edit: that’s my interpretation, at least)

[@RobarthVideo]

In the argument that |A| =/= |P(A)| when contructing the set B, it might have been better to mention, that such a set actually exists thanks to axiom schema of specification. But that might be a wee bit too technical. And with that contradiction, my guess is that constructing "a complete list of all infinite cardinalities" requires unrestricted comprehension and therefore leading to the paradox.

[@kenwalter3892]

Watch the video he suggested that we should watch, if we haven't already. That should probably prepare anyone better for this one.

[@yuvalpaz3752]

Do you know what classes are in set theory? if so, this is the solution to the paradox: A is not a set, it is a class, so the axiom does not let's us construct B.

[@CuulX]

Yeah I feel like there's a part missing where he proves the set B actually isn't empty for infinite sets. What if 0 maps -> to {} and then the next natural numbers (1) maps to all subsets you can make with 1 (injection so skip previously covered subsets). The following unused natural numbers maps to the subsets you can make with {1,2} while not breaking the injection criterion ([2,3] -> [{2}, {1,2}]). After that you take subsets up to 3 with the following natural numbers like so: [4,5,6,7] -> [{3}, {1,3}, {2,3}, {1,2,3}] and so on. This gives an bijection from N to P(N), which proves that |N| = |P(N)|. I realise the flaw in this construction however. No member of N maps to an infinite subset (like {5,6,7,...} or {6,7,8,...}) and I think no mapping could for similar reasons as to why you can't map naturals to reals. I can make a injection from Reals to P(N) and prove |P(N)| >= |R|. Something like this: Take any real number X and a list L = []. If X is positive append 1 to L, otherwise append 0. The number of digits before the decimal point in X is then appended to L. After that take the whole infinite list of digits in X and append them in order to L. Lsum is another infinite list where each member in it is Lsum[p]=Lsum[p-1]+L[p] (out of bounds is 0 for Lsum). This makes Lsum a list of increasing natural numbers from which L can be derived by subtracing the number before it. The list Lsum can be made into a subset of the natural numbers trivially (and converted back to a list trivially and sorted to get back the position information). This algorithm should be able to represent any real number with a set. Is |P(N)| = |R|? I haven't figured out yet but I'll post this anyways and maybe I'll get back to it?

[@nom654]

If B should happen to be the empty set, then that's just fine. It would still be a set in the power set that is not the image of any element.

[@CuulX]

nom654 Can you explain and not just affirm that it is so? B is a set of items not reachable by G. If B is empty then G is surjective unless B is not an exhaustive list of items not reachable by G. So if there are other ways to construct unreachable items and B can be empty then the proof given in the video would still be inadequate. Are you referring to my proof that |P(N)| >= |R| to claim it?

[@jarlsparkley]

0:56 "I know the right way to do things but I'm not going to because the wrong way is easier." Like a true physicist Gabe! ;)

[@11kravitzn]

The definition of comparing cardinalities isn't quite right. There can exist a mapping from A to B that is injective but not surjective when the sets have the same cardinality. For example, the identity map from N to Z is injective but not surjective, even though N and Z have the same cardinality. The important thing is that, if |A|

[@JoshuaHillerup]

Or for example the function f(x) = x + 1 on the natural numbers, which is also injective but not surjective. I think of it as "is there a bijection"? Partly because I still keep mixing up injective and surjective.

[@DrMikero]

Agreed, take f(x)=2x over the natural numbers, which is an injection but not onto. The rules at @3:27 would have you conclude that |N| < |N|

[@JoshuaHillerup]

He said all injective functions would not be surjective, not that there exists such a function (I just rewatched at your timestamp).

[@pbsinfiniteseries]

Hey all, Gabe here. Did I inadvertently misspeak? I kind of thought people would follow that what we're going for is "Can you find an injection that's also surjective, i.e. can you find a bijection?", not "Is every conceivable injection also onto?". But I get how the language in the graphic circa 3:27 could be confusing -- it inadvertently suggests that the cardinality comparison tests involve "check surjectiveness of the specific injective map you found in step 1", as opposed to "see whether *any* injective maps you can find in step 1 are also bijective". Fair point, and I should have been more lexically careful.

[@zairaner1489]

PBS Infinite Series yeah I believe the graphic is the only problem

[@fun_vandanrevanur]

Man such a cliffhanger ! When are you going to release the next episode ?

[@JoshuaHillerup]

Is the "thing" that is all infinite sets not a set maybe, and if you could describe it as having a cardinality greater that the largest possible cardinality of any set?

[@romajimamulo]

Joshua Hillerup I believe this is the problem. How do we know we've hit every infinite set? No matter what we do, we can keep unioning and powersetting and we'll get new ones. So there's no way, even with infinitely many steps to "get everything"

[@JoshuaHillerup]

Romaji same way we do it with creating the set of natural numbers. We introduce an axiom saying it exists. The difference here is the axiom we introduce can't make it be something that has all the properties of sets.

[@romajimamulo]

Joshua Hillerup actually, we only have Aleph Nul existing by axiom. You never actually build infinity, you just say it exists

[@JoshuaHillerup]

Romaji it's called the Axiom of Infinity. Without it (or some other such axiom) you only have finite sets.

[@romajimamulo]

Joshua Hillerup exactly what I'm talking about

[@StefanReich]

12:48 "The next time I see you"? What a euphemism.

[@792p]

The comment section is superb. By the way Gabe, all of us are supporting you guys. You really do us a favour by uploading these videos. I wish you could come back.

[@michaelkindt3288]

@12:25-12:35--It seems kind of like how The number of finite numbers Cab itself be a finite number and so it must therefore be an infinite number, The only problem is what does it mean to go beyond the infinite?

[@livingreason9751]

I believe the phrasing around 1:45 might be a bit misleading. I may be misunderstanding but I think it accidentally implies the existence of a non bijective mapping would prove the cardinalities weren't equal. Ex, N has the same cardinality as Z. But of course if we mappedevery n in N to itself in Z we would have an injection that isn't onto. Really appreciate the channel, great series of videos.

[@diribigal]

Very good episode. And shoutout to the Polynomial Method!

[@yulflip]

At 3:23, is there not a mistake in those definitions? The evens are a subset of the naturals, and there is the obvious injection from the evens to the naturals which is not onto. But their cardinality is the same. Would you not say, rather, that |A| < |B| if for EVERY injection A->B, it is NOT onto ?

[@itamaradoram4395]

In 11:12 as I understand it the set U is used and compared to the former U set even though they are different and contain different sets/subsets

[@mathymathymathy9091]

This is essentially equivalent to the Burali-Forti "paradox", although the Burali-Forti paradox as usually stated uses ordinals. If such a U (the cardinality of all infinities) existed, it would be Aleph(the largest ordinal) and P(U) would be Aleph(some strictly greater ordinal, which is the largest ordinal +1 assuming GCH) and so the supposed largest ordinal cannot exist as it would have to be larger than a strictly greater ordinal. The way to solve it is simply that the "set" of all infinities cannot be an actual set. It's known as a proper class, which is essentially some collection of sets that isn't itself a set and thus it cannot have a cardinality.

[@huysavage6883]

maybe the collection of all infinite sizes is not a set but a class, and we haven't defined cardinality for classes?

[@voteforno.6155]

Huy Savage Yes, in NBG set theory (an extension of ZFC) the paradox is resolved by simply concluding that the class of all cardinal numbers is a proper class.

[@Patsoawsm]

Thank you so much for this comment, you've untied a knot in my mind.

[@wihatmi5510]

This videos answers a question I've had for more than 6 years.

[@Nulono]

What's the difference between the cardinality of a list of sets and the cardinality of a list of those sets' cardinalities? Each set only has one cardinality, right?

[@General12th]

Doctor: "Stick your tongue out." Patient: "Aaaaaaaaaaah-leph naught."

[@JoshuaHillerup]

What I find really fun is that even though the size of the real numbers is larger than the size of the natural numbers, the size of all numbers a mathematician would likely work with (the computable numbers) is the same size as the natural numbers. Although it's unknown if all the constants in physics fall in that set.

[@SmileyMPV]

Joshua Hillerup It's even more crazy. Since any mathematical statement must ultimately consist of a finite list of fundamental logical deductions, of which only finitely many exist, one can deduce there are only countably many possible mathematical statements as a whole. In particular, there are only countably many definable objects as a whole. Even more particular, only countably many real numbers are definable. In measure theory we could literally say "almost all numbers are undefinable" and it would be a well-defined true statement :D

[@JoshuaHillerup]

SmileyMPV what does "definable" here mean? If I say "let there be a distinct funky object with these properties for every real number" I've just definined uncountably many funky objects, no?

[@zammer990]

Any number you give a unique description for, say "the square root of two" requires some finite list of instructions that uniquely define it. So the amount of numbers definable uniquely, rather than in some group eg. the set of real numbers, so there are unaccountably many real numbers for which no definition can be made. blog.ram.rachum.com/post/54747783932/indescribable-numbers-the-theorem-that-made-me is a good blog post that goes into it

[@SuperMerlin100]

Uniquely defined. "square root of 2" is the positive solution to x^2-2=0.

[@SmileyMPV]

Joshua Hillerup Can you be more concrete? Do you mean for example "let a>0"? Because that is not a definition. Also, you can define the set of real numbers, but that does not mean you defined every individual element. Also, a definition must not leave room for ambiguity. For example "define x by x^2=2" is an invalid definition as multible objects satisfy the equation. But "define x as the positive real number satisfying x^2=2" is a valid definition, because it has been proven there is exactly one such x.

[@ReaperUnreal]

My 3rd year university math classes suddenly make a whole lot more sense. I should've spent more time just thinking about consequences of what I was shown. If only I'd had more time.

[@YY-wu7et]

This is like numberphile if they didn't assume we were all 12. You guys deserve more subs.

[@lucanalon1576]

11.03 The strongest version of Choice I've ever seen. You were savage man.

[@michaelsommers2356]

Starting at around 6:20, when proving that |A| < |P(A)|, given the mapping of A to P(A), wouldn't it be enough to just note that the empty set is a member of P(A) but is not mapped to by any element of A? Am I missing something?

[@pbsinfiniteseries]

Someone else asked about this, and here's the reply I gave. I think you're saying "The empty set is a member of P(A) but not a member of A, so there's at least 'one extra' element in P(A), so P(A)'s cardinality must be greater, no?". I'd give two responses: (1) Adding 'one extra' element doesn't necessarily change cardinality. For instance, take the naturals and now just append -1 in there. Does the new set {-1, 0, 1, 2, ...} have a larger cardinality than just the set of naturals {0,1,2,...}? They're actually the me (because you can find a bijection between them). (2) Your argument also breaks down if the empty set is actually an element of A (element, not subset) to begin with. For instance, in the von Neumann formulation of the natural numbers that I discussed in the "What are Numbers Made of" episode, 0 is identified with the empty set {}. So {} is actually a member of N *and* of P(N). Make sense? Demonstrating that |A| < |P(A)| really does turn out to require a more careful argument than the one you asked about.

[@michaelsommers2356]

_"Someone else asked about this, ..."_ Sorry. I looked, but didn't see it. _" I think you're saying ..."_ Yes, basically. _"(1) Adding 'one extra' element doesn't necessarily change cardinality."_ But isn't that how your proof works? You create the subset B to which nothing in A maps to. I realize that your proof is more general, and works for any mapping, not just the one used in the first part of your proof, but is that important? Is it possible for one one-to-one mapping to be onto but another one not? _"(2) Your argument also breaks down if the empty set is actually an element of A (element, not subset) to begin with."_ I don't think so. If {} is in A, then {{}} is in P(A), and, using the mapping from the first part of the proof, {} in A would map to {{}} in P(A), leaving {} in P(A) unmapped-to. Thanks for taking the time to answer, and I apologize for being so obtuse.

[@michaelsommers2356]

+Jonathan Love _"No - because if I pair them up in a different way, I can get a bijection."_ Okay. I guess. It still seems weird that you can have one one-to-one mapping of A to B that is onto, but another that is not onto. But I guess infinities _are_ weird. Thanks.

[@levipoon5684]

Do we need the axiom of choice to pick the sets at 11:03? Is this paradox still valid in ZF?

[@MikeRosoftJH]

You can't use the axiom of choice on the class of all sets of a particular cardinality. But assuming AC, there is a way to get a cardinal number for every set: 1) From the axiom of choice it follows that every set can be well-ordered (so that every subset has a minimum); 2) For every well-ordered set there exists a corresponding ordinal number; 3) A cardinal number is an ordinal number for which there doesn't exist a smaller with the same cardinality (it can be proven that ordinal numbers themselves are well-ordered, so the set of ordinal numbers less than or equal to x but of the same cardinality as x has a minimum). Without axiom of choice, you can't prove that all sets can be well-ordered, but I believe that the paradox still works if you restrict yourself to the sets that can.

[@MikeRosoftJH]

On the other hand, we can skip the problem with cardinal numbers and the axiom of choice, and just use this formulation: 1) Assume that set X contains sets of all cardinalities; meaning that every possible set can be 1-to-1 mapped to some set in X. 2) Take X' = union of all sets in X. 3) Take P(X') = set of all subsets of X. By Cantor's proof, P(X) can't be put in 1-to-1 correspondence with X'; nor with some subset of X', and every member of X is a subset of X'. 4) So the set P(X') can't be put in a 1-to-1 mapping with any set in X, contradicting our assumption; therefore, X doesn't contain sets of all cardinalites.

[@mathymathymathy9091]

At 10:07, you said that the cardinality of U was greater than or equal to the cardinality of any of the sets in the chain. Wouldn't this imply that the cardinality of U is strictly greater than that of any set in the chain (as each set in the chain is strictly smaller than another set in the chain)? Then P(U) would not be necessary.

[@pbsinfiniteseries]

Hmmm. Consider the following finite chain: the naturals, and then the reals. The cardinality of the reals is strictly greater than that of the naturals. But the union U of those two sets is just the reals, and its cardinality would be equal to that of the reals.

[@mathymathymathy9091]

In that specific case, it is true. But the infinite chain is different. Let X be a member of the chain. Then P(X) is also a member of the chain. Since the cardinality of U is greater than or equal to that of P(X), it is strictly greater than that of X. This is true for all X so the cardinality of U is strictly greater than that of any member of the chain. The difference seems to be if the chain has a maximal element. If a chain of infinite sets has a maximal element, the cardinality of the union is equal to the cardinality of this maximal element. Otherwise, the union is strictly larger.

[@joeybf]

Mathymathymathy's argument is correct. In fact, if we assume the generalized continuum hypothesis, the set U has cardinality aleph_omega.

[@pbsinfiniteseries]

No, no, you two are misunderstanding me. I know that +Mathymathymathy's argument is a correct alternate argument that obviates the need for introducing P(U), and it's a good observation. Another commenter made the same observation. Rather, I'm saying that I made a *presentational* choice to use an argument based on P(U) in part because of additional caveats that doing it Mathymathymathy's way would have required me to introduce, e.g. for some finite chains it's a little different, plus some other caveats that I mentioned in a reply to another commenter who made the same observation as Mathymathymathy. After wrestling with different presentations, I ultimately settled on the P(U) way b/c, though still a bit long and not easy to follow in a single viewing, it required fewer caveats to cover objections. And I said the same thing to the other commenter who raised this point (and whose name is escaping me right now). Anyway... know what I mean? For context into my "process", when I put together episodes, my target audience is educated smart lay people (and this includes teenagers) who have some but not necessarily tons of math training and want higher level stuff made accessible via a presentation that touches undergrad level or early grad level material but with a bit more scaffolding than a typical undergrad textbook would provide, just to get them to over a critical activation energy where they can become more autodidactic on the topic. The exception to this is puzzles/challenges, which I just throw out since everyone digs puzzles. My view (and I'm speaking only for myself Gabe here, not putting any words in Tai-Danae's mouth) is that the more expert folks in our audience are here less b/c they will learn stuff (occasionally those in this category all may get a "Huh, hadn't thought of that", but most of what we cover is known material to these folks) and more b/c they enjoy seeing how math they know gets translated to this KZbin format and/or they just enjoy math, even if it's stuff they know and/or they dig the community aspect, in particular helping plug invariable gaps (or correcting our errors) for less knowledgeable viewers. So while I definitely want to keep the math-uminati happy and engaged, and while I rely on their comments to help plug holes and correct lapses, I'm at the same not so much trying to _reach_ you per se. The math-uminati are already steeped in math; reaching not required. So my mindset with the show is to build math appreciation and empower/inspire people who may not otherwise do so to go further with their own math (and related) explorations. Of course, if more expert folks feel I'm off the mark with that audience-targeting philosophy, I'm open to feedback.

[@joeybf]

I just thought you weren't convinced by mathymathymathy's argument. And I will agree with you that the argument for P(U) is easier from a presentational perspective. Thanks for the insight in the show's process.

[@ruskibeaner5983]

for the proof at 7:30ish, why can't the set B be the empty set? ie, what allows us to claim that such a set has elements?

[@MuffinsAPlenty]

B absolutely could be the empty set. The empty set is a subset of A, and hence, is an element of P(A). Regardless of whether B is the empty set or not, B is still an element of P(A) not hit by G, showing G is not surjective.

[@omeryehezkely3096]

Given the set of all positive integer numbers, calculate S - the sum of all the numbers, then S is larger than any number in the set yet it is a positive integer number... It is easy to define non-constructable/non-computable operations.

[@thecaneater]

Yeah.... I'm gonna need you to go ahead and make an ELI5 for this video. That'd be greaaaat.

[@letstalkaboutmath2121]

i loved this episode, good job gabe

[@davidhopkins6946]

Your flaw in logic was that the infinities that you listed (aleph0, aleph1, aleph2, aleph3, and so on forever) formed a countably infinite set, i.e. a set of cardinality aleph0. When you added the extra infinity that was the infinity to describe the cardinality of the power set of the union of all these levels of power sets of natural numbers, you added only one extra infinity that was bigger than all the infinities that you listed, but aleph0 infinities plus one extra infinity is still aleph0 infinities. It's like when Hilbert's hotel was full, and the extra guest wanted a room, and the hotel manager moved everyone down one room and was able to accommodate the new guest anyway.

[@MikeRosoftJH]

I don't follow what you are claiming. Okay, you have a countably infinite set of infinite sets: {Aleph_0, Aleph_1, Aleph_2, ...} What follows from there? And sure, you can take the union of all these sets. That yields Aleph_ω. (It can be easily seen that Aleph_ω has strictly greater cardinality than Aleph_n for any natural number n.) And because for any infinite set there exists a set of strictly greater cardinality, it follows that you can get Aleph_(ω+1), and so on.

[@davidhopkins6946]

@@MikeRosoftJH True. Aleph_(omega+1) is a larger infinity than Aleph_(omega), but it's an addition of only one more kind of infinity, not uncountably many more kinds of infinity. That's why the number of infinities we have after taking the power set of this infinite union is still a countably infinite number of infinities.

[@MikeRosoftJH]

@@davidhopkins6946 All right. But it can be shown that there are more than countably many different types of infinite sets, *and* more than uncountably many different types of infinite sets. Given any set (collection of sets) X, there is a set with strictly greater cardinality than any element of X. Let X be any set. Let X' be the union of all sets in X. Let Y be the set of all subsets of X'. Now observe: X' is a superset of every set in X, and so it has no lesser cardinality than any element of X. And Y has strictly greater cardinality than X', and therefore than any element of X. It follows that no set can contain sets of all cardinalities. Now take any countable set, and assume that no two have the same cardinality (such as {Aleph_0, Aleph_1, Aleph_2, ...}; then this set can't be the set of all cardinalities. That doesn't a priori prove that there's an uncountably infinite set of different cardinalities. But such set can be explicitly given. Aleph is a function which for every ordinal number gives some cardinal number (and cardinal numbers are defined as initial ordinals). Now consider the set of all Aleph_n for n being all countable ordinal numbers. There are uncountably many countable ordinals; so this yields a uncountable set of sets, none of which have the same cardinality. If we take the union of all these sets, we get Aleph(Aleph_1). (This construction depends on the axiom of replacement.) For that, see the video 'How To Count Past Infinity'.

[@__nog642]

You can define cardinalities formally by defining cardinal numbers as sets. Each cardinal can be represented by the lowest ordinal with that cardinality. Each ordinal can be represented by a set containing all previous ordinals. So for all cardinals K, |K|=K.

[@hklausen]

ok, what is the practical use of all these different sizes infinities? Can they be useful in physics or engineering?

[@hklausen]

Many thanks for the enlightenment. For me, the first two types of infinities is important for philosophical reasons :-) And in trying understanding tailor series, the Basel problem and things like that. But I have only taken highschool math.

[@sudharshantr8676]

at 4:22 , it is told that the set algebraic irrationals is the same size as set of natural numbers. I believe that this is wrong, as if there exists a table which maps all irrationals with all natural numbers, I can generate an irrational using the diagonal method which is not in the table.

[@MikeRosoftJH]

There are countably many algebraic numbers. That can be seen: An algebraic number is a solution of a polynomial equation with integer coefficients. There are countably many polynomials (each polynomial has a finite degree, and so it is defined by a sequence of finitely many integers; and it can be shown that there are countably many finite sequences of integers). And each polynomial equation has finitely many solutions (for a polynomial of degree n there are no more than n solutions). You can apply the diagonal procedure to a sequence of algebraic numbers, and get a number not in the sequence. But if the sequence contains all algebraic numbers - and such a sequence exists - then the resulting number cannot be algebraic.

[@onuktav]

brain... hurts...

[@matt-stam]

Watching this channel makes me feel like an idiot.

[@LudwigvanBeethoven2]

Brain farts

[@t3st1221]

It's exactly the same issue you get when looking back at "the biggest positive integer" (biggest element of N). They are different integer of different "bigness" (=some are bigger than other) If n is an integer, then I can get a bigger one by doing n+1 If I take all the integer I know, I can still construct a bigger one by taking the biggest and adding one So if I try to get "the biggest integer" I just get nonsense because from the set of all integers, the biggest integer is in it but I can still build a bigger one. The only answer is that there are no bigger integer with just these axioms.

[@vexrav]

What is the cardinality of B if G is infinite, I believe that it must be strictly larger than G, but could be less than or equal to P(G). is there a class of infinite cardinality between G and P(G) where the size of B lies?

[@jacksparrow440]

My best guess on what's going wrong : Let E be a set. It turns out that P(E) contains E: E is an *element* of P(E). However, when trying to take their union, you don't get anything different from P(E) itself: every single element of E is already contained in P(E). Thus, the union of all the powersets is already on that list. So at the limit, the union U would be the omega_0-th set from the list (omega_0 being the first infinite ordinal).

[@MikeRosoftJH]

Careful: the union of all sets in P(E) is precisely the set E. But we don't do a union of all sets in P(E); we do a union of all sets in E and then do a powerset (set of all subsets) of the result. Of course, Aleph(Aleph-0) is not the greatest cardinal; the set of all subsets of this set has even greater cardinality (by Cantor's theorem).

[@cpttomdodge3596]

It’s the All. To count you have to be an observer, you’re trying to observe yourself. It’s pure infinite consciousness. No matter how hard you look it with continued to paradox itself. We lack the language capabilities to reach understanding.

[@oida10000]

Question: Is there a way to rule out that B is just the empty set and we could get |A|=|P(A)-emptyset|?

[@lucasmoser7483]

oida10000 If B were empty then every element would be mapped onto itself, thus the injection wouldn‘t be onto, since P(A) contains elements that aren‘t in A

[@therealAQ]

7:23 to be in B or not to be in B?

[@TGC40401]

When I extend the resolution of the paradox inwards, I would expect Natural numbers to be a non-onto subset of Integers. I'm imagining: #N U #{-1,-2,-3,...} = #I * U #{0} applied as necessary Additionally, what happens when you map the Integers to the power set of the natural numbers? Sadly, my understanding of mathematics is less complete than I wish it to be.

[@grantweekes3280]

After a little consultation to my old Michael J. Schramm, I think the contradiction presented at the end (edit: i.e. Cantor's contradiction) is tied to Russell's Paradox. Just as R={S s.t. S not in S} is not a set in traditional set theory, so is X={the collection of all cardinalities} since both the statements "Cardinality(Union(selections from X)) is [not] in X" lead to a contradiction. Looking forward to the next episode!

[@MikeRosoftJH]

Cantor's paradox is a stronger result than Russel's paradox. Russel's paradox only proves that there can't be a set of all sets (i.e. given any set X, it is possible to create some set Y which is not in X); Cantor's paradox proves that no set can contain sets of all cardinalities (i.e. given set X it is possible to create set Y which has a strictly greater cardinality than any set in X), from which it immediately follows that there can't be a set of all sets.

[@grantweekes3280]

Thanks for the clarification and confirmation that this is an issue of "This kind of set actually can't exist! [in traditional set theory]" I was looking for a problem with the logic when applied to a set of all cardinalities or defining a set like that for example, but couldn't find one and figured an analogy to Russel's seemed likely. What exactly do you mean by Cantor's paradox is a stronger result than Russel's? Usually I would interpret this to mean "Cantor's implies Russel's, but Russel's does not imply Cantor's", but neither works with the ZF axioms so either could imply anything. Is one "stronger" in the sense I am bringing up if you work with some alternate (sensible) set of axioms?

[@MikeRosoftJH]

Once again: from Russel's paradox it follows that given any set X there is some set Y which is not in X (i.e. the set of all sets cannot exist). From Cantor's paradox it follows that given any X set there is a set Y with cardinality strictly greater than any set in X (i.e. no set can contain sets of all cardinalities). Of course, such a set can't be in X, because every set has the same cardinality as itself (or, using the negative formulation, the universal set would contain sets of all cardinalities, because it contains all sets, so Cantor's paradox precludes its existence as well).

[@jasonschuchardt7624]

Noticing a major issue with the statements of the cardinality comparison rules. They say |A| B and |A| < |B| if the injection is not onto. That should be |A| < |B| if *every* injection/function from A to B is not onto. Otherwise you have |naturals| < |integers|, since the natural map is not surjective, which is not right.

[@docopoper]

Wait, what's to stop B being the empty set? Surely the empty set would have to be possible as the power set of a single element set or the empty set would be itself, right?

[@pbsinfiniteseries]

If you're talking about the subset B from 6:51, and you want to handle the case where the original set A contains only element -- say, A = {x} -- then B could very well be the empty set, and that's ok. Watch... a singleton set like A has only two subsets: A itself, and the empty set. So P(A) = { {}, {x} }. A function G from A to P(A) either maps x to {} -- in which case B would have to be {x} (yes, same as A in this case, which is cute) -- or it maps x to {x} -- in which case B is in fact the empty set. Either way, though, the map G isn't onto, so you can't have a bijection b/w A and P(A).

[@docopoper]

Oh right, ok. Thanks for clarifying. I honestly just missed that the empty set was part of the power set.

[@IlTrojo]

Just a little nitpickery: at 08:10 it looks like the B set is the same whatever G is, while it rests on G itself for its own definition.

[@abdelarmstr5173]

IlTrojo yes B depends on G

[@pbsinfiniteseries]

Yeah, this whole animation was supposed to be just one representative example, not a depiction of the general case for all G and all A.

[@sebastianelytron8450]

Cantordinal numbers be transfinite though?

[@twistedsim]

is that a pun?

[@sebastianelytron8450]

is the pope catholic?

[@twistedsim]

Is the cat alcoholic?

[@fullfungo]

In 9:23 you we’re trying to order cardinalities of P(...N...) which correspond to some א’s and you show that we have a set P(U) that has a bigger cardinality (א_ω). So we have a number ω which differs from any natural number (not necessarily bigger). We can conclude that we are using ordinal as indices. Then you ask a question about the cardinality of the set of א’s. But it has to be the same as cardinality of the set of ordinals which doesn’t exist. That’s why there’s no such thing as the set of all cardinals (and in particular infinite ones)

[@pbsinfiniteseries]

Small comment (and this is something that I'll briefly touch on in my next episode in two weeks) -- technically, the cardinalities of the P(...N...) *aren't* the א’s , which are indeed defined by cardinal assignment (defining cardinality in terms of minimum compatible ordinal / order type). The (generalized) continuum hypothesis states that corresponding entries in the sequence of P(...N...)'s and in the sequence of א’s all agree (i.e. that those are the same sequence), but working as if they're different sequences turns out also to be compatible with ZFC. Fun fact.

[@pongesz2000]

moreover, it is far from trivial, that if you have f:X->Y and g:Y->X injections than there exists a X->Y bijection (in other words if you have an injection from A to P(A) and not have a surjection than A

[@pbsinfiniteseries]

Agreed, and I didn't want to get into Cantor-Schroder-Bernstein stuff in this video, which is why I stuck to speaking just about maps in the A-to-B direction. As for your other comment, other people have also raised it over the past few days and I've replied there -- I did not mean to suggest that, once you've found one injection f (which establishes |A|

[@enlightedjedi]

Best topic since infinite series birth. One virtual beer for everybody involved :)!

[@colinjava8447]

Gonna have to watch the 2nd half again, I always assumed it was just aleph0 infinities (based on CH being true). Hard to take in one go, amazing though, maybe I can figure it out, or maybe not!

[@PennyAfNorberg]

I think... we might need some set which are greater than |A| but less then |P(A)|, ie contium since |p(a)|= 2^|a| for finite set, perhaps a smaller function could be used.

[@MikeRosoftJH]

This is the generalized continuum hypothesis. In general, it cannot be proven that such a set (given an infinite set A) exists; on the other hand, it cannot be proven that such a set doesn't exist.

[@macronencer]

10:30 I knew it was gonna get this crazy when you started :)

[@JxH]

@11m41s: It's like kids... "Name a number bigger than infinity." "Infinity + 1 !!"

[@MikeRosoftJH]

Great, we are now talking about Dedekind-finite infinities (an infinite set which doesn't have a countably infinite subset).

[@r75shell]

10:30 There is at least one infinite cardinality, the size of P(U) that our original list missed, which means that list wasn't complete, and the cardinality of the true collection C of all infinite cardinalities, whatever that is, has to be grater than aleph naught." Nope. countable list + one element in it, is still countable. Aleph naught plus one element is still aleph naught.

[@MikeRosoftJH]

But the same argument can be used on the set with that additional element added. It can be easily shown that given any set U, there exists a set with a cardinality strictly greater than any member of U. * Take any set U. I claim that the set does not contains sets of all cardinalities; in other words, there exists a set X for which there does not exist a 1-to-1 mapping between X and some element of U. * Take U' = union of all sets in U. * By Cantor's diagonal proof, P(U') (the set of all subsets of U') has a strictly greater cardinality than U'. (There cannot be a 1-to-1 mapping between P(U') and U' or some subset of U'.) * Every element of U is a subset of U'. * Therefore, P(U') has a strictly greater cardinality than any element of U. * Therefore, P(U') is not in U. (Otherwise, I would have to conclude that there is no 1-to-1 mapping between P(U') and itself; of course, such a mapping indeed exists: x -> x.) * Therefore, there doesn't exist a set containing sets of all cardinalities (and indeed, there doesn't exist the set of all sets).

[@r75shell]

You see, this is completely different proof. Anology to proof in video is similar to situation if you would proving that real numbers are uncountable just by selecting some particular countable set and finding real number which is missing. { 1/n } for example. Valid proof instead saying that for *any* countable set of real numbers, some real number is missing. Also, stressing out "has to be grater" is making viewer to assume the reason is lack of missing cardinal, instead of contradiction with definition of C - set of all infinite cardinals. Proof in video: we don't know whether guessed set is what we looking for, and what size of it. Proof above: we assume that set is what we want, and have contradiction.

[@MikeRosoftJH]

I'd say that the proof in the video is not wrong, just incomplete. The video correctly proves - using essentially the same argument as I did - that {N, P(N), P(P(N)), ...} is not the set of all possible cardinalities. He then says that from this it follows that the set of all cardinalities (with an unsaid assumption: "if it exists") cannot be countable. This is correct, but it misses one step; namely, that the same proof can be applied to any countable set (and indeed to any set). But he does finish the proof by taking any set U (this time making no assumption about its cardinality or contents), applying the same argument, and reaching a conclusion that the assumption that this is the set of all cardinalities leads to a contradiction. (As I have demonstrated, we don't really need to assume that U is the set of all cardinalities, because the conclusion that it is not will directly follow from the construction.)

[@r75shell]

"is not wrong, just incomplete" Incomplete may be assumed as wrong, but it is controversial. If incomplete is considered to be ok then "how much incomplete is ok?" - also controversial. If "in the end" proof is right, I still don't like mistakes in "draft" proofs in between. Mistakes for me: conclusions that is true but wrong reasoning, simply wrong reasoninig, wrong assumtions etc. Tai-Denae in last episodes just keeps saying "in fact THAT'S TRUE" - and, no need explanations :D

[@davidwilkie9551]

Singularity intersection-connection of infinities requires a total unique infinity. (?) This lecture is an observation of the topological significance of the integration of sets of time rates, typical of a spectrum analysis, or reading pulses of resonance within pulses, etc etc, the natural sequences of dimensional everything.., mathematically speaking.

[@FrankAnzalone]

Where can I find a list of the hollowed out n r and which means what

[@efulmer8675]

N, R, and a couple others are the first letter of the set that they are named for: N for Naturals, R for Reals, C for Complex numbers. There are others that have to do with the word in German: Z for "Integers" I think is this way where the word for "integers" in German starts with a Z, so the convention stuck even though it is no longer a nice rule like "N for Naturals".

[@FrankAnzalone]

@@efulmer8675 thanks

[@beenaalavudheen4343]

Could have included the Schroeder-Bernstein theorem but amazing video!

[@pbsinfiniteseries]

Originally had it. Ended up in the script jetsam. As-is, this video ended up longer than we can normally accommodate given the budget.

[@beenaalavudheen4343]

PBS Infinite Series I thought that was the reason. Can't wait for the next video!

[@felipem4900]

@@beenaalavudheen4343 where's the follow-up video? I've been searching for it

[@felipem4900]

@@pbsinfiniteseries fantastic video, please share where's the follow-up, thank you

[@trevormarshall2817]

ayy its old spacetime dude. Good to see him again

[@gunhasirac]

For anyone didn’t get the second part of power set has larger cardinality than itself, here’s my explanation. (His wording were quite hard to understand even I know the proof.) The way the subset B is chosen is known as “Cantor’s diagonalization”. It has became a standard proof technique. You can find this process in proving real number is uncountable, and similar arguement in proof of Arzela-Ascoli Theorem. In this case, it is a systematic way to find a element (sbuset) that is not on the list. Starting from A with size 3. Let A={1, 2, 3}, and f : A→P(A). Say f(1)={1,3}, f(2)={}, f(3)={1}. For clarity let’s express this in the following way. f(1)={1, ,3}, f(2)={ , , }, f(3)={1, , }. Now to construct the subset B, we look at the diagonal of the right part of the array. If the element is in the diagonal, we exclude it. If it is not, we include it. In this case, we end up with B={2,3}. Note that there’s no element in A that can map to B through f. No matter what f you start with, you always end up with a subset B that is not on the list, since B is differ to any f(x) by at least one element. Now for a general set A, finite or infinite, countable or uncountable. We choose B in similar way. Given a function f : A→P(A) (remember P(A) is a collection of sets, hence values of f are sets), if x is an element of f(x), then we exclude x. If x is not an element of f(x), we include it. Then there doesn’t exist x such that f(x)=B, since B is differ to f(x) at x. Hence whichever f we choose, we will always miss an element in P(A), and f is never surjective.

[@vojtechsejkora1554]

At 6:53 you select some subset and than arguing about it. But you do not proove existence of that specific subset. If I recall correctly this kind of process lead to braking math. All it start with simple set - lets have set, which contains every set, which do not contains itself. For that reason we cannot create sets from nothing, but we can select from existing set some another set with property \fi. But it has to exist in original set.

[@MikeRosoftJH]

This set is guaranteed to exist by the axiom of separation. The proof that no set can be mapped one-to-one with the set of all its subsets goes as follows: * Let A be any arbitrary set. P(A) is the set of all subsets of A. * Let f be any function from A to P(A). (For any x∈A, f(x) is a subset of A, and exactly one of the two is true: x∈f(x), or x∉f(x).) * Let D = {x: x∈A and x∉f(x)}; in other words, D is the set of all elements x of A, such that the function f maps the element x to a subset of A which doesn't contain the element x. This set exists by the axiom of separation. * By construction, D differs from f(x) for any x∈A. Let x be any element of A; then if x∈f(x), then by construction x∉D, and if x∉f(x), then x∈D. Either way, f(x) differs from D by (at least) the inclusion of x itself, and so f(x) and D cannot be the same set. * Therefore, for no element of A is f(x)=D. In other words, the function f does not cover all elements of P(A) (in particular, it doesn't cover the set D). * Therefore - because I haven't made any assumption about the function f (or the set A) - the same holds for every set and every function from it to the set of all its subsets. QED. What doesn't exist is the set of all cardinal numbers; no set can contain sets of all cardinalities. Given any set A (i.e. set of sets - in set theory all objects are sets), there exists a set B with strictly greater cardinality than any element of A.

[@sp17studioo]

3:09 Isn't it true only for finite sets?

[@badhbhchadh]

11:00 How do we now that such a set does always exist?

[@MikeRosoftJH]

By axiom of separation. (It might be an empty set, or it might contain all elements of the original set.)

[@twentytwentyoneishvkmemory7430]

So if you use "naught" or "nol" to say the 0, is it ok to say "zero"? I'm a bit new to this infinite stuff

[@pbsinfiniteseries]

It's less common in this context, but honestly, I think you can say whatever you want, as long as the meaning is clear.

[@JorgeMartinez-dp3im]

New to the channel. Very interesting topics. Plus Tai-Denae is suppa cute. 😁

[@AstralStudioGames]

also, what about negative aleph null, would it be called aleph -1?

[@MikeRosoftJH]

There's no such thing as Aleph(-1). Aleph-0 is the set of all natural numbers, and it can be proven that it is the smallest infinite set; any set with a smaller cardinality must be finite. In particular, any subset of the natural numbers is either countably infinite (and has the same cardinality as Aleph-0), or finite. (The function Aleph(x) is only defined for x being an ordinal number, and -1 is not an ordinal number.)

[@rkpetry]

*[**05:50**] "distinct", meaning what for infinity which is an open-definition: there is 'any' but no 'last', "each"....* *[**06:44**] but your original G: x→{x} had-no-exceptions... so G: (?)→{}*

[@pbsinfiniteseries]

To answer your 1st question, "distinct" for two sets (infinite or otherwise) means "not having an identical roster of elements". 2nd question -- in step 1, we just show that an injection exists, period, in order to establish |A| {x} for that purpose. Now, in step 2, we throw away that G and start over, asking "among all possible injections G, are any of them onto?" and show that the answer is 'no'. Now, even though it has no bearing on the argument in step 2, you are correct that the function G from step 1, i.e. G:x --> {x} does not map anything to the empty set, i.e. {} is not in the image of G:x --> {x}.

[@rkpetry]

*PBS Infinite Series Okay, I can use that definition, but, it does recall two classical dilemmas, 1: X=0.999... ≈ Y=1.000... but-there-is-no-last-digit, nor infinitesimal-carry, to make X,Y distinct; and 2: if we take the-one-distinctive element from an arbitrarily-scrambled infinite set and throw it back in, we may never see again because there-is-no-last, (in fact we may never see 99% of infinitely-many)-so, whatever we do in an infinite set, changes 'the rules' of what we're trying to prove... **_Now back to infinitely many distinct finite, subsets, except of course infinitely many of them have only finite-fewer elements and are therefore maybe-in/distinct..._*

[@pbsinfiniteseries]

1) Remember, the definition of "identical" I gave is specific to _sets_ , whereas you're talking about decimal representations of real numbers. 2) Again, when you talk about "seeing it again", you seem to be describing a physical process of procurement. This is math -- if the element is in the set, we can talk about selecting it from that set.

[@rkpetry]

*Yes, [KZbin/IE11 CRASHED] 1. **_infinitely-'distinct' elements to make infinite subsets 'distinct';_** 2. **_finding by iterating a cardinal index e.g. x→{x}; Sets are by Rules, and by Rosters, but we've not-yet secured the Rules (ibid, 'arbitrary' throwback is a 'rigorous' attempt to untie some rule(s)) in 'no-last-selection'..._*

[@youtubeuser8232]

This video is excellent! Maybe too good for KZbin!

[@voteforno.6155]

A closely related result for ordinal numbers is the Burali-Forti paradox.

[@gorgolyt]

These videos are usually great, but surprisingly this one is wrong is some quite basic ways. At 3:22 the second claim is wrong. If an injection from A to B is not onto, that doesn't prove A is smaller than B. Consider for instance if A was the natural numbers 0, 1, 2..., and B was also the natural numbers, and consider the function from A to B which adds 1 to a given number. This function is injective, and it is not onto (because 0 is not mapped to), but clearly this does not prove that A is smaller than B; A and B are the same and thus the same size.

[@rb1471]

So we assumed the cardinality of U is somehow greater than or different than the cardinality of any other power set of A. I believe the cardinality of U is equal to the cardinality to the "largest" of the sets in your list, so taking the power set of U just gives us the next element in the list. It's similar to the idea that inf + inf = inf. If inf1 > inf2, then inf1 + inf2 = inf2. So the cardinality of the union of all power sets of A is equal to the cardinality of the largest power set of A.

[@MikeRosoftJH]

This is not necessary the case for an infinite set. Take ω, the set of all natural numbers. (In set theory, a natural number is defined as a set of numbers less than itself, so 0 is the empty set, 1 is {0}, 2 is {0,1}, 3 is {0,1,2}, and so on.) The powerset of each element in ω is finite, but the union of all powersets of the elements in ω is countably infinite (in fact, it is precisely ω), and the powerset of the union of all sets in ω is uncountable infinite (it is precisely P(ω), because the union of all sets in ω is precisely ω). We don't assume anything about the cardinality of sets in U. The proof is the following: * Take any set U. I claim that the set does not contains sets of all cardinalities; in other words, there exists a set X for which there does not exist a 1-to-1 mapping between X and some element of U. * Take U' = union of all sets in U. * By Cantor's diagonal proof, P(U') (the set of all subsets of U') has a strictly greater cardinality than U'. (There cannot be a 1-to-1 mapping between P(U') and U' or some subset of U'.) * Every element of U is a subset of U'. * Therefore, P(U') has a strictly greater cardinality than any element of U. * Therefore, P(U') is not in U. (Otherwise, I would have to conclude that there is no 1-to-1 mapping between P(U') and itself; of course, such a mapping indeed exists: x -> x.) * Therefore, there doesn't exist a set containing sets of all cardinalities (and indeed, there doesn't exist the set of all sets).

[@rb1471]

Well I'm mostly confused about this: The question itself asks to assume (or is shown that) P(U) has a different cardinality than any of the power sets of A. And then we determined this was weird because the list of power sets of A was thought to be a complete list of infinities. So assuming P(U) has a different cardinality, this would imply U has a different cardinality than any power set of A. Proof would be by contradiction: * Assume U had the same cardinality as one of the power sets of A, * Call it P'(A) where P' is one of the power sets of A (like P(P(P...(A)))). * Then P(U) would have the same cardinality had P(P'(A)). * But P(U) must have a different cardinality ## * So our assumption is wrong about U having the same cardinality as P'(A) So I'm back to U having a different cardinality than any power set of A, yet how can we assume this is true in the first place like I inquired about in my original comment?

[@MikeRosoftJH]

Again, we don't need to assume anything about the cardinality of U. The proof in the video is just a specific example of the proof I have given above. Take U as the set {N,P(N),P(P(N)), ...}. Take U' as the union of all sets in U. It can be shown that P(U') has a strictly greater cardinality than U'; therefore, it also has a greater cardinality than any subset of U'; therefore, it has a greater cardinality than any element of U (because U' is the union of all sets in U, so any element of U is a subset of U'). I am not sure what you don't understand - is it the concept of a proof by contradiction itself? (The proof that there can't be a set of all cardinalities is not really a proof by contradiction; I can directly show that for every set U there is a set with a greater cardinality than any element of U.)

[@b3z3jm3nny]

How do we know there even is a subset B for an arbitrary function G? If G just maps each element to a singleton just containing that element, as was just proposed, then how can there be a B? Every element is a member of the subset G maps it to, right? What am I missing that guarantees the existence of B? EditL oh, yeah that makes sense -- thanks

[@willnewman9783]

This means that no element maps to the empty set, so the map is not surjective

[@greifhippo]

In that case B is the empty set that is also contained in the power set

[@abdelarmstr5173]

But what if A contains the empty set ? Wouldn't then B be defined as the empty set which is already linked to {} in A ? Or wait, is {} distinct from {{}} ? If so, never mind :)

[@greifhippo]

Abdel Armstr Yes, {{}} and {} are two different sets

[@bautibunge737]

Suppose B={a€A/a€\G(a)} (where €\ means it's not in). Suppose G biyective and B=Φ . Take a€A, b€A then you have the elements of P(A) {a}, {b}, {a,b} if G(a)={a} and G(b)={b} there exists c€A such that G(c)={a,b} and then c€B. Suppose now that G(a)={a} and G(b)={a,b} then there exists c€A such that G(c)={b} so c€B so B=\Φ

[@eHcOZaX]

a set containing all possible set is larger than all of his subsets, but a set containing all possible set must contain itself, and has therefore a cardinality larger than itself

[@elliott614]

Whats the cardinality of the set of all sets that don't contain themselves

[@MikeRosoftJH]

Undefined - in ZF or ZFC there isn't a set of all sets (or set of all sets which don't contain themselves). It's a proper class, and so it's in a sense bigger than any set.

[@amine2684]

Does the theorem " A A

[@MikeRosoftJH]

Define: Set A has the same cardinality as set B, if there exists a one-to-one mapping between them. Set A has a no greater cardinality than set B, if there exists a one-to-one mapping between A and a subset of B. (Note that any set is a subset of itself.) Now the following can be proven without axiom of choice: If A can be mapped one-to-one with a subset of B, and B can be mapped one-to-one with a subset of A, then sets A and B have the same cardinality (there also exists a one-to-one mapping between them). Can the cardinality of any two sets be compared? That is: is it true that given any two sets A and B, either A has no greater cardinality than B, or B has no greater cardinality than A? (In other words: either A has a strictly lesser cardinality than B, or the same cardinality as B, or a strictly greater cardinality than B.) Assuming axiom of choice, yes; even more, the proposition is in fact equivalent to axiom of choice.

[@bmbirdsong]

I've always thought of it in terms of density. The infinite set of whole numbers is very dense, much more dense than the set of prime numbers, but not as dense as the set of irrational numbers. Density being a measurement of the average distance between two adjacent members of the set.

[@eggynack]

What's the density of the rational numbers? How does that density compare to that of the real numbers?

[@bmbirdsong]

Oops! I didn't mean to use the term irrational numbers in my first comment. I meant fractions. The set of fractions is much more dense than whole numbers. Sorry about the confusion.

[@eggynack]

The question wasn't really related to that. My issue is with the general construction that is density, because it has no means of making a distinction between rationals and reals.

[@MuffinsAPlenty]

I think you're using the word "density" to talk about how "tightly packed" certain sets are vs. how "spread apart" other sets are. In this sense, density is a relative term. In order for it to make sense, you need to be working in a universe (i.e., every set you're dealing with is a subset of some set U - so you're always working inside U) and that universe has to have some sort of topology or measure on it. To describe the density of a subset S of U, you would need to compare S to U itself. When people bring up the notion of density when talking about comparing sets, they are usually only thinking about subsets of the real numbers. So you're comparing each subsets to the real numbers in order to think about density. But the goal here is to be able to compare arbitrary sets - for any two sets S and T, how do they compare? We don't just want to compare subsets of the real numbers. Here, your notion of density breaks down. There is no universal set in which all sets are contained. So we can't compare each set to a universe. For example, consider the power set of the real numbers, which I will denote P(R). This is the set of all subsets of the real numbers. So one element of P(R) is the set of irrational numbers. Another is the set of natural numbers. Another still is the set {π,e}. Now, we can look at subsets of P(R). For example, you could let S be the subset of P(R) which consists of all elements of P(R) whose members are all irrational. In other words, S is the set of subsets of R in which every element of each subset in S is irrational. Now, how tightly packed is S in P(R)? I have no sort of intuition for how any such notion could/should be defined.

[@baburamprasad926]

Welcome back..!

[@mindstormmaster]

In terms of Cantor's theorem - can someone explain to me why the first part of the proof isn't sufficient? If you have a function G that maps each element of set A to the subset of A which is just that element, then you already have at least one subset of A (the subset of every other element, for example) to which G doesn't map an input (presuming A is an infinite set). Is the second part even necessary?

[@pbsinfiniteseries]

Because it doesn't suffice to show that a single injection isn't onto -- you have to show that no conceivable injection could be onto. There was some confusion about the criteria summarized at 3:22 , with people thinking you only have to test the single injection you've stumbled on for establishing |A|

[@lyrimetacurl0]

I thought of another infinite set that I'm not sure what the size is: "The set of all possible type matchups". Because for every new type, there needs to be as many new types added that include that type, as the total number of types already is. So the set infinitely explodes.

[@__nog642]

2:42 No, just because there is an injection A -> B doesn't prove that |A| < |B|. You also need to show that there does not exist a bijection A -> B. Otherwise we could prove that |N| < |Z| < |Q|.

[@ativjoshi1049]

In step 2, what ensures that such a set B exists. In other words, what prevents me to make a similar argument as in step 2 for two infinite sets with same cardinality (e.g N and Z).

[@alissonalmeidabueno250]

very good project

[@YYYValentine]

Wow a mind blowing video in my favourite topic :) power sets of power sets. When Vsauce was good, it made an also awesome video about infinity. I am looking forward to your next video!

[@FistroMan]

I can't see which is the next video... but is so funny. I have SO MANY answers for THIS video...

[@PhillipChalabi]

Miss you on Spacetime, guess I need to subscribe here as well....

[@gregoryfenn1462]

Ay 10:30 we show that |P(U)| is a strictly larger cardinatlity than any in the list "|N|, |P(N)|, |P(P(N))|, ...", where U is the union of {N, P(N), P(P(N)), ...}. This is fine. But we could have easily just shown that |U| itself is a strictly larger cardinarilty too, than any in our first list. This is because, for any S in U, P(S) is in U too, which implies the cardinality of U is strictly greater than S.

[@pbsinfiniteseries]

True, but I think the argument is a little bit slipperier and has more caveats, ergo I opted against it for presentation purposes. Here's what I mean. To lay it out that way, we'd need to remember that things in U aren't the sets N, P(N), P(P(N)), ... per se -- the things in U are all the *elements* of all those sets. Now in the late 19th century, elements of N aren't necessarily thought of as sets in the Zermelo or von Neumann sense -- the math world isn't there yet. But elements of all the power sets are of course sets, so say S is one of those elements of a power set. That means S is a subset of one of the previous sets in the chain. Since every set is a subset of itself, you could always pick S to be one of the actual power sets in the chain. Carrying the above argument forward one extra step, then P(S) is in U as well, which means |U| >= |P(S)| > |S|, where the 1st inequality is due to the rule that A subset-of B ==> |A|

[@gregoryfenn1462]

You're right! Your approach makes for much smoother teaching ^_^

[@abdelarmstr5173]

You defined the subset B as such. But how do you know it exists ?

[@pbsinfiniteseries]

This has been asked by a couple of other commenters, and I gave some detailed answers there (as did some other views). Browse the comments -- you'll find it.

[@abdelarmstr5173]

Found it. Thanks.

[@Uriel238]

I got lost at the point where *B* is defined as all elements of *A* that are not members of the subsets to which each one is mapped by the function *G* At what point is it made certain such subsets exist? Why can not *¦B¦* = 0?

[@pbsinfiniteseries]

Others have asked this same question, which I answered on other comments, but in short... there can certainly be functions G for which the subset B ends up empty. That doesn't mean it doesn't _exist_ -- it just means it's empty (the empty set is a perfectly valid entity in set theory, and it exists). And if B turns out to be empty, the argument in Cantor's theorem still works just fine (try working it out, and consult my answers to others' similar comments if you get stuck).

[@jeffirwin7862]

What about the power set of the empty set? Isn't it just the empty set? Their cardinality is the same.

[@voteforno.6155]

Jeff Irwin The power set of the empty set is the singleton set { { } } that has just the empty set as its only element. This set is not empty and fits the rule 2^0 = 1.

[@General12th]

VSauce mentioned an "aleph fixed-point", where the cardinality of the number aleph(aleph(aleph ...)) -- an infinite chain of alephs -- was equal to the cardinality of the next number in this list, aleph(aleph(aleph(aleph ...))) -- the same infinite chain but inside one more aleph(). But since both chains are infinitely long, and aleph(0) is indistinguishable from aleph(0) + 1, that means these two infinities are the same size and also not. Was Michael completely off his rocker? (Or did I explain it horribly?)

[@General12th]

It's a function, W, such that W(N) = aleph(N). Suppose N is zero, then W(N) = aleph-null, and W(W(N)) = aleph(aleph-null). We know W(0) is greater than 0, and we know W(W(0)) is greater than W(0), then that's proof by induction. Therefore, if we iterate W -- meaning we take the output of W(N) and put it back into W, then take *that* output and put it into W, and so on -- we will always generate an aleph whose cardinality is larger than the aleph that came before it. (If I'm misusing terms and sounding like an idiot, it's because I don't have a math degree, and maybe I should shut up forever and leave this stuff to the experts. Anyway.) Now we get to VSauce's video on fixed points. Suppose we iterate W infinitely: W(W(W ... )) -- an infinite chain of W's. Let's call the output X. Now suppose we iterate W one more time: W(X) = W(W(W(W ... )))). We add one more W to the chain of W's. But that chain is infinitely long. Infinity plus one is still infinity. Therefore, we can replace the current output, W(W(W(W ... )))), with the previous output, W(W(W ... )), and have the same result. But that would mean X = W(X). Granted, my proof of induction was not at all rigorous, but I can't imagine any point where N isn't less than W(N), until this fixed point. Is there any meaning here? Or am I just being stupid?