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?

[@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

[@josephcohen734]

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

[@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

[@schlaier]

All I keep hearing is "Olive Knot"

[@dlevi67]

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

[@gibsonman507]

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

[@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! ;)

[@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)

[@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

[@StefanReich]

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

[@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).

[@wihatmi5510]

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

[@fun_vandanrevanur]

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

[@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

[@YY-wu7et]

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

[@General12th]

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

[@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.

[@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.

[@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.

[@diribigal]

Very good episode. And shoutout to the Polynomial Method!

[@lucanalon1576]

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

[@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.

[@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

[@thecaneater]

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

[@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.

[@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.

[@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|

[@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.

[@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.

[@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.

[@letstalkaboutmath2121]

i loved this episode, good job gabe

[@__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.

[@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.

[@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).

[@pedraumbass]

The "size" of this tower is not a set (if you suppose it is, you'll get a contradiction, that size is "too big" to be considered a set), it's a proper class.

[@TheManxLoiner]

It's a proper class in ZF(C), but it is actually a set in alternative set theories (e.g NF set theory).

[@johnkentzel-griffin2855]

The notion of proper class is not defined in ZF. ZF only deals with sets.

[@pedraumbass]

The objects in ZF are indeed only sets, but you can still talk about classes in this theory, check the book Introduction to Axiomatic Set Theory ( G. Takeuti ), chapter 4.

[@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?

[@alessandropizzotti932]

It's not even that subtle. The axiom schema of specification can only produce subsets of a given set, thus there's no reason the "set of all cardinals" should exist. In fact, this very argument proves, by contradiction, that it doesn't.

[@enlightedjedi]

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

[@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

[@dasmartretard]

my only gripe with this video is with the term "1-1 correspondence" in relation to bijective. Not to be confused when one sees a "1-1 mapping" which is an injective map. The distinction of "correspondence" over "mapping" should not be over looked.

[@FistroMan]

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

[@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).

[@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.

[@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

[@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 ?

[@fullfungo]

We know that א_i>i. So we have to find number α such that א_α=α. That looks exactly like our contradiction. That’s why the answer is א_א_א_..... (where α_β means α with index β)

[@MikeRosoftJH]

That's just a limit (=union) of the sequence of sets Aleph-0, Aleph-Aleph-0, Aleph-Aleph-Aleph-0, ... And this is a well-defined cardinal number, a union of countably many sets (and yes, it is a fixed point of the Aleph function). If we label this set as Ω, then P(Ω) certainly has a greater cardinality.

[@MikeRosoftJH]

But an inaccessible cardinal cannot be proven to exist in ZFC. Worst, you can't even prove that its existence is consistent with ZFC (as long as ZFC itself is consistent).

[@martonlovas4583]

There is a simpler argument to show that |A| ≠ |P(A)|: Map all elements of A to the subset of A that only contains that element. Now, we have "used up" all of the elements of A, but we haven't mapped anything to the empty set.

[@MuffinsAPlenty]

This is not good enough to show that |A| ≠ |P(A)|. This is because |A| = |B| is _there exists_ a bijection (one-to-one correspondence) between A and B. It doesn't matter if you can create a particular function (or even infinitely many functions) which aren't bijections. In order to show that |A| ≠ |B|, you have to show that *no bijection can possibly exist* between A and B. For example, you could have functions from the natural numbers to the even natural numbers. Or to the prime numbers. Or to the odd natural numbers. Or to the natural number multiples of 5. All of these functions are injective but not onto the entire set of natural numbers. But clearly, the set of natural numbers has the same "size" as itself. So just because a particular function exists which isn't a bijection does not imply that no such bijection exists.

[@onuktav]

brain... hurts...

[@matt-stam]

Watching this channel makes me feel like an idiot.

[@LudwigvanBeethoven2]

Brain farts

[@skinzfan8926]

How many times does one have to dislike an ad to never see it again? As far as I'm concerned, Veridesk is the worst

[@macronencer]

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

[@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?

[@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).

[@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.

[@sulferx6370]

There’s a number called “Absolute Infinity”, which means no number before it can be grouped any amount of times to get it.

[@MikeRosoftJH]

I think that what you're talking about is the (strongly) inaccessible cardinal, which cannot be obtained by taking either a powerset of a set smaller than itself; nor by taking a union of a collection of sets smaller than itself, where the collection is also smaller than the cardinal number itself. Absolute infinity is the class of all sets, which is a proper class and can be proven not to exist as a set. (On the other hand, a set of all sets with a rank less than an inaccessible cardinal - assuming such a cardinal number exists - is a model of set theory. So what is a proper class - class of all sets - from the point of view of the inner model, is a set - von Neumann universe up to an inaccessible cardinal - from the point of view of the outer model.)

[@AlejandroBravo0]

8:40 - A child - NO

[@__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|.

[@Deguiko]

I think the problem here is that if you try to formalize this reasoning you would need to define a Universe of Discourse to quantify over. When you say "All things with *this* property" you need to specify a Universe set for where to look for these objects, otherwise the definition is not strictly objective and the resulting "set" might not satisfy the *Well Deifined* axiom. When a Universe of Discourse is defined then you have a restriction to the possible cardinal numbers that might show up, this is obvious since the Universe has a cardinality and the Power Set of the universe cannot be bigger as it is a subset of the Universe, so this Power Set is surely not the Full Unrestricted usual Power Set.

[@Deguiko]

This happens because the Power Set of the universe is the set of "Elements of the Universe that are subsets of the Universe". Of course, by Cantor's theorem, some "subset" of the Universe is not an element of the universe .

[@jaydeepvipradas8606]

Number's value going towards infinity and infinity itself are two different things. Infinity is singularly. If you assume that set of all integers and set of all real numbers has infinite elements, then it's still same infinity, not different. You can still always map a real number to integer. It's just that smaller value will get mapped to bigger value, which is ok. It's our interpretation of small and big. In some world, in some maths, fractional value of 1.5 could be bigger than value 2, just because of dimensions and space time arrangements. E.g. if someone focuses on reaching, instead of counting and reaching 2 from 1 could be energy plus time efficient, as compared to reaching 1.5 from 1. Hence, 1.5 could be higher potential than 2. In such case, bigger value is mapped to smaller value, it's how you look at things and arrange things. When considering infinity, count and mapping is important than value, infinity remains the same, it's same concept.

[@MikeRosoftJH]

And if we assume that my aunt has a beard, we conclude that she is my uncle. The set of all natural numbers by definition doesn't contain any infinite numbers - a set is finite, if its number of elements is equal to some natural number.

[@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

[@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.)

[@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 ^_^

[@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.

[@douglaswilliams8336]

There is an infinitesimal of numbers between 01-2.0,but also an infinity of numbers between 1 and 2 even though the infinity between 1 and 2 is obviously larger. Get it y'all

[@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!

[@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.

[@voteforno.6155]

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

[@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.

[@baburamprasad926]

Welcome back..!

[@sebastianelytron8450]

Cantordinal numbers be transfinite though?

[@twistedsim]

is that a pun?

[@sebastianelytron8450]

is the pope catholic?

[@twistedsim]

Is the cat alcoholic?

[@trevormarshall2817]

ayy its old spacetime dude. Good to see him again

[@JorgeMartinez-dp3im]

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

[@6iaZkMagW7EFs]

What a question. What a title.

[@alissonalmeidabueno250]

very good project

[@Winium]

"That sounds highly problematic" pls im dying

[@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

[@GUIHTD]

Infinities are so much fun!

[@sulferx6370]

Yes! THE BIGGEST POSSIBLE NUMBER IS “ABSOLUTE INFINITY”, SO NOW I’M GOD!

[@danpost5651]

If all chosen cards are from the same deck, then you can only choose one before you obtain a set. Even if not from the same deck, once you choose the second card, you have a set of two cards.

[@johnalexanderpaiva]

I believe the answer to this question is 42.

[@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?

[@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?

[@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.

[@pongesz2000]

2:40-3:20: f:N->N, f(n) = n + 1. injective, but not surjective, still the cardinality is the same.

[@MikeRosoftJH]

Remember, two sets have the same cardinality if there *exists* a one-to-one function between them. Your function doesn't prove that the set of natural numbers has the same cardinality as itself (function x->x does the trick); it proves that the set of natural numbers and the set of positive natural numbers have the same cardinality. It can also be proven that if there exists a one-to-one function between set A and a subset of B; and there also exists a one-to-one function between set B and a subset of A; then the sets A and B have the same cardinality.

[@pongesz2000]

If you read my two comments, surely will realize that this is what I have written.

[@TheWyrdSmythe]

“God made the integers, all else is the work of man.” (Leopold Kronecker) I think there’s countable infinity and uncountable infinity. Anything after that is a mathematical game. The question may be the validity of the idea of a power set on an uncountable infinity. How can there be an uncountably infinite input to a power set function?

[@MikeRosoftJH]

Sure there can: the set of all sets of real numbers. (The set of all real numbers is uncountable, so its own powerset, or a set of its subsets, has an even greater cardinality than the set of real numbers.)

[@roxanneedl5484]

Now that I think about it, you may get larger and larger infinite sets starting with the natural numbers and doing power sets upon power sets, but no matter how large it gets, the real numbers will always be even larger than that, because you could just take as many numbers between 0 and 1 as you need to map them and still have unused numbers between 0 and 1 AND all the numbers that AREN'T between 0 and 1 unused. This isn't related to solving the issue you encounter, just a thought about how crazy large is the set of all real numbers(you can still do power sets and get bigger infinities once you get there, you just won't get there with power sets if you start with a set that as a smaller or equal cardinality to any power set you can obtain from that infinite chain of power sets on the naturals.)

[@voteforno.6155]

Evildark Link Actually, no. It can be proven that the power set of the natural numbers has exactly as many elements as the set of all real numbers. HINT: think of expansions of real numbers in binary digits 0 and 1. For example, 1/2 = 0.1, 1/4 = 0.01, 5/8 = 0.101, etc.

[@roxanneedl5484]

What? I was pretty sure I had this stuff down, but I guess I overlooked something... Either that or you're messing with me... I have no idea why I should go with binary except if I wanted to be even more confused...

[@pbsinfiniteseries]

You're not being messed with -- showing that |P(N)| = |R| was one the "big deal" things that Cantor proved. In case it helps provide some clarification, there's a side comment about this in my next episode (not Tai-Danae's this Thursday, but _my_ next episode the first week of April).

[@roxanneedl5484]

Alright, great, hopefully I'll understand better after that video.

[@astaroth0316]

I think that the contratiction is produced because of loosely defined concepts, similarly to the problem of an almighty being creating an immovable rock: if the being can move the rock, he couldn't create an immovable one, so he isn't almighty; on the other hand if he couldn't move it, he also isn't almighty; all that beacuse of a vaguely defined concept of "all"

[@bk1168]

I came up with this by using the cartesian product instead of the union. Both work, but the union is more to the point.

[@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.

[@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.

[@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.

[@KohuGaly]

my guess is that the dreaded axiom of choice is the culprit. The collection of all sizes is a non-measurable set.

[@MikeRosoftJH]

No. The collection of all sizes is not a set at all. (Otherwise, if a set X' exists, then either there exists a one-to-one mapping between X' and P(X') - but that would contradict Cantor's result that such a mapping cannot exist; or the mapping doesn't exist - but this would contradict how the set X was defined: if X is a set of all cardinal numbers, and X' is a union of all sets in X, then if P(X') has a greater cardinality than X', that means that it can't be put in one-to-one correspondence with X', and of course neither with some subset of X', such as with some of the original elements of X. [Of course, P(X') can't have a smaller cardinality than X'.] The only way out is that our initial assumption was incorrect: X is not a set of all cardinal numbers.)

[@youtubeuser8232]

This video is excellent! Maybe too good for KZbin!

[@abstractnonsense3253]

The problem is that the axiom of comprehension is for classes, not for sets, which use the axiom of separation.

[@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).

[@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

[@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.

[@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.