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

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.

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

yeah, I thought it was not right

@@yuvalpaz3752 היי גם אתה רואה את זה

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

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

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

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

Is that a JOHN Greene reference?

all the way up

Brief history of time?

That's why mathematicians drink coffee

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

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

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

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

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

agreed

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.

And infinitely empty.

But which infinity?

is it infinitely infinite

Go home, Bertrand. You're drunk!

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

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.

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.

No set can have itself as an element

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

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

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.

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?

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.

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?

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

lol Interesting reasoning

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.

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

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)

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...

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?)

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

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.

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

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.

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|

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

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.

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

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

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?

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

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

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.

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"

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.

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

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

Joshua Hillerup exactly what I'm talking about

Watching this channel makes me feel like an idiot.

Brain farts

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.

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

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

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

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

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

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|

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.

_"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.

+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.

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).

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.

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.

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

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

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.

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.

Yeah, this episode is really hard. Not intended to be easily digested on just one viewing, for sure.

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

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.

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?

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).

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.

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).

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

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

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

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.

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.

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.

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.

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).

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.

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.

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.

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

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 RU-vid 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.

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.

Антоша Пушкин what? Sets can't be formed because they're too big??

Yes. For example a collection of all sets can't be a set (it's called Russell's paradox), also a collection of all cardinal numbers can't be a set. For example in NBG set theory there are entities called "classes", some classes are "small enough" to be sets and some are too big to be sets and are called "proper classes"

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|

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

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

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).

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?

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.)

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.

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 .