I've heard of some other paradoxes involving sets (i.e., "the set of all sets that do not contain themselves") and I understand how paradoxes arise from them. But this one I do not understand.

Why is "the set of all sets" a paradox? It seems like it would be fine, to me. There is nothing paradoxical about a set containing itself.

Is it something that arises from the "rules of sets" that are involved in more rigorous set theory?

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Justin L.
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    Intuitively, you can put many things in a basket -- but one thing you can never put in that basket is *the basket itself*. – littleO Aug 26 '14 at 14:55
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    @AirMike: Please avoid making trivial edits to old posts. Let alone stylistic edits like this (which, frankly, doesn't make any sense). – Asaf Karagila Sep 20 '20 at 19:16
  • @littleO the set of all sets, though, would not be akin to a basket. A more appropriate concept might be a conception of God as being all things. In a sense God thus /contains/ all things: the set of all sets, but is not /larger/ than it... Applying, for example, 2 to the power God would presuppose that in the new paradigm we are working in 2 could even be conceived of as being raised to the power God, there was a higher power which was hithertofore unconceived... – Peter David Carter Oct 19 '21 at 07:38

13 Answers13


Let $|S|$ be the cardinality of $S$. We know that $|S| < |2^S|$, which can be proven with generalized Cantor's diagonal argument.


The set of all sets does not exist.


Let $S$ be the set of all sets, then $|S| < |2^S|$. But $2^S$ is a subset of $S$. Therefore $|2^S| \leq |S|$. A contradiction. Therefore the set of all sets does not exist.

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Chao Xu
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    What does 2^S mean? – Casebash Jul 21 '10 at 00:51
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    @Casebash: 2^S is standard notation for the [power set](http://en.wikipedia.org/wiki/Power_set) of S: That is, the set of all subsets of S. If S were {1, 2}, 2^S would be {{}, {1}, {2}, {1, 2}}. – Larry Wang Jul 21 '10 at 00:58
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    I like this answer because it shows that the Universal Set is a paradox in itself, and not that "it must contain paradoxical sets". – Justin L. Jul 21 '10 at 05:36
  • This answer seems to be saying that the powerset of S is a subset of S. This is so wrong I don't know where to start. {S} is an element of 2^S... Every element of 2^S is a SUBSET (not an element) of S. – Seamus Jul 21 '10 at 12:08
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    @Seamus Here S is defined as the set of all sets. Every element of 2^S is a set, implies they are also elements of S. – Chao Xu Jul 21 '10 at 12:22
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    This is incorrect, as one of the other, lower-rated, answers notes, citing Quine’s New Foundations. My preferred theory is Church’s (first) Set Theory with a Universal Set (presented 1971, published 1974). For more details, see the Wikipedia article on Universal Set: http://en.wikipedia.org/w/index.php?title=Universal_set#Set_theories_with_a_universal_set – Flash Sheridan Feb 25 '11 at 17:49
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    How is that incorrect. We always consider the standard framework when we are not presented other ones. – Chao Xu Mar 06 '11 at 07:55
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    @JustinL. Sure... but it requires Cantor's theorem to prove. You can prove that it must contain a "paradoxical set" with a diagonalization argument directly (of course that argument is essentially the one used to prove Cantor's theorem anyways). I should mention that the reason I have a natural disinclination for these proofs is because in my set theory course we were asked to prove a lot of theorems/lemmas without the assistance of stronger theorems. – roliu Nov 23 '12 at 03:46
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    Cardinality is a definition based on set, not a logical consequence of set. Based on the contradiction, I can also say maybe something is wrong with the cardinality of the set of all sets. _I.e._, neither cardinality, nor power set, nor "less than" has been exonerated. Why is the set of all sets alone being implicated for causing the contradiction? – George Chen Aug 25 '14 at 03:55
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    If you know the bizarre arithmetical properties of $\aleph$ numbers, you should be prepared that the cardinality of the set of all sets, if it exits, may be an entirely different type of number. Maybe greater and less than another such number is no contradiction at all. Schröder-Bernstein Theorem has shown many wonders. Again, the root of all these doubts is that cardinality is not implied by set. If you want to use Reductio ad absurdum, make sure the chain of implication exists. – George Chen Aug 26 '14 at 03:02
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    @GeorgeChen The notion of cardinality is not at fault here. You can carry out the proof without talking about cardinal numbers as follows: Observe that there is an injective function $S\to 2^S$ given by $x \mapsto \{x\}$. Observe that there is an injective function $2^S \to S$ given by $x\mapsto x$, since $S$ is the set of all sets. Now use the Cantor-Schroeder-Bernstein argument to show that there is a bijection between $2^S$ and $S$. Finish with Cantor's diagonal argument showing that the existence of such a bijection leads to a contradiction. – Alex Kruckman Aug 26 '14 at 16:56
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    @FlashSheridan That fact that you prefer an alternative foundation doesn't give you the authority to call valid arguments in the standard foundation incorrect... – Alex Kruckman Aug 26 '14 at 16:59
  • Thanks, @AlexKruckman. While going down that path, I realized that I can imitate Cantor's theorem and dispense with the power set also, because the power set is both a member and a subset of $S$. See my answer http://math.stackexchange.com/a/909719/119110 – George Chen Aug 26 '14 at 22:41
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    What bothers me is that the notion of cardinality and less than involve many other notions that would be invalid unless the notion of set is put right first. – George Chen Aug 26 '14 at 23:25
  • Cmiiw, but I think this is supposed to be every set in $2^S$ is a subset of $S$? – BCLC Dec 06 '15 at 10:54
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    Note that this is actually the exact same argument as Russell's paradox; the set that Cantor's diagonal argument produces to get a contradiction to the inclusion map $2^S\to S$ being surjective is exactly the Russell set $\{x:x\not\in x\}$. – Eric Wofsey Apr 17 '16 at 03:55
  • If all we have assumed thus far is that sets exist, then what does one mean by "cardinality?" I feel that the question has more to do with the historical principles that lead to contemporary set theory than it does with the set of all sets in modern context. In and of itself, the set of all sets is NOT paradoxical. – j0equ1nn Aug 27 '16 at 05:58
  • I think the paradox does exist because it came about when Whitehead tried to formally prove complex things. Basically the standard logics don't work with such a statement. – marshal craft Mar 09 '17 at 13:05
  • How does this apply if we say that $F(x)= \left\{ \begin{array}{ll} \aleph_0 & x<1 \\ \aleph_{F(x-1)} & x\ge 1 \\ \end{array} \right. $ and we define $S$ such that $|S|=\lim_{x\to\infty} F(x)$? $2^{S}$ is still $S$ – Mark Oct 01 '19 at 17:25
  • @ChaoXu Really... So let S be the empty set, then the empty set does not exist. Q.E.D. – Cam White Mar 22 '20 at 02:53
  • @AlexKruckman Since $P(V)=V$, there is no need for cantor berenstein. – Violet Flame Feb 27 '22 at 03:39
  • Or even $P(V)\subset V$, and hence there is a surjection from $V$ to $P(V).$ – Violet Flame Feb 27 '22 at 03:40
  • @VoiletFlame Indeed, good point. – Alex Kruckman Feb 27 '22 at 03:45

Just by itself the notion of a universal set is not paradoxical.

It becomes paradoxical when you add the assumption that whenever $\varphi(x)$ is a formula, and $A$ is a preexisting set, then $\{x\in A\mid \varphi(x)\}$ is a set as well.

This is known as bounded comprehension, or separation. The full notion of comprehension was shown to be inconsistent by Russell's paradox. But this version is not so strikingly paradoxical. It is part of many of the modern axiomatizations of set theory, which have yet to be shown inconsistent.

We can show that assuming separation holds, the proof of the Russell paradox really translates to the following thing: If $A$ is a set, then there is a subset of $A$ which is not an element of $A$.

In the presence of a universal set this leads to an outright contradiction, because this subset should be an element of the set of all sets, but it cannot be.

But we may choose to restrict the formulas which can be used in this schema of axioms. Namely, we can say "not every formula should define a subset!", and that's okay. Quine defined a set theory called New Foundations, in which we limit these formulas in a way which allows a universal set to exist. Making it consistent to have the set of all sets, if we agree to restrict other parts of our set theory.

The problem is that the restrictions given by Quine are much harder to work with naively and intuitively. So we prefer to keep the full bounded comprehension schema, in which case the set of all set cannot exist for the reasons above.

While we are at it, perhaps it should be mentioned that the Cantor paradox, the fact that the power set of a universal set must be strictly larger, also fails in Quine's New Foundation for the same reasons. The proof of Cantor's theorem that the power set is strictly larger simply does not go through without using "forbidden" formulas in the process.

Not to mention that the Cantor paradox fails if we do not assume the power set axiom, namely it might be that not all sets have a power set. So if the universal set does not have a power set, there is no problem in terms of cardinality.

But again, we are taught from the start that these properties should hold for sets, and therefore they seem very natural to us. So the notion of a universal set is paradoxical for us, for that very reason. We are educated with a bias against universal sets. If you were taught that not all sets should have a power set, or that not all sub-collections of a set which are defined by a formula are sets themselves, then neither solution would be problematic. And maybe even you'd find it strange to think of a set theory without a universal set!

Asaf Karagila
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  • Can you expand a bit on "If A is a set, then there is a subset of A which is not an element of A." should it not be "If A is a set, then there is a set of subsets of A which is not an element of A."so suppose A = { 1,2,3,{1,2}} then i mean the set { {1,3} , {2,3} , {1, {1,2}} , {3,{1,2}} } – Willemien Jan 17 '15 at 15:30
  • Yes, it *could* be also there is a set of subsets which is not an element of $A$. But there is also a subset of $A$ which is not an element of $A$, why is that a problem? There are also sets that has nothing to do with $A$ which are not elements of $A$. – Asaf Karagila Jan 17 '15 at 15:37
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    sorry I am just trying to understand it, could you give a me the subset you mean given the set A I gave? – Willemien Jan 17 '15 at 15:45
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    Well, $A$ itself is that subset in the case of $A=\{1,2,3,\{1,2\}\}$. It is true, $\varnothing$ could also work, and so could any singleton, or so on. But given any set, $A$ the set $\{x\in A\mid x\notin x\}$ is either not a set, or not an element of $A$. In presence of comprehension axioms which do not restrict the formulas, it is always a set, and therefore not an element of $A$. In Quine's New Foundations, and other more "type theoretic" set theories, we restrict the formulas that can define a set, which mean it might not be a set at all. – Asaf Karagila Jan 17 '15 at 15:54
  • Sorry, still not sure why the set $ \{ x \in A \land x \not\in x \} $ cannot be an element of A. (now I seem to think that is is the same as set A) – Willemien Jan 17 '15 at 17:22
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    In all the sets that you can write down, this set will be $A$ itself, and $A\notin A$. This is because naively we cannot construct a set $x$ such that $x\in x$ (otherwise we can naively disprove the axiom of foundation; but that's not true). If $B=\{x\in A\mid x\notin x\}\in A$, then we ask either $B\in B$ or $B\notin B$. If $B\in B$, then $B\in A$ and $B\in B$, so $B\notin B$; and if $B\notin B$ then $B\in A$ and $B\notin B$, so $B\in B$. Either case a contradiction. So the only solution is that $B\notin A$. The question is whether or not $\{x\in A\mid x\notin x\}$ is even a set. [...] – Asaf Karagila Jan 17 '15 at 17:26
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    [...] In the presence of separation axioms, like the ones in the $\sf ZF$ system, it immediately follows that $\{x\in A\mid x\notin x\}$ is a set, and therefore it has to be that case that this is a subset of $A$ which is not an element of $A$. In set theories like Quine's New Foundations, this is not a set since the formula is not stratified and therefore doesn't have to define a set. It follows, therefore, that in $\sf ZF$ there is no set of all sets, since $\sf ZF$ proves that given a set, it doesn't contain *all* sets. – Asaf Karagila Jan 17 '15 at 17:28

The existence of universal set is incompatible with the Zermelo–Fraenkel axioms of set theory. However, there are alternative set theories which admit a universal set. One such theory is Quine's New Foundations.

François G. Dorais
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    Possibly also useful: *[Elementary Set Theory With a Universal Set](http://math.boisestate.edu/~holmes/holmes/head.pdf)*, by M. Randall Holmes. – MJD Mar 18 '12 at 18:12
  • Late to the party, but how does this relate to Cantor's proof about cardinals? – Guido Mar 10 '17 at 23:56
  • The stratified formulas used in New Foundations prevent the direct comparison of a set's size with that of its power set. If you try to make a comparison, the best you can do is that the size of the set of all singletons of S is less than the size of P(S), so there are fewer singletons than there are sets of any size. Interestingly, NBG https://en.wikipedia.org/wiki/Von_Neumann-Bernays-Gödel_set_theory states that the proper class of singletons is equinumerous to the proper class of all sets, which is the opposite of what NFU says. – Zemyla Jun 17 '19 at 11:18

The "set of all sets" is not so much a paradox in itself as something that inevitably leads to a contradiction, namely the well-known (and referenced in the question) Russell's paradox.

Given any set and a predicate applying to sets, the set of all things satisfying the predicate should be a subset of the original set. If the "set of all sets" were to exist, because self-containment and non-self-containment are valid predicates, the set of all sets not containing themselves would have to exist as a set in order for our set theory to be consistent. But this "set of all sets" cannot exist in a consistent set theory because of the Russel paradox.

So the non-existence of the "set of all sets" is a consequence of the fact that presuming it's existence would lead to the contradiction described by Russel's paradox.

This is in fact the origin of Russel's paradox.

In his work "The Basic Laws of Arithmetic", Gottlob Frege had taken as a postulate the existence of this "set of all sets". In a letter to Frege, Bertrand Russell essentially blew away the basis of Frege's entire work by describing the paradox and proving that this postulate could not be a part of a consistent set theory.

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  • This is how I had understood it before and I always found it to be the simplest answer, which is quite neat. – Justin L. Jul 21 '10 at 05:35
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    *If the "set of all sets" were to exist, because self-containment and non-self-containment are valid predicates, the set of all sets not containing themselves would have to exist as a set in order for our set theory to be consistent* - Not true, as Quine's New Foundations, cited by François, shows. It has a universal set, but not the Russell set. – Charles Stewart Jul 21 '10 at 11:05
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    I guess this depends on what one considers a "set theory". The usual image of the universe of sets, given by the cumulative hierarchy, doesn't allow for the set of all sets. – Michael Greinecker Mar 08 '12 at 09:13
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    I don't see how you could not consider Quine's New Foundations a set theory; it has the usual universe of sets in it. It may in fact have a superset (superclass?) of ZF's sets, though it's possible stratified comprehension excludes some sets ZF includes. It is akin to NBG set theory; some definite weirdness if you're used to ZFC, but basically the same thing for practical purposes. – prosfilaes Oct 17 '12 at 06:48
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    NF doesn't necessarily have all of ZF's sets in it. While it's consistent that the von Neumann version of $\omega$ exists in NF, I believe it's not provable that it does. – Malice Vidrine Jun 23 '14 at 10:12
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    Stratification is the key difference. PM has universal set too, but its members are typically homogeneous. They are all from the type next below. When OP said the set of all set, his universe was flat. – George Chen Aug 31 '14 at 01:30
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    All the individuals that won't disappear after analysis reside in a flat universe. But sets are stratified, and there is a universal set in very type (layer or sphere). This universal set has totally different meanings from what the OP refers to as the set of all sets. – George Chen Aug 31 '14 at 03:26
  • Well, when I say "sets are stratified," I'm all ready referring to the set of all sets which is meaningless. What a slippery field of philosophy! – George Chen Aug 31 '14 at 03:38
  • But you know what I'm talking about. Could it be true that there are things we understand but cannot speak of? – George Chen Aug 31 '14 at 03:56
  • Yes,unfortunately this is an incorrect answer because it assumes a form of comprehension powerful enough to obtain Russel sets. Quine's NF avoids that problem by having a weaker form of comprehension that is still sufficient to give the set of all sets (and indeed set complement). The problems with a universal set are potentially different from those with a Russel set. – Francis Davey Dec 01 '14 at 20:08

Russel's paradox arises if you consider the set $U=\left\{x:x\not\in x\right\}$. Ask yourself if $U\in U$. If you suppose so, then by the definition of unrestricted set comprehension $U\not\in U$. You have a contradiction, so it must be the opposite of what you supposed, that is, $U\not\in U$. But this is the same as saying $U$ belongs to the complement of itself, that is, $U\in U$. You now have another contradiction, but this is far worse, since you have no hypotheses. The whole theory is logically inconsistent.

In set theory there are two ways for getting rid of the Russel's paradox: either you disallow the set of all sets and other similar sets (see for example the Zermelo-Fraenkel set theory), or you allow them, but you also restrict the way they are used (see for example the Morse-Kelley set theory).

In the first case, set comprehension says if you have a set $A$ you can have $\left\{x\in A:\phi\left(x\right)\right\}$ (notice: writing $\left\{x:\phi\left(x\right)\right\}$ is just wrong in this case, because you have to have an initial set). If you now define $U=\left\{x\in A:x\not\in x\right\}$ and you repeat the same passages as before, it only follows that $U\not\in A$. There's no contradiction and the theory is consistent.

In the second case, you consider classes, not just sets. Sets are classes that belong to some other class, while proper classes are classes that belong to no class. Set comprehension, in this case, says you can have $\left\{x:\phi\left(x\right)\right\}$, but all its members are sets by definition. If try to reproduce Russel's paradox, you get that $U\not\in U$. If you then suppose that $U$ is a set, then you have a contradiction, so $U$ must be a proper class. This is all you get. No contradictions. The theory is consistent.

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  • There are other ways to avoid Russell's paradox, too. For example, the axiom schema of comprehension in [NF](http://en.wikipedia.org/wiki/New_Foundations) doesn't allow you to construct $\{x: x \notin x\}$ (or, AIUI, even $\{x \in A: x \notin x\}$ in general) because $x \notin x$ is not a [stratified formula](http://en.wikipedia.org/wiki/Stratified_formula). It _does_ let you construct the universal set $\{x: x=x\}$, though. – Ilmari Karonen Mar 08 '12 at 08:57
  • Another way, don't assume LEM on $\in$. – DanielV Mar 10 '18 at 01:48
  • Or you could forbid ∉ and other negations in the axiom of comprehension, like in [positive set theory](https://en.wikipedia.org/wiki/Positive_set_theory). The universal set exists in this theory, as does { x : x ∈ x }, but not { x : x ∉ x }. – Zemyla Oct 11 '20 at 03:49

The set of all sets is not a paradox.

When set theory was invented, it was assumed that given any predicate, there was a set containing all the things that satisfied the predicate. This assumption is called naive comprehension. Unfortunately, this allowed paradoxes like the set of all sets not containing themselves.

So people invented restricted axioms of comprehension. These are rules that say that only certain predicates give rise to sets. There are different kinds of set theory with different comprehension restrictions.

One idea for limiting comprehension is to say that if the class of things satisfies the predicate is too big, the class is not a set. All the commonly used set theories use this idea. The set of all sets is very large indeed, so it is not a set in these theories.

There are other ways of restricting comprehension that get rid of the paradoxical sets but do allow a universal set. One such set theory was invented by Quine and called New Foundations. There is a book by Holmes, Set Theory with a Universal Set.

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Roger Witte
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Perhaps the following image will help take from here:

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Sniper Clown
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Some of the comments in the previous answers make a subtle mistake, and I think it may be worth clarifying some issues. I am assuming the standard sort of set theory in what follows.

Cantor's diagonal theorem (mentioned in some of the answers) gives us that for any set $X$, $|X|<|\mathcal P(X)|$. Unlike what some comments claim, this really has nothing to do with cardinalities. All it says is that no map $f\!:X\to\mathcal P(X)$ is onto. The usual proof proceeds by noting that $A=\{x\in X:x\notin f(x)\}$ is not in the range of $f$, because if $A=f(a)$, then $a\in A$ if and only if $a\notin f(a)=A$.

The usual argument for Russell's paradox (also mentioned in some of the answers) proceeds by considering $A=\{a:a\notin a\}$. If $V$ is a set, $A$ would be a set as well (by comprehension, if you wish), and we reach a contradiction by noting that $A\in A$ if and only if $A\notin A$.

I think it is misleading to think (as some of the comments suggest) that the two proofs are (fundamentally) different. They are essentially the same.

The point is that if there is a set of all sets (let's call it $V$), then $\mathcal P(V)$ is a set as well (by comprehension) and in fact $\mathcal P(V)=V$ because, on the one hand, any subset of $V$ is a set, and therefore a member of $V$, and on the other hand, any member of $V$ is itself a subset of $V$ (since the members of any set are sets themselves), and therefore a member of $\mathcal P(V)$.

Now, the identity function is a map from $V$ to $\mathcal P(V)$. Let's call it $f$. The set not in the range of $f$ given to us by the standard proof of Cantor's diagonal theorem recalled above is $A=\{x\in V:x\notin f(x)\}$, which in this case reduces to $\{x:x\notin x\}$, which in turn is precisely the set given by the standard proof of Russell's paradox.

[What to make of this result from a foundational point of view is another matter. In $\mathsf{ZF}$ the conclusion is just that there is no set of all sets, although it is perhaps more accurate to say that the result is used to justify dismissing unrestricted comprehension and adopting instead the version of bounded comprehension used in $\mathsf{ZF}$. In $\mathsf{NF}$ the solution is instead to limit unbounded comprehension by only allowing stratified instances. Other foundational solutions may and have been adopted as well.]

Andrés E. Caicedo
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This turns out to be the same as @donroby's answer. The scheme is inspired by the proof of Cantor's theorem, the use of which is implicit in the accepted answer.

As the following example shows, if all is construed as commonsense does, the set of all sets contains a member that simultaneously is and is not a member of itself.

Let $T = \{ \alpha \ |\ \alpha $ is a set and $ \alpha \notin \alpha \}$,

$T \in T \Rightarrow T \notin T$ --- (1)

$T \notin T \Rightarrow T \in T$ --- (2)

(1).(2) $\Rightarrow$ Paradox.

Note: This answer dispenses with such higher notions as "power set," "cardinality" and "less than" whose properties cannot be ascertained before the properties of set are ascertained.

Further analysis leads to Type Theory which uses meaning as a guarding criterion. In Type Theory, all sets, as commonsense understands it, is meaningless; propositions like $T\in T$ and $T \notin T$ are also meaningless.

George Chen
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  • @Elliott - Cleaned up. Thanks for the feedback. – George Chen Oct 16 '20 at 14:28
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    You're welcome. I read lots of the answers here, but it wasn't until I read yours that I really understood it - so thank you. =) – Elliott Oct 16 '20 at 18:42
  • Re: the last paragraph, type theory isn't the *only* place further analysis can lead to. [$\mathsf{ZFC}$](https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory)-style set theory is also a response, in which "$T\in T$" is meaningful but doesn't hold of any object, as are things like [$\mathsf{NF(U)}$](https://plato.stanford.edu/entries/settheory-alternative/#NewFounRelaSyst) or [positive set theories such as $\mathsf{GPK^+_\infty}$](https://plato.stanford.edu/entries/settheory-alternative/#PosiSetTheo) in which there *is* a universal set (and hence some sets do contain themselves!). – Noah Schweber Oct 16 '20 at 19:47
  • Different people will prefer different approaches. For myself, ultimately the $\mathsf{ZFC}$-style approach seems most faithful to intuition. – Noah Schweber Oct 16 '20 at 19:49
  • @NoahSchweber - Sure I don't think this is a good answer now but I can't recall what epiphany I had in mind back then. There is a more direct reason against the conception of the set of all sets: a totality is not determined until each of its constituents are determined; if one of the constituents is the totality itself, a vicious circle ensues. A set of all sets contains itself and thus involves a vicious circle. – George Chen Oct 17 '20 at 00:14

How about a set that contains everything with the exception of itself?

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    You can still make the cardinality argument that the set cannot contain all its subsets. – Michael Greinecker Mar 08 '12 at 09:11
  • Insightful! @MichaelGreinecker. In type theory, the set of all the subsets resides on upstairs. – George Chen Aug 22 '14 at 23:50
  • A set cannot have any of its subset as members because a subset of a set involves the totality also, and therefore must not be a member of the larger set; otherwise there will be a vicious circle. – George Chen Aug 23 '14 at 00:42
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    @GeorgeChen There is something in set theory called a *transitive set*, and for such a set, every element is a subset. Here is an example: $\{\emptyset\}$. – Michael Greinecker Aug 23 '14 at 07:25
  • This is wild. In type theory, the subclass stays on the same floor; the member goes to downstairs. They are totally different things. – George Chen Aug 23 '14 at 09:02
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    Suppose $S$ is such a set. Then $S \cup \{ S \}$ is the set of all sets. –  Aug 27 '14 at 09:10

An informal explanation is Russel's Paradox. The wiki page is informative, here's the relevant quote:

Let us call a set "abnormal" if it is a member of itself, and "normal" otherwise. For example, take the set of all squares. That set is not itself a square, and therefore is not a member of the set of all squares. So it is "normal". On the other hand, if we take the complementary set that contains all non-squares, that set is itself not a square and so should be one of its own members. It is "abnormal".

Now we consider the set of all normal sets, R. Attempting to determine whether R is normal or abnormal is impossible: If R were a normal set, it would be contained in the set of normal sets (itself), and therefore be abnormal; and if it were abnormal, it would not be contained in the set of normal sets (itself), and therefore be normal. This leads to the conclusion that R is both normal and abnormal: Russell's paradox.

Edan Maor
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  • I mentioned this paradox in my question; it however is not the same one that I am asking about. – Justin L. Jul 20 '10 at 23:17
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    You're right, sorry. Probably you can say that, since "the set of all sets" is not so well-defined, it must necessarily contain the set from Russel's paradox. Since the set from Russel's Paradox can't exist, neither can any set that contains it. But I feel this is cheating somehow. – Edan Maor Jul 20 '10 at 23:37
  • That's not cheating, that's exactly correct. I always use Russel's Paradox for the "set of all sets" question. The main point in both cases anyway is remembering that Cantor's definition of a set is a little to vague if you dig deep enough. – balpha Jul 21 '10 at 05:10

"Why is "the set of all sets" a paradox? It seems like it would be fine, to me."

The lesson of this and other set theory paradoxes is that when you define a set in a proof, you are obliged to demonstrate that the set exists and is well-defined. To me, a paradox of Russell's Paradox is that he fell into this trap since he pretty much wore himself out trying to generate a rigorous foundation for Mathematics.

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    The set of all sets *is* well-defined in the set theory Cantor used: to any logical predicate there is a set of all objects satisfying it. The problem is that Cantor's set theory is inconsistent. –  Mar 08 '12 at 09:36

I think this paradoxon is not so complicated and can be explained simply.

If you put all existing sets S(1) to S(k) together in a set S(all), you create a new set S(k+1)={S(1),...,S(k)} and this set has to be in the set of all sets as well (because it should contain all sets). Thus, you put the new set S(k+1) in the set of all sets S(all). But this in turn creates a new set S(k+2)={S(1),...,S(k+1)} wich has to be inserted in the set of all sets as well. Putting S(k+2) in S(all), however, creates a new set and so on and so on.

In conclusion, everytime you put the new set into the set of all sets you create a new set. As a consequence, you can never build a set S(all) that contains all sets without creating a set which is not contained in S(all).

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    Building a set inductively like this is not paradoxical. The natural numbers are built in exactly this way starting with the empty set. You can't talk about 'putting a set in' to the set of all sets. – Dan Rust May 03 '15 at 22:56
  • The building process is not the paradoxon, but it shows that a set cannot contains all sets without producing a new set that is not contained in this set and hence a set that is assumed to contain all sets actually does not contain all sets. The latter is the paradoxon. – user237596 May 04 '15 at 01:05
  • But ok, I am not a mathematician and this solution was the first that comes to my mind. Maybe, it is completely bullshit ;-) – user237596 May 04 '15 at 01:11
  • I am just curious. You mentioned the natural numbers. In my opinion, you have a similar case if you say there is a natural number that is maximal. You can simply show that this number cannot be maximal because you can add +1 to this number. Isn't this case similar? – user237596 May 04 '15 at 01:16
  • with "there is a natural number that is maximal" I meant the assumption that there is a natural number that is greater than all other natural numbers. 1) set that contains all set => this set is larger than all other sets 2) maximal number => this number is greater than all other natural numbers. I see there some similarities. – user237596 May 04 '15 at 02:23