# The set of all sets of two elements does not exist, but the set of sets of two elements of a given set does?

I'm trying to wrap my head around this concept of extracting sets from a given set to show that a subset is indeed a subset.

Suppose I want to show that the set of all sets of two elements does not exist. I approach it like this.

Suppose such a set $V$ exists of all elements containing exactly the sets of two elements. Then for any nonempty set $X$, the set $\{X,\emptyset\}$ is a set, by the pairing axiom, and since $X\neq\emptyset$, $\{X,\emptyset\}$ has two elements, so $\{X,\emptyset\}\in V$ for all sets $X$. Then by the power set axiom, $\bigcup V$ is the set of all sets, since $\emptyset\in\bigcup V$, and any nonempty set $X\in\bigcup V$ as well. But the set of all sets does not exist, a contradiction, so such a set $V$ does not exist. Does this approach work?

However, if I'm given a nonempty set $A$, is the following a proper formulation that shows that the set of all sets of the form $\{a,b\}$ for $a,b\in A$ is a set?

If $a,b\in A$, then $\{a,b\}\subset A$, and $\{a,b\}\in\mathscr{P}(A)$. Assuming that $a$ and $b$ need not be distinct, consider $$\{t\in\mathscr{P}(A)\ |\ \exists x\exists y(x\in A\land y\in A\land t=\{x,y\}\}.$$ This is a set by the subset schema. Have I correctly specified the type of sets I want, or do I need to include something about if $z\in t\implies z=x\vee z=y$?, or is that unnecessary? Thanks.

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 Why doesnt "The set of all sets of two elements does not exist" ? – Holowitz Feb 5 '11 at 10:30

## 1 Answer

The problem is that $V$ is not a set, it is in fact a proper class. A good rule of thumb when it comes to these sort of problems is that if you can map each set into your collection - it's a proper class.

If, on the other hand, you started with a set $A$ (which you already know that it is a set) then you can have the collection of all pairs from the set. The formula which you wrote above is indeed the collection of all singletons and pairs from the set $A$, if you want to exclude the singletons you have to add $x\not= y$ as well.

This is a result from the Subset Axiom/Replacement Axiom - both are schemes and not actual axiom - which state that if you have a formula in one free variable, then the collection of elements which satisfies the formula coming from a set $A$ is a set itself, namely - if it's a subcollection of a set then it is a set.

The replacement axiom is slightly more complicated in that matter, but it says that the range of a function whose domain is a set is also a set, from this the subset axiom can be deduced easily.

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 Thank you Asaf. I gather that I at least did the second part correctly. Is my proof by contradiction of the first part correct, by assuming $V$ is a set, and then showing that I prove that there exists a set of all sets, which is of course wrong? I'm not sure if you're saying it's wrong, or just not as elegant as it could be. – user6655 Feb 5 '11 at 10:00 @youser: The first proof is fine, yes. – Asaf Karagila Feb 5 '11 at 10:28 "A good rule of thumb when it comes to these sort of problems is that if you can map each set into your collection - it's a proper class." — In NBG set theory, this is more than just a rule of thumb: all proper classes are the same size, so you can show that a class is proper by mapping it onto any other proper class. – Tanner Swett Aug 30 '11 at 22:24 @Tanner: I think that your using Axiom E, also known as Global Choice. – Asaf Karagila Aug 30 '11 at 22:26 Well, Wikipedia says that global choice is a consequence of the axiom of limitation of size, which is part of NBG. – Tanner Swett Aug 30 '11 at 22:55
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