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.
Best 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.