# Equinumerosity without ordered pairs

If the two sets are disjoint, you can just define it via an equivalence relation in which every equivalence class has an element of $A$ and an element of $B$ and nothing else. (specify that if $a,a'\in A$ and $a\sim a'$ then $a=a'$, that if $b,b'\in B$ and $b\sim b'$ then $b=b'$; that for all $a\in A$ there exists $b\in B$ with $a\sim b$, and that for all $b\in B$ there is $a\in A$ with $b\sim a$. But you will encounter difficulties when $A$ and $B$ are not disjoint. You find similar difficulties in the usual development of cardinality (cont) –  Arturo Magidin Oct 16 '11 at 23:57
(cont) e.g., when defining the sum of two cardinals, you have to refer to a "disjoint union". Unfortunately, the only way I know to deal with these issues is by "disjointizing" the two sets, which is achieved through the use of ordered pairs: instead of considering $A$ and $B$, one consider something like $A\times\{1\}$ and $B\times\{2\}$... –  Arturo Magidin Oct 16 '11 at 23:58
Arturo's suggestion works fine if you handle it in two steps. Say $A$ and $B$ are equinumerous if there's a set $C$ disjoint from both such that $A, C$ and also $B, C$ admit equivalence relations as described. Then you have to prove a lemma: given any sets $A$ and $B$, there's a set $A'$ disjoint from both $A$ and $B$ such that $A, A'$ admit the described equivalence relation. This is very similar to the approach originally taken by Zermelo, since he didn't have access to a good ordered pair. –  user83827 Oct 17 '11 at 0:15
The trick is to use the Russell paradox to find a set $x$ not in $\bigcup A \cup \bigcup B$ and define $A'$ by adding $x$ to each element of $A$. It's a bit awkward to state what "tools" you're allowed in a historically honest way, since ordered pairs clean everything up so much and don't really take any work to define (in retrospect). –  user83827 Oct 17 '11 at 0:42