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After some discouraging comments to this question let me ask straight ahead:

Can the concept of equinumerosity be defined basically without the concept of ordered pairs (in any of its disguises, e.g. Kuratowski's)?

(Just like the concept of equivalence (relation) can be defined basically without the concept of ordered pairs.)

EDIT: The question was for an example. The other way around: Is there a proof that there cannot be a definition of equinumerosity without ordered pairs? (Like the proof that there cannot be a first order definition of being finite.)

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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
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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
    
@ccc: But that lemma is precisely the problem. It isn’t even clear to me what ‘tools’ you’re permitted, let alone how the proof might go. –  Brian M. Scott Oct 17 '11 at 0:34
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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

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