Show that there exists only one finite structure (up to isomorphism), where the axiom of pairing is valid.
Axiom of Pairing: To every two sets x,y there exists one set $c=\{x,y\}$
$\forall x \, \forall y \, \exists c : ~ \forall z (z \, \epsilon \, c \iff z = x \lor z=y)$
We consider arbitrary structures (M,E), where E is an arbitrary binary relation. Empty structeres are not allowed.
My toughts:
If the axiom of pairing holds, I can build up sets with every finite cardinality through iterated using of the pairing axiom and the axiom of union. And I can do this for every finite structure. So if I use $\subset$ for ordering I should be able to find an isomorphism, namely sets of the same cardinality are isomorphic.
So my question, am I on the wrong track?
The only finite non-empty structure satisfying that axiom is $\big\langle\{0\},\{\langle 0,0\rangle\}\big\rangle$, a structure with one element that is related to itself. If $M$ has distinct elements $a$ and $b$, it must have distinct elements $\{a\}^M,\{b\}^M$, and $\{a,b\}^M$, so it must have at least three elements. You can easily convert this to an argument showing that if $M$ has at least two elements, it must have at least $n$ elements for each $n\in\Bbb Z^+$.