# Enderton's isomorphism type arithmetic

On page 222 in Enderton's Elements of Set Theory, there's a remark which is then justified as an exercise:

Show that for any order types $\rho$ and $\sigma$ there exist structures $\langle A,R\rangle$ and $\langle B,S\rangle$ of types $\rho$ and $\sigma$, respectively, such that $A\cap B=\varnothing$.

Essentially the isomorphism type of a partially ordered set $\langle A,<\rangle$ is the set of all partially ordered sets which are order isomorphic to $\langle A,<\rangle$ and has the least rank.

Enderton claims that we can find disjoint members from any pair of isomorphism types. I find it hard to prove this (which is a foundational for defining order type arithmetic).

NOTE: the isomorphic equivalence class is too large to be a set but Enderton used Scott's trick to make a set. Now Enderton is working with it for isomorphism type arithmetic, but the detail about finding disjoint members from any two iso-types is too elusive for me.

Can I get insight as to the proof?

1. If $A$ and $B$ are two sets, then there is some $A'$ such that $|A|=|A'|$ and $A'\cap B=\varnothing$.
2. If $f\colon X\to Y$ is a given bijection (any given bijection), and $R$ is a relation on $X$, then there is some $R'$ such that $f$ is an isomorphism between $\langle X,R\rangle$ and $\langle Y,R'\rangle$.
• If $(A,\prec_A)\in it(A^*,\prec_0)$ and $(B,\prec_B)\in it(B^*,\prec_1)$, then I agree that I can find $(A',\prec_{A'})$ which is order isomorphic to $(A,\prec_A)$ and disjoint from $B$. But is $(A',\prec_{A'})$ in the isomorphism type of $(A^*,\prec_0)$? They have equal isomorphism type (because they're isomorphic) but that's not what's important. I.E. how to show it has the least rank for which it's isomorphic to $(A^*,\prec_0)$. – Alberto Takase Apr 19 '15 at 21:53
• This means on page 222 Enderton made a typo? "Any member of an order type $\rho$ is said to be a linearly ordered structure of type $\rho$.'' where "An order type is the iso-type of some linearly ordered structure...We will use Greek letters $\rho,\sigma,\ldots$ for order types." – Alberto Takase Apr 19 '15 at 22:27