As I have been,once again reading through the book of Herbert Enderton (which I am giving up due to unnecessary complications of later material) I came upon an ecercise which claims that

For any two order types $\alpha$ and $\beta$ there exist structures $\langle A,R \rangle$ and $\langle B , S \rangle $ members of order types $\alpha$ and $\beta$ respectively such that $A \cap B = \emptyset$

I can not seem to prove it,and it is all confusing given the definition that an order type is an isomorphism type of linearly ordered structure,and that isomorphism type of structure X is set of structures such that they are isomorphic to X,and they are nor isomorphic to any structure of lesser rank



If $X$ and $Y$ are sets, and $f\colon X\to Y$ is a bijection, then any (say binary) relation $\leq$ on $X$ can be transferred into a relation on $Y$ via $f$.

Define $\leq'$ on $Y$ as the relation $\{\langle f(u),f(v)\rangle\mid u\leq v\}$. Then we have that $u\leq v\iff f(u)\leq' f(v)$, and $f$ is a bijection. Therefore $\langle X,\leq\rangle$ and $\langle Y,\leq'\rangle$ are isomorphic.

Now show that given two sets $A$ and $B$, there are $A'$ and $B'$ such that $|A|=|A'|$ and $|B|=|B'|$ and $A'\cap B'=\varnothing$.

  • $\begingroup$ I can get sets of same cardinality but I can not make them belong to order type,since it requires to have the same rank as given set,basically I would take a cartesian product of A and {0} and cartesian product of B and {1},thus I would obtain disjointed sets,but then their rank increases and then they do not belong to the given order types according to given deffinition $\endgroup$ – Vanio Begic Sep 5 '14 at 21:17
  • $\begingroup$ But this is not a question of rank. It's a question of structure. You need to use the first part of the hint for that. $\endgroup$ – Asaf Karagila Sep 5 '14 at 21:19
  • $\begingroup$ I do not understand what you mean,if you could furnish full proof please? $\endgroup$ – Vanio Begic Sep 5 '14 at 21:21
  • 1
    $\begingroup$ I've added more to the hint, not a full proof, but I think it should be enough. $\endgroup$ – Asaf Karagila Sep 5 '14 at 21:24
  • $\begingroup$ Oh it has hit me as if it came out of the sky,now we have same isomormphism types since constructed structures are isomorphic thus they belong to same order type.Brilliant $\endgroup$ – Vanio Begic Sep 5 '14 at 21:27

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