1
$\begingroup$

Maybe this is too trivial a question to be posted anywhere, but anyway. I am reading Poizat's "A Course in Model Theory". In page 4 he defines the notion of two $k$-tuples, each in the universe of some relation, being $p$-equivalent.

He gives two conditions:

  1. $a_i = a_j \leftrightarrow b_i = b_j$.
  2. The function $s$ defined by $sa_1=b_1,...,sa_k=b_k$ is a $p$-isomorphism from $R$ to $R'$.

I guess the first of them is just a remark, because it seems redundant given the second. If the function $s$ is a $p$-isomorphism then it is, in particular, a bijection between the finite subsets $\{a_1,...,a_k\}$ and $\{b_1,...,b_k\}$, and hence it is impossible to falsify condition 1. Am I missing anything?

$\endgroup$
  • 1
    $\begingroup$ Don't worry, we've seen more trivial questions :-) $\endgroup$ – joriki Oct 19 '11 at 10:00
1
$\begingroup$

Yes, I was missing something pretty obvious. Condition 1 is what enables him to talk about the function $s$ in condition 2. Otherwise he should have talked about the relation $s$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.