Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In Admissible Sets and Structures, page 101, theorem 5.8, Barwise introduces a weird form of his compactness theorem in which there are two theories $T$ and $T'$ both $\Sigma_1$, such that every $\varphi \in T$ is a pure set (while sets (or formulas) of $T'$ are allowed to be not pure sets, so that they may involve urelements). Also, he assumes that the theory $T$ is a set of finitary formulas (because of the key assumption $o(\mathbb{A}_{\mathfrak{M}})=\omega$).

Then assuming that for every finite $T_0 \subset T$, $T_0 \cup T'$ has a model then necessarily $T\cup T'$ has a model.

It seems to me that this leads to a contradiction because of the following: Consider the set $\omega$ (set of finite ordinals) and assume it is the pure part of $\mathbb{A}_{\mathfrak{M}}$.

So clearly $o(\mathbb{A}_{\mathfrak{M}})=\omega $ (where by $o(...)$ we mean the ordinal rank of the set of pure sets (i.e. not involving urelements) of a set). Also, let $\mathfrak{M}=(M )$. $M$ is set of urelements containing a copy of $\omega$. Assume $T'=\lbrace {\rm There \; exists \; a \; surjection \; from \; a \; finite\; ordinal \; to \;} N \subset M \rbrace \cup T''$.

Then $T'$ is a set of one infinitary sentence plus $T''$ where $T''$ is a set of formulas specifying that there exists a map from $\omega$ into $N$, plus a set of sentences specifying that two elements of $\omega$ map to the same element of $N$ only if they are congruent by some equivalence relation $\equiv$. Also, let $T$ be the set of (finitary!) formulas specifying that there exists a more than $n$ distinct equivalence classes for $\equiv$, a sentence for each $n \in \omega$.

Then every $T_0 \cup T'$ has a model where $T_0 \subset T$ finite, (since $N$ is assumed finite) but $T\cup T'$ does not have a model. How could that be?

Thank you

share|improve this question
    
@jeq: this kind of editing of an answered and perfectly understandable/legible question does not seem very reasonable. See meta.mathoverflow.net/questions/169/… –  Yemon Choi Jul 9 '13 at 0:45
    
@YemonChoi. Understood. I will refrain. –  jeq Jul 9 '13 at 0:49
    
@jeq Thanks, but judging by your activity over the last two hours, it is not clear to me you are doing much refraining... –  Yemon Choi Jul 9 '13 at 2:51
add comment

1 Answer 1

up vote 4 down vote accepted

This is a good question. The issue though is you have made several assumptions on your model $\mathcal{M}$ which cannot all hold simultaneously. To be precise lets enumerate the assumptions you have made:

(1) $\mathcal{M}$ is a model whose underlying set consists of urelements

(2) There is a relation $\lt$ in the language of $\mathcal{M}$ where $(\mathcal{M},<^{\mathcal{M}})$ is isomorphic to $(\omega, \in)$.

(3) There is an admissible set $A$ such that $\mathcal{M} \in A$ and $o(A) = \omega$.

To see that these three things can't all happen simultaneously suppose $A$ is any admissible set containing an $\mathcal{M}$ that satisfies (1) and (2) and let

$\varphi :=(\exists m \in \mathcal{M})(\exists x$ an ordinal $) [m'\in M:m' < m] \cong x$

This is a $\Sigma_1$ formula which holds in $A$ and hence by reflection there is a set $a \in A$ such that

$\varphi^a :=(\exists m \in \mathcal{M})(\exists x\in a$ an ordinal $) [m'\in M:m' < m] \cong x$

which also holds in $A$.

But then by assumption every finite ordinal is in $a$ and so by $\Delta_1$ separation we have $\omega \in A$ and $o(A) > \omega$, thus contradicting (3).

share|improve this answer
    
Thank you very much. –  user16974 Mar 30 '13 at 8:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.