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



Barwise compactness theorem
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). 


