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