$\Sigma_{1}$-embedding in models of bounded theories

Question

Let language L contains $ \leq $ and countable set of constants symbols $c_{1}, c_{2}, ... $. Is there a bounded theory T in language L, With the following property? For every model A of T, there is model B of T such that $B \preceq_{\Sigma_{1}} A $ and $ c_{i}$'s are cofinal in B. Bounded Theory means a Theory that can be axiomatized by bounded formulas. A quantifier free formula is a bounded formula and if $\varphi(x)$ is bounded, then so are $\exists x ( x\leq t \wedge \varphi (x))$ and $\forall x (x \leq t \to \varphi (x) )$, Where t is a term in language L.

Answer 1

Edit: I assumed in the answer below that a "bounded theory" is axiomatized by a set of bounded sentences. But after reading the quote from Buss in your comment below, it's clear that this is not the intended definition. Rather, a bounded theory is axiomatized by a set of bounded formulas, and there's a convention in play that a formula with free variables should be interpreted as its universal closure.

So a bounded theory is one that's axiomaized by sentences of the form $\forall \overline{x}\, \psi(\overline{x})$, where $\psi$ is a bounded formula. In particular, any universal theory (like the theory of linear orders) is a bounded theory.

That means this is not an answer to the question. But I'll leave it here, since someone might find it interesting.

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It's implicit in your question (since you say "cofinal") that you want $\leq$ to be a (partial) order on every model of $T$. But no consistent bounded theory implies that $\leq$ is a partial order. So this answer won't really answer your main question, but it shows you at least need to reformulate the question.

Indeed, let $M$ be a model of $T$. Let $M'$ be any structure containing $M$ such that the elements in $M'\setminus M$ are not $\leq$-related to any elements of $M$. In particular, you can make the relation $\leq$ fail to be reflexive or transitive or both on these new elements.

Now you can prove the following claim by induction: for any bounded formula $\varphi(\overline{x})$ and any $\overline{a}$ in $M$, $M\models\varphi(\overline{a})$ if and only if $M'\models\varphi(\overline{a})$. In particular $M'\models T$.

The quantifier-free case is easy. The quantifier cases are handled by the observation that $M$ contains all the terms with parameters in $M$, so any witness for a bounded quantifier has to come from $M$ (since no new element is $\leq$-related to any of these terms).