Generalization of a previous model theory question regarding independent sets of sentences.

Question

This is a follow up to and a generalization of my previous model theory question, here: Can an independent subset of a theory be completed to axiomatize the theory while preserving independence?. Let $L$ be a first-order theory. Let $S$ be a consistent set of sentences, this time not necessarily deductively closed, unlike in the previous question. Let $I$ be an independent subset of $S$. (As before, independence means no redundant sentences). Does there exist a set $I'$ such that $I \subseteq I' \subseteq S$, $I'$ is independent, and the deductive closure of $I'$ is the same as the deductive closure of $S$? This time, I am inclined to think the answer is no. There should be a counterexample for some language $L$ and some small finite set $S$ and subset $I$.

Answer 1

You are correct that there is a counterexample to your claim. I don't have a finite set $S$, but here is a not-too-complicated infinite one.

Take $L = \{a_1, a_2, \ldots\}$ where the $a_i$'s are distinct constant symbols. For each $i \in \mathbb{N}$, let $\sigma_i$ be the sentence expressing that $a_1, \ldots, a_i$ are all distinct. Let $S = \{\sigma_i : i \in \mathbb{N}\}$.

Observe that the only independent subsets of $S$ are $\emptyset$ and singletons (i.e., sets of the form $\{\sigma_i\}$ for some $i$). This is because if $i < j$ then $\sigma_j$ implies $\sigma_i$. On the other hand, $\{\sigma_i\}$ does not have the same deductive closure as $S$, because $S \vdash a_1 \neq a_{i+1}$ while $\{\sigma_i\} \not\vdash a_1 \neq a_{i+1}$.

This shows that no subset of $S$ is both independent and has the same deductive closure as $S$; in particular, $S$ is a counterexample to your statement no matter what $I$ you choose.