Can an independent subset of a theory be completed to axiomatize the theory while preserving independence?

Question

Let $L$ be a first-order language, and let $T$ be a theory in $L$, that is, a consistent and deductively closed set of sentences. Let $I$ be a subset of $T$ which is independent. An independent set of sentences is one where no sentence is redundant. We are not assuming that the deductive closure of $I$ is $T$, merely that $I$ is a subset of $T$. My question is, is there always a set $I'$, where $I \subseteq I' \subseteq T$, such that $I'$ is also independent, and the deductive closure of $I'$ is $T$? Or, is there a counterexample for some language $L$ and theory $T$ and set of sentences $I$?

Answer 1

If I am not mistaken, this should be a counterexample:

The language contains the constants $\omega$ and a unary predicate $r(x)$.

Let $S=\{r(i):i\in\omega\}$.

Let $T={\rm cl}(S)$, the closure of $S$ under logical consequences.

Let $I=\{r(0)\vee r(i+1):i\in\omega\}⊆T$.

Clearly $I$ is independent

Let $I'$ be such that $I ⊆ I'⊆ T$ and $T={\rm cl}(I')$.

By compactness, there is some $I''⊆I'$ finite such that $I''\vdash r(0)$.

Then $I''\vdash I$, hence $I'$ is not independent.