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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 $I$?

user107952
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    If $T$ is countable, the usual strategy for giving an independent axiomatization also gives an affirmative answer to this question. If $T$ is uncountable, then even proving the existence of an independent axiomatization in the first place is hard. See https://math.stackexchange.com/questions/3289207/nonexistence-of-independent-axiomatization-of-a-theory-in-a-finite-language/3289290#3289290. – Noah Schweber May 15 '23 at 19:14
  • @NoahSchweber: I don't think the usual strategy for giving an independent axiomatization delivers the goods if it is required to give a result that contains some given subset of the theory. Primo Petrii's answer looks good to me. – Rob Arthan May 18 '23 at 20:52
  • @RobArthan You/he might be right, I'll take a look later today. – Noah Schweber May 18 '23 at 21:48
  • @RobArthan Belatedly looks like you're right. – Noah Schweber May 24 '23 at 01:20

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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.

Primo Petri
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