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 $I$?
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1$\begingroup$ 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 math.stackexchange.com/questions/3289207/…. $\endgroup$– Noah SchweberMay 15 at 19:14
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$\begingroup$ @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. $\endgroup$– Rob ArthanMay 18 at 20:52
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$\begingroup$ @RobArthan You/he might be right, I'll take a look later today. $\endgroup$– Noah SchweberMay 18 at 21:48
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$\begingroup$ @RobArthan Belatedly looks like you're right. $\endgroup$– Noah SchweberMay 24 at 1:20
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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.
