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

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 $$I$$?

• 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/…. May 15 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. May 18 at 20:52
• @RobArthan You/he might be right, I'll take a look later today. May 18 at 21:48
• @RobArthan Belatedly looks like you're right. May 24 at 1:20

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.