# Lindenbaum's lemma and the hypotesis of deductively closure

In my lectures, we define an Henkin set for a theory $$T$$ as a set $$H$$ of enunciates in a signature $$L$$ that is:

1. Deductively closed for $$T$$: if $$T \vdash A_1,...,A_n \Rightarrow B$$ and $$A_1,...,A_n \in H$$, then $$B \in H$$;

2. Rich: if $$\exists xB \in H$$, then it exists a ground term $$t$$ s.t. $$B(t/x) \in H$$;

3. Maximal: for all $$A$$ enunciate, $$A \in H$$ or $$\neg A \in H$$;

4. Consistent: with the following equivalent definitions: it exists an enunciate $$B$$ s.t. $$B \notin H$$, or for all enunciate $$B$$, $$B \in H$$ and $$\neg B \in H$$ is not acceptable at the same time.

And then we try to prove the Lindebaum's Lemma:

Suppose $$S$$ a set of enunciates consistent, it exists a set $$S^* \supseteq S$$ s.t. $$S^*$$ is consistent and maximal.

My question is: shouldn't we have the hypotesis that $$S$$ is deductively closed for $$T$$ in Lindenbaum's lemma? And shouldn't we use the hypotesis of closure for deduction in order to demonstrate that the two definitions of consistency for a set of enunciates are equivalent?

• Why do you think that you would need the hypothesis of closure for Lindebaum's lemma? The lemma is perfectly stated and proved without it. Commented Aug 12 at 18:58
• Because when we defined the consistency for a set of enunciates in class, we used two definitions that we claimed are equivalent: "it exists an enunciate B s.t. B∉H , or for all enunciate B, B∈H and ¬B∈H is not acceptable at the same time", where I think we are using that this set is already deductively closed Commented Aug 12 at 19:47
• I think my problem is related with this different definition of consistency using ∈, instead of the usual definition of consistency using ⊢ Commented Aug 12 at 19:57
• I suspect you are right. To define consistency as the existence of a statement $\phi$ that does not belong to the set assumes the set corresponds to the set of formulas deducible from a theory. Commented Aug 12 at 20:03