I'm working through various presentations of Henkin-style completeness proofs for first-order logic, where the underlying first-order language need not be countable. I understand most of the technicalities without any problems, but for one minor point.

Let a theory in a first-order language L be a set of L-sentences that is closed under derivability. Let a theory be consistent, if no contradiction can be derived from it. Finally let a theory be maximally consistent, if it is consistent and it is identical to each of its consistent supersets.

Now in the course of proving Lindenbaum's lemma (each consistent theory is contained in a maximally consistent theory) the usual move is to use the followong claim: If T is a set of theories in L linearly ordered by inclusion, then T's union (the set of those things that are members of some member of T) is a theory.

I'm not sure how to prove this claim. Any pointers?


Let U be the union of T. Any derivation U ⊢ A uses a finite number of assumptions from U. Since the set T of theories is linearly ordered by inclusion, there is a theory T' ∈ T containing all these assumptions, so that T' ⊢ A. Since T' is a theory, A ≠ ⊥. Also since T' is a theory, A ∈ T', which gives A ∈ U. Thus U is a consistent and closed under derivations. Thus U is a theory.

EDIT: Above I assumed that a theory is consistent, which the question did not require. Allowing inconsistent theories, the proof still works (with the obvious changes).

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