# If $T$ is a maximal $L$ theory such that every finite subset of $T$ is consistent, then either $\sigma \in T$ or $\lnot \sigma \in T$.

Prove without using compactness that if $$T$$ is a maximal $$L$$ theory such that every finite subset of $$T$$ is consistent, then either $$\sigma \in T$$ or $$\lnot \sigma \in T$$. Here, by maximal, we mean maximal with respect to this property.

Attempt:

Let $$T$$ be an $$L$$-theory with the given property and fix an $$L$$ sentence $$\sigma$$. Suppose both $$\sigma, \lnot \sigma \in T$$. Then $$S = \{\sigma, \lnot \sigma \} \subseteq T$$ is a finite subset of $$T$$, so $$S$$ is consistent. Thus there exists a model of $$S$$, say $$\mathcal{S}$$ such that $$\mathcal{S}$$ satisfies both $$\sigma$$ and $$\lnot \sigma$$. This is impossible, so we conclude that for any $$L$$ sentence $$\sigma$$, at most one of $$\sigma, \lnot \sigma \in T$$.

Suppose $$\sigma, \lnot \sigma$$ are both not in $$T$$. Then consider $$T_1 = T \cup \{\sigma\}$$ and $$T_2 = T \cup \{\lnot \sigma\}$$. $$T$$ is contained in $$T_1$$ and $$T_2$$ and it is maximal, so it must be that $$T_1, T_2$$ each contain finite subsets which are not consistent. These subsets must contain $$\sigma, \lnot \sigma$$ respectively because otherwise they would be finite subsets of $$T$$, which are consistent by assumption. Thus we can write them as $$T' \cup \{\sigma\}, T'' \cup \{\lnot \sigma\}$$ resp.

I am not sure how to proceed from here. Any hints would be great!

• First part: Suppose both σ,¬σ∈T. Then S={σ,¬σ}⊆T is a finite subset of T tha is inconsistent. Contradiction. Nov 19, 2020 at 15:13
• Second part: the two are supersets of T: thus T is not maximal. But... you have to prove that at least one of them is consistent. Nov 19, 2020 at 15:15
• If $T'\cup\{\sigma\}$ and $T''\cup\{\lnot\sigma\}$ are finite, inconsistent theories, then $T'\cup T''\subseteq T$ is a finite, inconsistent theory. Nov 19, 2020 at 15:23
• Why is that theory inconsistent @spaceisdarkgreen?
– John
Nov 19, 2020 at 15:26
• @John Thought you were asking for hints! Nov 19, 2020 at 15:27

For the second part, we will prove that $$T' \cup T''$$ is a finite inconsistent subset of $$T$$. This contradicts $$T$$ having only finite consistent subsets, so we will be done.
$$T′∪T′′∪\{σ\}$$ is inconsistent because it is a superset of $$T′∪\{σ\}$$ and $$T′∪T′′∪\{¬σ\}$$ is inconsistent because it is a superset of $$T′′∪\{¬σ\}$$. Thus $$T′∪T′′$$ must be inconsistent, otherwise if a model existed, then it would satisfy either $$σ$$ or $$¬σ$$, so one of $$T′∪T′′∪\{σ\}, T′∪T′′∪\{¬σ\}$$ would be consistent.
Note : Supersets of inconsistent theories are inconsistent. If a model exists for $$T$$, a theory which contains $$Q$$, then this model satisfies all the $$L$$ sentences in $$T$$, so it also satisfies all the $$L$$ sentences in $$Q$$. Thus it must be a valid model for $$Q$$ as well, so $$Q$$ must be consistent too.