A sentence is a first-order formula without any free variables. Let $A$ and $B$ be sets of first-order sentences such that $A\cup B$ is unsatisfiable (i.e. there does not exist a valuation that satisfies it). Prove that there exists a first-order sentence $C$ so that
- every model that satisfies $A$ also satisfies $C$ (i.e. $A\vdash C$, and
- every model that satisfies $B$ also satisfies $\neg C$ (i.e. $B\vdash \neg C$).
I think this might relate to the consistency of $A$ and $B$; if a set of sentences $A$ is inconsistent, then there is a set of sentences $C$ so that $A\vdash C$ and $A\vdash \neg C$ (i.e. $A\vdash \perp$). Since $A\cup B$ is unsatisfiable, it seems this can be shown, but I can only conclude that there exists a set of sentences $C$ so that $A\cup B \vdash C$ and $A\cup B\vdash \neg C.$ It also seems that I may need to prove that unsatisfiability implies inconsistency.
Suppose $A\cup B$ is unsatisfiable. By the compactness theorem, there are finitely many sentences $A_1,\dots,A_n\in A$ and $B_1,\dots,B_m\in B$ such that $\{A_1,\dots,A_n,B_1,\dots,B_m\}$ is unsatisfiable. Let $C = \bigwedge_{i=1}^n A_i$ (and note that $C = \lnot \left(\bigwedge_{i=1}^m B_i\right)$ would work just as well).
Topologically, this exercise says that in the Stone space of complete $L$-theories, any two disjoint closed sets are separated by a clopen set and its complement.