Quantifier Elimination in the quantifier free type

323 Views Asked by At

Show that if in the theory of $T$ every quantifier-free type has a unique extension to a complete type, then $T$ has quantifier elimination.

1

There are 1 best solutions below

0
On

Hint for abstract proof approach: continuous bijections between compact Hausdorff spaces are homeomorphisms, and the Stone Representation Theorem is true.

Hint for concrete proof approach: you're going to need to use the compactness theorem somewhere. Probably twice. Start by considering, for a formula $\varphi(x)$, the set of quantifier free formulas that $\varphi(x)$ implies.