It is!? Does anyone know a proof of Compactness that doesn’t use completeness as a lemma?
Yes: instead of proving “A theory that has no model has a finite proof of a contradiction” you prove the equivalent converse “If every finite subset of a theory has a model, then the theory has a model” (which is why the theorem is named after compactness at all) by constructing a chain of model and showing that the limit of the chain has a model.
Also the original proof of Goedel used Compactness to prove Completeness.
Yes: instead of proving “A theory that has no model has a finite proof of a contradiction” you prove the equivalent converse “If every finite subset of a theory has a model, then the theory has a model” (which is why the theorem is named after compactness at all) by constructing a chain of model and showing that the limit of the chain has a model. Also the original proof of Goedel used Compactness to prove Completeness.