Let A be a countable system of axioms with an infinite model. Then A has a model of every lower infinite cardinality.

In particular, A has a countable model.

Method of proof:
For the countable case, here's how to "build" a model. The objects of the model will be based on the set of all finite combinations of functions (and constants, which are just 0-adic functions!) of the language of A. This is a countable set. But some of these objects might be equal (e.g. "(SS0)+(SS0) = SSSS0" in Peano arithmetic), so just check all these in the given model, and "divide" the set by this equivalence relation, to get our countable model's set of objects (it needs to be proved that this set remains countable).

Similar actions need to be taken for every predicate in the language, then for all combinations of quantifiers and formulae. Eventually you're left with a countable model; it's easy but tedious to prove it's indeed a countable model.

Note that to prove the case of intermediate (uncountable) cardinalities, more work is needed.

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