Is the set of sentences in peano arithmetic countably infinite?

Question

Im wondering if the set of all sentences, which would be wffs in which there are no free variables, is countable infinite or not. I imagine that since the number of axioms in PA is countably infinite, that the number of symbol strings might be uncountably infinite, as their set seems similar to the power set of a countably infinite set. However this doesnt seem to help me conclude whether or not the number of sentences is countable infinite. Thanks!

Answer 1

If $C_1$ and $C_2$ are countable sets, then $C_1\times C_2$ is countable too. Let $A$ be your countable alphabet of symbols. Let $A^1=A$ and define recursively $A^n=(A^{n-1})\times A$. For all $n$, $A_n$ is countable. The countable union of countable sets is countable, so the set of all finite strings from the alphabet, which is $$\bigcup_{n=1}^\infty A^n$$ is countable. Since every sentence is a string of symbols fom a countable alphabet, the answer is yes.