Question on maximum consistent theory

Question

Lindenbaum theorem states that any consistent set of sentences can be enlarged to a maximal consistent set of sentences. Thus, by choosing a consistent set of sentences in Peano arithmetic and by Lindenbaum theorem, they can be enlarged to a complete theory. However, this contradicts that Peano arithmetic is incomplete. Any explanation?

Answer 1

There's no contradiction at all: the completion is not the same as the theory being completed.

Peano arithmetic (by which I presume you mean first-order Peano arithmetic) $\mathsf{PA}$ is consistent and incomplete. Lindenbaum says that since $\mathsf{PA}$ is consistent, there is a consistent complete $T$ with $\mathsf{PA}\subseteq T$. This $T$ is not the same as $\mathsf{PA}$ itself, so there's no contradiction between $T$ being complete and $\mathsf{PA}$ being incomplete. (Also note that $\mathsf{PA}$ has many different complete consistent extensions.)

Now $\mathsf{PA}$ does have a "weak non-completeability" property: no computably axiomatizable consistent extension of $\mathsf{PA}$ can be complete (this result is Rosser's improvement of Godel's first incompleteness theorem, and this property is called "essential incompleteness"). However, this doesn't contradict Lindenbaum, it merely means that the above $T$s cannot be computably axiomatizable.