Is there a countable set that is not constructible?

Question

The title pretty much explains my question. I've been reading into basic set theory the past couple days and I was wondering if in ZFC we can prove the existence of a countable set that is non-constructible.

Answer 1

It depends what you mean by "constructible."

In set theory, constructibility has a very specific technical meaning according to which it is consistent with $\mathsf{ZFC}$ that every set (countable or not) is constructible, assuming of course that $\mathsf{ZFC}$ itself is consistent in the first place.

However, this notion of constructibility has nothing to do with constructive logic, nor with any sort of "computational" notion of construction. It's certainly the case that $\mathsf{ZFC}$ proves the existence of high-complexity (e.g. not computably enumerable) sets, but we shouldn't use the term "constructible" here.