Choice principle strong enough to defy $V=L$

Question

When using axiom of choice in proofs, people often say that this is non-constructive because AC gives us only proofs of existence, without giving explicit example. However, because in $L$ AC holds, we have that for every existence proof using choice there is an explicit example for that in L. So AC is non-constructive, but it sort of is constructible.

I have recently wondered if we could make up some stronger choice principle, which would allow us to make up sets so complex that they aren't constructible, that is, which could let us prove the existence of non-constructible sets, thus making $V=L$ false.

Answer 1

There are principles which imply the axiom of choice, and are inconsistent with $V=L$.

For example, the statement "Every well-ordered chain of cardinals between $X$ and $\mathcal P(X)$ is finite" implies the axiom of choice.¹ Couple that with "Every finite $n$ is embeddable in the cardinals between $X$ and $\mathcal P(X)$, for some $X$", and you get the axiom of choice, but also the negation of $\sf GCH$. Therefore this principle must imply $V\neq L$.

In this spirit we can in fact generate more than a handful of statements which will imply the axiom of choice, but will violate $V=L$.

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1. In fact we just want an ordinal which bounds these sort of chains globally. Of course, Easton showed that the axiom of choice is consistent with the statement that there is no such ordinal; so this sort of statement is in fact stronger than the axiom of choice.

   Truss, John. "On certain arbitrarily long sequences of cardinals." Z. Math. Logik Grundlagen Math. 19 (1973), 209–210. MR317941.