Is the universe conservative?

[Monroe Eskew]

> On 04.10.2020, at 23:28, Annatala Wolf <a.lupine at gmail.com> wrote:
> 
> V = L implies AC. Under ZF(C) (in which you appear to be working based on section 3), AC necessarily implies the existence of sets which cannot be constructed. Why then is AC in the context of ZF, and thus V = L as well, not non-apparent instead?

Instances of AC need not be non-constructive. For example, if we have a family of closed sets of reals, we can choose one member from each by picking the least element. Under V=L, the entire universe is so orderly that witnesses to AC can always be picked according to a rule.

On the other hand, is the non-existence of indiscernibles “apparent”?  This would seem to suggest that we have a method for discerning the ordinals. Another way of putting this is, given the elementary diagram of some L_\alpha, there is no alternative way of interpreting this theory in the same structure. The experience of modern set theory with large cardinals would seem to cast doubt on the existence of such rigid structure.