Is the universe conservative?

[martdowd at aol.com]

FOM

Monroe Eskew writes

The experience of modern set theory with large cardinals would seem to cast doubt on the existence of such rigid structure. 
 This is indeed a compelling argument on the "other side" of the question.  As a starting point it suffices to consider K^DJ, which equals L unless 0# holds, which is so iff some L_\alpha has uncountably many indiscernibles.  I titled the paper as a question, because as usual there's no methodology for "dispensing with" the question.  However it is of interest to try to understand why we can neither "see" indiscernibles, nor prove they don't exist.  Why should they "creep in" when the cumulative hierarchy is extended to an omega_1-Erdos cardinal?  Why aren't they apparent, independently of any large cardinal hypotheses?  K^DJ depends on 0#, so 0# seems to be an adequate "counter-principle" for trying to gain some further understanding of the situation.
Martin Dowd

 
 
-----Original Message-----
From: Monroe Eskew <monroe.eskew at univie.ac.at>
To: fom at cs.nyu.edu
Sent: Sun, Oct 4, 2020 11:09 pm
Subject: Re: Is the universe conservative?


> 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. 
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