Tag Info

New answers tagged

0

Different models of set theory have different sets of ordinals. In particular if $0^\#$ exists, the $V$ and $L$ have very different sets of ordinals. For example $\{\omega_n^V\mid n\in\omega\}$ is a set in $V$ but not in $L$. But you don't need you go as far as $0^\#$ to see that regularity, or even being a cardinal, is not absolute. Simply force with ...


1

Your first conclusion is the correct one, the point is that being a cardinal is not absolute. So it may exists $\alpha<\omega_1$, so that $L\models \alpha=(\aleph_1)^L$, simply because there is no bijection between $\alpha$ and $\omega$ in $L$. Even worst, it may exist a strictly increasing sequence of countable ordinals $\alpha_n$ such that $L\models" ...


3

First, let's clear up the easy part. Suppose $0^\#$ exists, then we can work in $L[0^\#]$, where it also exists. If there are any large cardinals let in that model (weak compact, inaccessible, even worldly cardinals) we can chop the universe at that cardinal, to have a large cardinals free universe where $0^\#$ exists just fine. For the first question, ...


7

Your theory is equiconsistent with what is known as the Levy scheme, or Ord is Mahlo, which is strictly weaker in consistency strength than the existence of a Mahlo cardinal. On the one hand, your theory implies Ord is Mahlo, since if $C$ is any proper class definable club (definable with parameters), then it will follow that $C$ is unbounded in the ...


8

No, this is not inconsistent, at least not relative to the existence of a Mahlo cardinal. Suppose that $\kappa$ is a Mahlo cardinal, then $V_\kappa$ has a club of ordinals $\alpha$ such that $V_\alpha\prec V_\kappa$. Therefore there is a stationary set of inaccessible cardinals satisfying this. Simply enumerate these inaccessible cardinals and let $\alpha$ ...



Top 50 recent answers are included