# Tag Info

0

Suppose $X$ is a metric space with a two valued diffused Borel measure. Let $\mathcal{U} = \bigcup \{\mathcal{U}_n : n < \omega\}$ be a basis for $X$ where each $\mathcal{U}_n$ is a disjoint family of open sets - See here. Let $\langle x_i : i < \kappa \rangle$ list $X$. For each $i$, let $B_i$ be a open ball around $x_i$ whose measure is zero. Let ...

0

Note that $\mathbf c$ does not appear at all in the problem statements. Problem 12 asks you to show that $\aleph_1:=|M|>\aleph_0$, Problem 13 asks you to show that there is no $m$ with $\aleph_0<m<\aleph_1$. The Continuum Hypothesis is concerned with the question if there exist cardinalities $m$ with $\aleph_0<m<\mathbb c$ where $\mathbb ... 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, ...

Top 50 recent answers are included