# Tag Info

## New answers tagged large-cardinals

1

My personal favorite - although it seems not to be very popular - is the inner model hypothesis (IMH) and its variants. IMH states that anything that can happen in an inner model, does: specifically, $$\mbox{Any parameter-free \varphi which holds in an inner model of some outer model of V,}$$ $$\mbox{already holds in some inner model of V.}$$ (Of ...

1

Carl's comments are exactly right. However, in a different sense there are indeed serious obstacles to finding "explicit" sets with intermediate cardinality: Every analytic set $X\subseteq \mathbb{R}$ satisfies the perfect set property - that is, either is countable or contains a nonempty closed set with no isolated points (and hence has size continuum). ...

5

Supercompactness is much stronger than measurability: Supercompactness of $\kappa$ states that for any $\mu$, we can find embeddings $j:V\to M$ of the universe of sets with $M$ transitive, critical point $\kappa$ and $M$ closed under $\mu$ sequences. This implies in particular that for any set $x\in V$, we can find one such embedding $j$ with $x\in M$ (...

8

Let $M$ be the model you start out with. If $M$ has no inaccessible cardinals, then you're done. Otherwise let $\kappa$ be the first inaccessible cardinal in $M$. Presumably you already know that the model's $V_\kappa$ is a model of ZFC, when viewed from $M$ itself and therefore also (why?) when viewed from outside the model. You then just need to prove ...

2

It is enough to show that actually $E_\alpha\cap\kappa$ is stationary by induction on $\alpha<\kappa$. The case $\alpha=0$ follows from the fact that $\kappa$ is weakly inaccesible, and thus the set of limit cardinals less than $\kappa$ is closed unbounded. There are two cases: Suppose $E_\alpha\cap\kappa$ is stationary. Then $(E_\alpha\cap\kappa)'$ ...

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