What is K?

The core model K is, among other things, great for providing lower bounds for the consistency strength of some interesting theory. For instance, every Jónsson cardinal is Ramsey in K, showing that the two are equiconsistent. Usually there are a lot of different K’s with different properties, sometimes denoted by things like K^{DJ}, K^{Steel} or something along those lines. This is my attempt at (briefly) explaining what K is, and what the differences between the various versions are. Here’s an overview of the various core models I’ll cover:

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From Determinacy to a Woodin I

In my previous posts I provided a sketch of how a measurable above a limit of Woodins implies that \textsf{AD} holds in L(\mathbb R). The “converse”, saying that \textsf{AD} implies that there is a model with infinitely many Woodin cardinals, is a lot more complicated. I will try to simplify a lot of these complications here, to give an idea of what is going on. I will only focus on showing the existence of a single Woodin (for now), where the Woodin in question will be \Theta^{L(\mathbb R)} inside of \text{HOD}^{L(\mathbb R)}. As always, I will be very sketchy in this blog post, but provide more details in my note.

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Hahn-Banach sans Zorn

The Hahn-Banach Theorem in functional analysis is the theorem saying more or less that normed vector spaces have many bounded linear functionals. The theorem is usually proven via Zorn’s lemma, giving the impression that Hahn-Banach uses the full power of choice. I’ll here give a proof based on the ultrafilter lemma, that every filter can be extended to an ultrafilter. One of the benefits of this approach, besides relying on a (strictly) weaker assumption, is that we get a little more information about how the functionals are constructed. The idea is that the ultrafilters allow us to “glue” functionals together. A latex’ed version can be found here.

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Choiceless non-free algebras

This post is a bit different, as it’s somewhat more of a curiosity I recently noticed. It’s a theorem of \textsf{ZFC} that every type of of algebraic structure has a free algebra, similar to the notion of a free group, free module and so on. How about without choice? It turns out that the theory

\textsf{ZF}+\textsf{F}:\equiv\textsf{ZF}+"\text{There is an algebraic type which does not admit free algebras}"

has considerable consistency strength. It consistency-wise implies \textsf{AD}^{L(\mathbb R)} and is consistency-wise implied by a proper class of strongly compacts.

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