116c- Lecture 18

We briefly discussed relative constructibility and compared the models L[x] where x is exclusively treated as a predicate with the models L(x) where x is an element. In particular, L[x] is a model of choice but L(x) may fail to be.

An amusing application of the fact that L[x]\models{\sf AC} is that the result of Exercise 3 from Homework 7 holds in {\sf ZF}, although the proof I wrote there uses choice. Namely, work in {\sf ZF} and consider two well-orderings of a set X. We can assume that X is an ordinal \alpha and the first well-ordering is \in. Let \prec be the second well-ordering. Then \prec\in L[\prec] (since \prec is a set of ordered pairs of ordinals). In L[\prec], where choice holds, (and therefore also in V) there is a subset of \alpha of the same size as \alpha and where \prec coincides with \in.

Question. Find a `choice-free’ argument for Exercise 3. 

The main example we will consider of a model of the form L(x) is L({\mathbb R}), due to its connection with determinacy.

We introduced the setting to discuss determinacy, namely infinite 2-person games with perfect information. We proved the Gale-Stewart theorem that open games are determined and discussed Martin’s extension to Borel games. A nice reference for the proof of Martin’s result (using the idea of `unraveling‘, which reduces any Borel game to an open game in a different space) is Kechris’s book on descriptive set theory.

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One Response to 116c- Lecture 18

  1. […] I present here a quick sketch of the solution of Exercise 3.(b). See Lecture 18, where it is shown that the result actually holds in , although the proof uses […]

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