## 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.