$\omega_1$ as seen by the reals

Question

Assume that there is no injection $\omega_1 \to 2^\omega$ (in a universe where choice fails, in particular in a universe where AD -the axiom of determinacy holds). Then one can see that for all $x\in \omega^\omega$, $\omega_1^{L[x]} < \omega_1$ (where $L[x]$ is -in this particular case- the smallest inner model that contains $x$ and all the ordinals). However, in Set Theory, Jech claims that for each $x \in \omega^\omega$, one can find $y \in \omega^\omega$ such that $x \in L[y]$ and $\omega_1^{L[x]} <\omega_1^{L[y]}$. He doesn't prove it and simply says it like that. However,I can't think of any reason why that would be obvious. Am I missing something ? Can anyone help ?

Answer 1

First, note that - although $\omega_1\not=\omega_1^{L[x]}$ for any real $x$ - we do always have $\omega_1=\sup\{\omega_1^{L[x]}: x\in\omega^\omega\}$. This is because if $\alpha$ is a countable ordinal, then there is a real $x$ coding a copy of $\alpha$ and so $\alpha<\omega_1^{L[x]}$.

As a corollary, we have:

For all $x\in\omega^\omega$, there is some $z\in\omega^\omega$ such that $\omega_1^{L[x]}< \omega_1^{L[z]}$.

OK, fix a real $x$ and let $z$ be as above; now set $y=x\oplus z$.

Answer 2

Assume otherwise, then there is some $x$ such that $L[x]$ computes $\omega_1$ correctly. But then there is an injection from $\omega_1$ into the reals.