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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 ?

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    Since $\omega_1^{L[x]}$ is countable, we can take $y$ which encodes a bijection between it and $\omega$. – 2017-01-18

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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 $y=x\oplus z$.

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    Can the coding of $\alpha$ by $x$ be $\Sigma_0$, so that $L[x]$ knows about it ? (Once I make sure I understand why $\omega_1$ is the sup, I get the rest of your argument, so I'll validate your answer afterwards, thanks !) – 2017-01-18
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    @Maxtimax Yes, you can. Let $R$ be a binary relation on $\omega$ such that $(\omega, R)$ is a well-ordering of ordertype $\alpha$. (Technically we need to assume that $\alpha$ is infinite for this; I leave it as an exercise to address the case of $\alpha$ finite :P.) Then consider the real $x\in\omega^\omega$ defined as follows: the $\langle m, n\rangle$-th bit of $x$ is $0$ if $mRn$, and is $1$ otherwise. (Here "$\langle\cdot,\cdot\rangle$" represents your favorite pairing function on $\omega$.) Note that, in a similar way, any countable structure (in a countable language) is coded by a real. – 2017-01-18
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    (I've replaced "$\mathbb{R}$" with "$\omega^\omega$" to more exactly match the question - however, there are nice bijections between each of the usual "versions" of the reals ($\mathbb{R}, 2^\omega,\omega^\omega, ...)$, so it really doesn't matter.) – 2017-01-18
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    Yes that I understood, but then you would have (letting $V$ denote a universe in which AD holds) $V\models x$ codes $\alpha$, but how do we know that $L[x]$ satisfies this as well ? Because as you define it, I don't see how "$x$ codes $y$" would be a $\Sigma_0$ relation. Maybe when you compute the ordinal that's coded by $x$ in $L[x]$, you obtain some $\beta$ far smaller than $\alpha$. Maybe there's something I'm getting wrong ? – 2017-01-18
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    @Maxtimax $L[x]$ thinks $x$ codes a well-ordering, since it's correct about wellfoundedness. So since $L[x]$ satisfies Replacement, $L[x]$ has an order-preserving bijection $f$ between (the structure coded by) $x$ and $\beta$, for some ordinal $\beta$. Now the property of being an order-preserving bijection is absolute between $L[x]$ and $V$ (exercise), so in $V$ we have that $x$ codes $\beta$. But this means $\beta=\alpha$. – 2017-01-18
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    Ok I managed to prove what was left ! Thanks again for your help ! – 2017-01-19
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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.

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    I already knew that $\omega^{L[x]} < \omega_1$ – 2017-01-18
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    Then if there is some $x$ such that for all $y$, $\omega_1^{L[y]}\leq\omega_1^{L[x]}$, then it has to be the case that $\omega_1=\omega_1^{L[x]}$. – 2017-01-18
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    That follows from Noah's argument, but it is clearly not immediate: how do you know this ? – 2017-01-18
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    All the $\omega_1^{L[x]}$ are comparable as ordinals; and $L[x]$ is always a model of choice, so there is an injection from $\omega_1^{L[x]}$ into the reals (of that model, and of $V$). But the working assumption is that the latter fails. – 2017-01-18
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    No what I mean is why does "$\forall y, \omega_1^{L[y]} \leq \omega_1^{L[x]}$" imply "$\omega_1^{L[x]} = \omega_1$" ? I know how to derive a contradiction from this, as I stated both in my question and in the comments. – 2017-01-18
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    Because if $\omega_1^{L[x]}$ is countable, this fact can be coded by a real $y$, in which case $\omega_1^{L[x]}<\omega_1^{L[y]}$. – 2017-01-19