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I many papers by Woodin, and on some answers here on MathOverflow (like the first answer of this question), I see the expression "$L(A,\mathbb{R})$" being used, but I have never seen it defined. I first assumed it meant "$L(A\cup\mathbb{R})$" but on the answer I linked it is precised that $A\subset\mathbb{R}$ so clearly that's not it as it would be $L(\mathbb{R})$ whatever $A$ is. A friend of mine suggested it might rather be $L(\{A,\mathbb{R}\})$ which seems likely (as $A$ might just be not constructible from the reals) but I would like a confirmation by someone who knows.

So, what does it mean?

François G. Dorais
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Julian Barathieu
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    I'd understand this to mean the smallest transitive proper class model of ZF that contains all the reals and $A$ as elements. More explicitly, start with $\omega\cup\mathbb P(\omega)\cup{A}$ (where I assume $A\subseteq\mathbb P(\omega)$), and iterate "definable power set" through all the ordinals (just as in the definition of $L$). – Andreas Blass Sep 28 '17 at 19:58
  • Defining it as $L({A,\mathbb{R}})$ would be exactly starting with (the transitive closure of) what you just said, I think that's it, but I haven't found any explicit statement of this – Julian Barathieu Sep 28 '17 at 20:04
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    Yes, that's it. – Andrés E. Caicedo Sep 28 '17 at 21:40

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