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vi21maobk9vp comments on Stupid Questions Open Thread Round 2 - Less Wrong
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Separate concern: Why constructible real numbers are only finitely higher than Q? Cannot it be that there are some elements of (say) 2^Q that cannot be pinpointed until a much higher ordinal?
Of course, there is still a formula that specifies a high enough ordinal to contain all members of R that are actually constructible.
I figured out the following after passing the Society of Actuaries exam on probability (woot!) when I had time to follow the reference in the grandparent:
The proof that |R|=|2^omega| almost certainly holds in L. And gjm may have gotten confused in part because L(omega+1) seems like a natural analog of 2^omega. It contains every subset of omega we can define using finitely many parameters from earlier stages. But every subset of omega qualifies as a subset of every later stage L(a>omega), so it can exist as an element in L(a+1) if we can define it using parameters from L(a).
As another likely point of confusion, we can show that for each individual subset, a<omega1 if it exists at all. That just means the union L(omega1) of their stages contains every element of 2^omega in L. Now for any ordinal b, |L(b)|=|b|. Since we can still show that |2^x|>|x|, this says if V=L then 2^omega must stay within or equal L(omega1). The same proof tells us that L satisfies the generalized continuum hypothesis.
Um. You might well be right. I'll have to think about that some more. It's years since I studied this stuff...