= 27d567bc2d48d9f40e9ccc20ea370b15)
Mark_Eichenlaub comments on Stupid Questions Open Thread Round 2 - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (208)
Yes. In ZF one can construct an explicit well-ordering of L(alpha) for any alpha; see e.g. Kunen ch VI section 4. The natural numbers are in L(omega) and so the constructible real numbers are in L(omega+k) for some finite k whose value depends on exactly how you define the real numbers; so a well-ordering of L(omega+k) gives you a well ordering of R intersect L.
I'm not convinced that R intersect L deserves the name of "the-real-numbers-as-we-know-them", though.
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...
Let's see. Assume measurability axiom - every subset of R has Lebesgue measure. As we can use the usual construction of unmeasurable set on L intersect R, our only escape option is that it has zero measure.
So if we assume measurability, L intersect R is a dense zero-measure subset, just like Q. These are the reals we can know individually, but not the reals-as-a-whole that we know...
Seems reasonable to me.
While some of the parent turns out not to hold, it helped me to find out what the theory really says (now that I have time).