= 27d567bc2d48d9f40e9ccc20ea370b15)
MrMind comments on Open Thread, September 30 - October 6, 2013 - 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 (295)
What Quinn said.
Throwing darts at immeasurable set was a technique used to 'prove' the continuum hypothesis.
Could you give a reference? Are you assuming choice?
Sorry, it was used to 'disprove' the continuum hypothesis.
It's the Freiling axiom.
Thanks! I think it's quite reasonable to reject choice and take as an axiom that all sets are measurable, so I'm interested in the consequences of it. I'd always been told that the continuum hypothesis is orthogonal to everything people care about, but that's only after assuming choice.
There's also the Axiom of Determinacy that rejects Choice and, when paired with the existence of a very strong measurable cardinal, gives a very broad class of measurable sets.
Could you give an example of a set whose measurability I might care about, other than subsets of R? something for random processes?
Could you give a reference for the combination?
Well, I guess this pretty much depends on the area you're working on. I'm interested in the foundation of mathematics, for which measurable sets are of big importance (for example, they are the smallest critical point of embedding of transitive models, or they are the smallest large cardinal property that cannot be shown to exists inside the smallest inner model). Outside of that area, I guess the interest is all about R and descriptive set theory.
Edit: It's not true that measurable cardinals are the smallest large cardinals that do not exists in L. Technically, the consistency strength is called 0#, and between that and measurables there are Ramsey cardinals.
Well, the definitive source is Kanamori's book "The higher infinite", but it's advanced. Some interesting things can be scooped up from Wikipedia's entry about the axiom of determinacy.
CH is orthogonal to ZF. CH is orthogonal to ZFC.
If ZFC is inconsistent, then ZF is also inconsistent.
AC is orthogonal to CH.
That doesn't actually answer Douglas's statement that the continuum hypothesis is orthogonal to everything people care about if one assumes choice. In fact Doug's statement is more or less correct. See in particular discussion here. In particular, ZF + CH implies choice for sets of real numbers, which is what we care about for most practical purposes.
A comment at your link baldly asserts that ZF+CH implies choice for sets of real numbers, but the link seems otherwise irrelevant. Do you have a better citation? In particular, what do you mean by CH without choice? In fact, the comment asserts that ZF+CH implies R is well-orderable, which I don't think is true under weaker notions of CH.
CH in that context then is just that there are no sets of cardinality between that of R and N. You can't phrase it in terms of alephs (since without choice alephs aren't necessarily well-defined). As for a citation, I think Caicedo's argument here can be adopted to prove the statement in question.
I said that I doubt your claim, so blog posts proving different things aren't very convincing. Maybe I'm confused by the difference between choice and well-ordering, but imprecise sources aren't going to clear that up.
In fact, it was Caicedo's post that lead me to doubt Buie. Everything Caicedo says is local. In particular, he says that CH(S) and CH(2^S) imply that S is well-orderable. Buie makes a stronger specific claim that CH implies R is well-orderable, which sounds like a stronger specific claim, unlikely to be proved by local methods. I guess it is not exactly stronger, though, because the hypothesis is a little different (CH=CH(N), not CH(R)).
Alephs are defined without choice. They are bijective equivalence classes of ordinals. In any event, ℵ_1 is the union of countable ordinals. Sometimes they are called cardinals.
It is widely reported that the (weak) CH is the that every uncountable subset of the reals is bijective with the reals, while strong CH is that the reals are bijective with ℵ_1. I think you and Buie are simply confusing the two statements.
Also, sometimes people use "weak continuum hypothesis" to mean 2^ℵ_0 < 2^ℵ_1; I think it is strictly weaker than the statement that there are no sets between ℵ_0 and 2^ℵ_0.
Hmm, that does make it seem like I may be confused here. Possible repaired statement would then use GCH in some form rather than just CH and that should go through then since GCH will imply that for all infinite S, CH(S) and CH(2^S), which will allow one to use Caicedo's argument. Does that at least go through?
I think you are likely correct here.