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Topic:
wellordering the reals
Replies:
2
Last Post:
Jan 28, 2013 10:44 PM




Re: wellordering the reals
Posted:
Jan 28, 2013 10:44 PM


On Jan 28, 2:34 pm, FredJeffries <fredjeffr...@gmail.com> wrote: > On Jan 28, 1:50 pm, Paul <pepste...@gmail.com> wrote: > > > It's often said that "you can't find an explicit wellordering of the real numbers." Is this just an informal way of saying that you can't prove that the reals can be wellordered using ZF (without C)? > > > What is a reference for proofs of this and related results? > > I believe that the standard reference is: > Saul Feferman "Some applications of the notions of forcing and generic > sets"http://matwbn.icm.edu.pl/ksiazki/fm/fm56/fm56129.pdf > > "No settheoretically definable well ordering of the continuum can be > proved > to exist from the ZermeloFraenkel axioms together with the axiom of > choice > and the generalized continuum hypothesis."
Then, is it so that Feferman has that iff ZFC + GCH + "V = L" that there is a "definable" wellordering of the reals? How about ZFC + "V = L", and, is "V = L" independent of ZF, or for that matter are large cardinals, if they decide features?
If ~CH then there would be cardinals between those of N^n and N^N. If so they would have initial ordinals and those, ordinal structures, defined by their elements, but then CH couldn't be independent of ZF, because they would be ordinals in ZF, and already exist (or not) in the cumulative hierarchy, of ordinals.
The range of this equivalency function or "EF" is of reals in their normal order.
Yes, a wellordering of any set, including the reals, exists in ZFC. Yet, then the transfinite case of nested intervals, in plugging the gap, if it's simply over limit ordinals, it's simply over limit ordinals. Where it's not a countable limit ordinal, then there would be at least uncountably many: in their normal order.
The reals, the real numbers of the continuum of real numbers, are drawn in a line.
Regards,
Ross Finlayson



