
Re: On the infinite binary Tree
Posted:
Dec 13, 2012 2:47 PM


On Dec 13, 3:56 pm, WM <mueck...@rz.fhaugsburg.de> wrote: > On 13 Dez., 11:49, Zuhair <zaljo...@gmail.com> wrote: > > > On Dec 13, 9:56 am, WM <mueck...@rz.fhaugsburg.de> wrote: > > > > It is nonsensical because the same could be assumed for Cantor's > > > diagonal. It would be undefinable and it would be impossible to prove > > > that it differs from all lines of the list  in particular if > > > undefinable reals exist and are members of the list. > > > Of course the diagonal in some cases can be nondefinable, that is > > well known. > > Also the elements of the list could be nondefinable, if nondefinable > real numbers existed. > > > That doesn't mean that we cannot prove it is different from all reals > > in the list, on the contrary we don't need parameter free definability > > in order to determine that the diagonal is different from the reals in > > the original list, we can do that without it, as Cantor did. > > Cantor did not accept nondefinable reals. If he had, he would have > seen that his proof fails. > > No Cantor's proof survives non parameter free definability. We don't need every real to be definable by a parameter free formula in order for Cantor's proof to go through. That's your simple mistake, you think Cantor's proof requires that all reals must be parameter free definable, but this is not the case. Cantor's proof works in a flawless manner even if MOST of the reals are non parameter free definable. Actually Cantor's proof mounts to the conclusion that MOST reals are non parameter free definable reals, of course he saw that, this is obvious really.
Zuhair > > > Your error is that you think too much of non definability. It is not > > so destructive as you think. > > No? Nearly every real number is undefinable. The measure of definable > reals is 0. If most reals are nondefinable, why must all reals of > every Cantor list always be definable? If all reals of the list are > definable, then they belong to a countable set. Then we cannot prove > uncounatbility. Or can we prove that the set of definable reals is > uncountable  because it is countable but there are, somewhere else, > undefinable "reals"? > > Regards, WM

