On Monday, May 5, 2014 3:07:48 AM UTC-7, muec...@rz.fh-augsburg.de wrote: > On Sunday, 4 May 2014 20:53:17 UTC+2, Zeit Geist wrote: > > > > I use the well-established theorem of ZF that every infinite subset of a countable set is countable. > > > But ignore the Theorem that Not All Subsets of a Definable Countable Set are Definable! > > No I show a contradiction in set theory.
You do no which thing.
Produce the Bijection that shows that the Set is Countable.
> > You might Not be able to, and in fact CAN'T Define that Bijection needed to show the Set is Countable. > > But I can prove countability of that set since every definition is a finite expression.
That is Not possible.
> > There is NO Formula that Defines ALL the Reals that are Definable by a Formula! > > Of course. There are not *all* reals. But if someone accepts finished infinity, then all finite expressions exist, are countable, and contain as a subset all finite definitions.
If you say. We know what "all reals" means.
Actually, if you assume that there are No Underinable Reals and that the is a Definable List of All the Definable Reals, then you Prove Anything. For, that System is Inconsistent!
> > However, then we can Prove that there Exists an Undefinable Real Number. > > No, you cannot, and nobody has ever done it. Everything you "can prove" is defined at least by the space-time coordinates of you when doing this "proof".