In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 6 Apr., 21:54, Virgil <vir...@ligriv.com> wrote: > > > > You were right. The reason is: There are at most countably many finite > > > definitions like e = SUM1/n!. That is undisputed. So if there should > > > be uncountably many reals, most of them cannot be defined - or can > > > only be defined by infinite sequences. > > > > Pi and sqrt(2) are both reals defined other than by infinite sequences. > > Of course. They exist as elements of the countable set of finite > names. > > > > > But that means the same as > > > being undefined, because none of those sequences defines a number > > > unless you know the last digit - which is impossible. > > > > Thus WM here claims that neither pi nor sqrt(2) can be defined! > > You err.
It is you who have claimed that numbers whose complete decmal expansions cannot be know are undefined, so that is your error.
> > > Now WM claims that neither pi nor sqrt(2) cannot be used in math because > > they cannot be communicated. > > You err.
Nope! YOU DO! > > > > The complete tree contains all infinite paths. The structure of the > > > Binary Tree excludes that are any two initial segments, B_k and B_(k > > > +1), which differ by more than one infinite path. > > > > Actually, there are infinitely many, uncountably many, > > You err.
Wrong again, dimwit! There are unaccountably many infinite paths that contain any such B_k but not B_(k+1), all of which differ from the uncountably many infinite paths containing B_(k+1). > > > > (In fact no B_k does > > > contain any infinite path > > > > But each , as a FISON, finite initial sequence of nodes, is contained > > You err. B_1 contains 0 and 0.0. B_2 contains 0 and 0.0 and 0.1.
Is 0.0 a terminal node for B_1 so B_1 does not pass through both 0.0 and 0.1? > > > > Hence either there are only countably many infinite paths.
Which is provably not the case
> > > Or uncountably many infinite paths come into the tree after all finite > > > steps of the sequence have been done.
To do all "finite" steps without in finitely many steps would require that every path have a terminal node
Consider unary trees. Either they are finite with a unique terminal node or infinite without any terminal node. So the only ones are either like FISONs or like |N.
Binary trees in which all paths are order-isomorpic to each other (complete trees) have all paths being isomprphic as unary trees to each other, i.e., all paths order isomorphic to some fixed FISON or all order isomorphic to |N.
Tertium non Datur.
> > > WM is great as assuming things, but abominable at proving things, > > You err. It looks so to you, because you cannot even understand the > most simple proofs like this one: > > Cantor's argues that the diagonal will differ from each entry at some > digit n. > I have proved that this is false, since for every n there are > infinitely many entries with the same FIS d_1, ..., d_n as the anti- > diagonal.
Infinitely many of an infinite set is not ALL. So claimed proof false!
That at each position a diagonal agrees with infinitely many entries still allows it to disagree at that same position with infinitely many other entries, and we only need it to differ from any p[articular entry and one out if infinitely many places.
So that WM's hand waving achieves nothing but stirring the air. --