In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 28 Feb., 22:54, Virgil <vir...@ligriv.com> wrote: > > In article > > <45c34b52-d039-4aed-9b42-5932bdb5a...@j29g2000vby.googlegroups.com>, > > > > > > > > > > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > > On 27 Feb., 22:18, Virgil <vir...@ligriv.com> wrote: > > > > > > > The sum of two real numbers of the unit interval need not be a real > > > > > number of the unit interval. > > > > > > Then the set of reals in the unit interval do not form a commutative > > > > group under addition and thus cannot be a linear space, and thus cannot > > > > be either the domain or codomain of any linear mapping. > > > > > It can and it is. > > > > > > > Nevertheless we have the same structure for reals, their > > > > > representation as binary strings, and paths of the Binary Tree. > > > > > > WM claimed a linear mapping between the set of binomial sequences and > > > > the set of paths of a Complete Infinite Binary Tree. > > > > > > Thus requires, among other things, that both sets have the structure of > > > > linear spaces, > > > > > You are in error. > > > Every sum and every product that is possible in the reals of the unit > > > interval is possible in the Binary Tree and vice versa. And that is > > > all that is required. > > > > Then WM need to show, among many other things > > (1) show us how to add any two infinite binary sequences to gat another > > binary sequence from the same set of binary sequences (which is not as > > easy as it appears), and show that this addition on the set of all such > > binary sequences creates an additive group (which the standard addition > > does not so) > > (2) show us how to add two paths in a Complete Infinite Binary Tree to > > gat another path, and show that this addition on the set of all such > > paths creates an additive group. > > > > Then show that that allegedly linear mapping between binary sequences > > and paths preserves addition. > > If you believe that the real unit interval together with + and * is > not isomorphic to the real unit interval with + and * then you may do > so . I call them isomorphic.
What I said was that the real with interval with + is not a group.
> If ax + by is in the unit interval, then f(ax + by) is in the tree.
But ax + by is NOT always in the unit interval, so accordingly f(ax + by) need not be in the tree.
> If ax + by is not in the unit interval, then f(ax + by) is not in the > tree.
But for a mapping to be linear on the unit interval requires that for any x and y in the unit interval and any a and b in the field of scalars ax+by also be in the interval. Otherwise the interval is not a linear space at all and there cannot be any linear mappings from it to anything.
Linear mappings only operate on linear spaces (vector spaces) and only have linear spaces (vector spaces) as their image and codomains.
Thus WM is WRONG AGAIN AS USUAL!
> This is enough to consider unit interval and tree isomorphic (= having > the structure).
Only inside that wild weird world of WMytheology, where the definition of "linear map" means only what WM wants them to mean, and may change from moment to moment.
Outside of Wolkenmuekenheim, there are bijections between the unit interval in binary and the set of paths of a cibt, but such bijections are definitely not linear, at least anywhere outside of Wolkenmuekenheim. --