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. If ax + by is in the unit interval, then f(ax + by) is in the tree. If ax + by is not in the unit interval, then f(ax + by) is not in the tree. This is enough to consider unit interval and tree isomorphic (= having the structure).