In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 18 Mrz., 13:50, William Hughes <wpihug...@gmail.com> wrote: > > > So you take a set of lines that contains an unfindable line > > remove all the findable lines and end up with a set > > that contains an unfindable line, but no findable line ?!? > > If you remove every findable line, there cannot remain a findable > line, can it? > > But the more pressing question is: You construct a list such that > every line contains all preceding contents. You get ready, i.e., the > list contains all that it can contain. Nevertheless there is no line > that contains everything that the list contains.
WM claimed: > The isomorphism is from |R,+,* to |R,+,*. Only in one case the > elements of |R are written as binary sequences and the other time as > paths of the Binary Tree. Virgil is simply too stupid to understand > that.everal flaws in WM's claim that the identity map on induces a linear map on 2^|N.
WM's flaws in making that claim work include, but are not necessarily limited to:
(1) not all members of |R will have any such binary expansions, only those between 0 and 1, so that not all sums of vectors will "add up" to be vectors within his alleged linear space, and
(2) some reals (the positive binary rationals strictly between 0 and 1) will have two distinct and unequal-as-vectors representations, requiring that some real numbers not be equal to themselves as a vectors, and
(3) WM's method does not provide for the negatives of any of the vectors that he can form.
On the basis of the above problems, and possibly others as well that I have not yet even thought of, I challenge WM's claim to have represented the set |R as the set of all binary sequences, much less to have imbued that set of all binary sequences with the structure of a real vector space or the showed the identity mapping to be a linear mapping on his set of "vectors". --