In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 8 Mrz., 22:14, Virgil <vir...@ligriv.com> wrote: > > In article > > <6b761426-151b-43b7-adbb-3841e48fe...@y9g2000vbb.googlegroups.com>, > > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > > On 8 Mrz., 11:05, William Hughes <wpihug...@gmail.com> wrote: > > > > > To make a change: Do *you* agree with the statement: It is silly to > > > claim the existence of a set of natural numbers that has no first > > > element? > > > every NON-EMPTY subset > > An empty set does not contain natural numbers. Therefore it is not a > subset of natural numbers but at most a subset of the set of all > unicorns.
According to what definition?
The definition accepted by everyone outside Wolkenmuekenheim is:
set A is a subset of set B if and only if for all x, if x is a member of A then x is a member of B, Which is true for any set B when A is empty. > > < of it is also > > well ordered, and thus has a first element, > > Fine. Why do you sometimes appear to have forgotten this elementary > wisdom?
What appears to WM is much of the time only delusion caused by the astigmatism inherent in WMytheology. > > > > And where is WM's proof that some mapping from the set of all binary > > sequences to the set of all paths of a CIBT is a linear mapping? > > I told you already EOD with respect to this isomorphism because you > are too stupid to understand this fact.
You have yet to show ANYONE that either the set of binary sequences or the set of paths is a linear space, and unless both of them are, there is no possibility of having a linear mapping between them.
And where is WM's proof that some mapping from the set of all binary sequences to the set of all paths of a CIBT is a linear mapping? WM several times claimed it but cannot seem to prove it. --