In article <84bc987e-e71b-4cb3-a631-25e8a31cd5bf@u20g2000yqj.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 23 Feb., 22:08, Virgil <vir...@ligriv.com> wrote: > > > > > If WM means they are of equal cardinality or biject with each other , > > > > true, but to establish an isomorphism, as WM is claiming, one must > > > > specify the structure that is being preserved by the bijection, which WM > > > > has NOT done. > > > > > The mapping is bijective and linear.
The function in question was the mapping between the set of all paths (as binary strings) of a Complete Infinite Binary Tree to the set of all subsets of |N in which for each path the function produces the set of all naturally numbers of those levels in the tree at which that path branches left.
This produces a bijection , though not a linear mapping, between the uncountable set of paths and the uncountable set of subsets of |N. > > > > There is certainly no meaning of "linear" in English > > mathematics that is appropriate. > > Then use German mathematics. There it is. > > f(ax + by) = af(x) + bf(y)
With suitable interpretations for f, a, b, x and y, this would makes f a linear function.
But if f is to be a mapping between the set of all paths of a Complete Infinite Binary Tree and the set of all subsets of |N, which is the only sort of mapping under consideration when WM claimed linearity, I defy WM to come up with an appropriate definition of a,b,x and y which will make such an f a linear mapping.
> > f(string) = path > > > > > This shows a contradiction > > > - at least in case someone accepts > > > Hessenberg's trick as part of mathematics. > > > > To which Hessenberg, Karl or Gerard or some other one, does WM refer? > > That one who "proved" the uncountability of P(|N), Gerhard that is. > > > > And to what alleged "tricks"? > > To look for a set that cannot exist.
What prevents it from existing?
Remember, WM, that the vast majority of the world does not force itself to work under the artificial constraints of your WMytheology .
For small enough sets, S, and functions f from S to its power set P(S), one can easily, for any such function, f, find an explicit member of P(S) not in the image of f.
And as the size of S increases, so does the number of necessarily non-image sets for any f:S -> 2^S.
For any finite set, S, of cardinality n, the cardinality of its power set, P(S), is 2^n, so there will always be 2^n-n sets in P(S) not in the image of any function from S to P(S).
When S is allowed to be actually infinite, as it is in all standard math, like ZF or NBG, then P(S) becomes actually uncountable. --
