Re: Cantor Confusion

In article <1166096247.875380.67080@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

Virgil schrieb:

In article <1166035958.305158.282870@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

*** T. Winter schrieb:


> > > That one which goes left from the root is not engaged, because
> > > we
> > > need
> > > only half of the set of edges.
> >
> > So your construction is not a surjection.
>
> Of course it is! It is a surjection from the set of edges onto the
> set
> of paths.

Than again. To what number maps the edge that goes left from the root?
If it is a surjection, there should be one.

A surjection onto the paths covers all elements of the *range* = every
edge. It is not necessary that every edge is mapped on a path, to show
that there are not less edges than paths. It is only necessary that
there is no path without edge mapped on it.

Which necessity WM has not accomplished. All Wm has done is to map
infinite sequences of edges onto paths, but that is trivial since an
infinite sequence of edges is essentially what a path is.

You had said that you had constructed a surjection. A surjection is a
mapping with specific properties. It is necessary to know what a
mapping
is in order to construct a surjection, which you did not do. And I
have
*no* idea what you mean with the second half of your sentence.

If someone makes an asserted list of all real numbers and hands it out
to Cantor, then Cantor can construct a diagonal number which is not in
the list, contradicting the assertion.

If someone makes a binary tree of all real numbers and hands it out to
Cantor, then Cantor cannot construct a diagonal number which is not in
the tree, so no contradiction of completeness is possible.


Actually anyone can. One does it by dealing with successive pairs of
binary digits to make the representation equivalent to a base 4
representation, in which the Cantor diagonal method works quite nicely.

Not in the binary (or any other) tree.

Wm seems to be up a tree. What have any of Cantor's proofs of
uncountability to do with trees?



The problem with Cantor's theorem aleph0 < 2^aleph0 is that the
function f(x) = 2^x must be discontinuous:

Since the domain and codomain of that function are neither of them
topological spaces, or even metric spaces, the notion of continuity is
impossible to consider.

Not in the tree with countably many beginnings of separated parts of
paths.

Until some definition of continuity other than the usual one is
presented, continuity does not apply to trees at all.

By means of the binary tree I can exclude any discontinuity.

What is your topology or metric on the binary tree?

Absent metrics, or at least a topologies, one cannot even speak of
continuity or discontinuity of a function.

Look at the tree. Continuity is guaranteed by the edges between two
nodes.

Until some definition of continuity other than the usual one is
presented, continuity does not apply to trees at all.
.

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