Re: Cantor Confusion

On 12 Mai, 22:54, William Hughes <wpihug...@xxxxxxxxxxx> wrote:
On May 12, 4:28 pm, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

<snip>

You should know:
The union (as I defined it for trees) of all finite trees is an
infinite tree with countably many nodes and countably many finite
paths. (If you like, you can union the nodes and the paths
separately.)

If we switch to the CIBT, then there is not a single node added. Only
the number of paths and their lengths changes from countable to
uncountable and from finite to infinite, respectively.

This is nothing more than the statement.

Let R be a countably infinite set.

the set of finite subsets of R is countable
the set of infinite subsets of R is uncountable.

Note the number of elements in R does not change.

The number of elements in any infinite subset is greater than the
number of elements in any finite subset, or not?.

When you change the size of the sets (the lengths of the
paths) you change the number of sets (the number of paths)

But in order to increase the length of the paths you have to add
nodes, haven't you?

Regards, WM

.

Relevant Pages

  • Re: Cantor Confusion
    ... > from the union of all rational trees. ... In the union there is *no* infinite path. ... So the union of all finite trees has ony a finite number of nodes? ...
    (sci.math)
  • Re: Cantor Confusion
    ... The union of all finite trees and the complete infinite tree are ... uncountably many limits. ...
    (sci.math)
  • Re: Cantor Confusion
    ... It consists only of finite initial segments. ... The union of all finite trees and the complete infinite tree are ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... > infinitely far from the root node in any binary tree, ... or infinite binary trees. ... > consecutive naturals. ...
    (sci.math)
  • Re: Cantor Confusion
    ... The union of all finite trees contains the union of all finite paths ... there are uncountably many infinite paths having ... The set of finite subsets of a countably infinite N is countable. ... So counting only the finite subsets doesn't count. ...
    (sci.math)
Quantcast