Re: Cantor Confusion

mueckenh@xxxxxxxxxxxxxxxxx wrote:
No. What I am using is this:
1) The union of all finite trees contains the union of all finite paths
and this set is countable.
2) The union of all finite trees contains all nodes.
3) You cannot add another tree to the set of all finite trees without
adding at least one node.
(The paths are subsets of the set of nodes. Two different sets must be
distinguished by at least one element.)
4) But there is no node to add.
5) Therefore the union of trees contains all possible paths.
6) We know that the set of all paths contained in his union is
countable.

How does 6 follow from 1-5?

--
David Marcus
.

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