Re: Cantor Confusion

In article <1163068762.173696.311700@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
*** T. Winter schrieb:
....
How wrong you are. In the case of the edges you show that for finite trees
of order n:
#edges(n) = f(n)
with some function n. From that you conclude:
#edges(oo) = f(oo)
which requires transfinite induction. To prove that Q is countable you do
not need infinity at all. See the construction of a bijection I gave
from N to Q+, using simplified continued fractions. This shows that
every finite n maps to a finite q>0 and the reverse. So we have a
function g: N -> Q+ that has an inverse. And that is all that is needed
to show countability. Where does the tranfinite induction come in?

There is no difference between enumerating all the rational numbers in
the well-known scheme, starting at the corner of an infinite square
matrix

There is. You can assign numbers to edges that terminate at nodes. And
*that* numbering is different. It numbers only those edges that are finitely
far away from the root. But that way you do not number all edges in the
infinite tree, because that contains edges that are *not* finitely far
away from the root.

Indeed. To prove countability you do not need transfinite induction.
But my remark "you are wrong" was to your statement:
"The mapping N -> {edges} has been established".
See J. H. Conway, On Numbers and Games, for a clear exposition about the
difference between the union of all finite trees and the infinite tree.

Whatever Conway may say, there is no difference between enumerating all
rational numbers in the well-known scheme, starting a the corner of an
infinite square matrix, and the edges of the tree. In both cases you
have a system with a limit which is countably infinite.

There is a difference, see above. You can only number edges that are
finitely far away from the root, but in that way you will only number
edges that terminate at nodes finitely far away from the root. As in
the edges there are nodes infinitely far away from the root you will
never number the edges that terminate there.

The situation with the rationals is quite different, because in the
matrix *each* rational is finitely far away from the root.

Perhaps you see a difference between the union of all finite numbers n
and the set N?

Apparently you are not interested in what Conway did write, otherwise you
would understand how ridiculous this comment is.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
.