Re: Attempts to Refute Cantor's Uncountability Proof?



Hatto von Aquitanien wrote (to kunzmilan):

So your method of generating all the members of an infinite
set failed.

Try this: Since I am told that it is meaningful to talk about
generation ad infinitum, why not start by something simple.
I have an algorithm which generates n.0, n.1,..., n.9, on the
first pass. On the second, it generates n.01, n.02, ..., n.09,
n.10, n.11, ..., n.19...,n.99. Where n is the non-negative
integer being visited. First visit 0, and make one pass,
then visit 1 and make one pass, return to 0 and make a second
pass, then to 1 and make a second pass, then visit two, etc....
Eventually, you will construct every number representable in
decimal notation.

Dave L. Renfro wrote:

At what point in your list will 1/3 be reached? A rough estimate
would be acceptable.

Hatto von Aquitanien wrote:

When I reach countable infinity.

The context is a list that has a first element, a second element,
a third element, and so on, for each positive integer. There is
no "countable infinite" position on such a list.

To show that the positive rationals have the same cardinality
as the positive integers, you need to assign (in a unique way)
a certain positive rational number to '1', a certain positive
rational number to '2', a certain positive rational number
to '3', and so on, in such a manner that every positive
rational number is used up. There is no "countable infinity"
among the numbers '1', '2', '3', etc. -- you have to stay
with the numbers '1', '2', '3', etc.

Besides, even if there were a "countable infinity" among the
positive integers (and there isn't), you'd then have to tell us
what corresponds to the infinite decimal for 1/6, and then what
corresponds to the infinite decimal for 1/9, and for many other
(infinitely many other!) positive rational numbers as well.
Since you've already used up all the numbers '1', '2', '3', ...
and "countable infinity", you can't use any of these again
for 1/6, 1/9, etc.

Dave L. Renfro

.



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