Re: Cantor Confusion


jpalecek@xxxxxx schrieb:

mueckenh@xxxxxxxxxxxxxxxxx napsal:
Dave L. Renfro schrieb:

Peter Webb wrote (in part):

This is a complete red herring. There is no question that
the Real generated by Cantor's proof is computable (r. e,)
if the original list is, [...]

mueckenh@xxxxxxxxxxxxxxxxx wrote (in part):

Of course. That's why the diagonal proof only proves the
existence of numbers which belong to a countable set i.e. the
set of constructible reals. This proof proves in essence that
the countable set of constructible real numbers is uncountable.
A fine result of set theory.

You're overlooking Peter Webb's hypothesis "if the original
list is". You need to have a list (x_1, x_2, x_3, ...) such
that the function given by n --> x_n is computable. Thus,
before you can conclude what you're saying (which sounds like
a metalogic "proof by contradiction" to me, but no matter),
you need to come up with a computable listing of the computable
numbers (or at least, show that such a listing exists).

One cannot compute a list of all computable numbers. By this
definition,
(1) the computable numbers are uncountable.
(2) There is no question, that the computable numbers form a countable
set.
This is a contradiction. It is not necessary to come up with a list of
all computable numbers.

There is no contradiction.

The fact that you cannot compute a list of all computable reals does
not mean that there is no list of all computable numbers. There is one,

and it is not computable.

The fact that you cannot compute a list of all reals does not mean that
there is no list of all reals. There is one, but it is not possible to
publish this list.

Every list of reals can be shown incomplete in exactly the same way as
every list of computable reals can be shown incomplete.

Regards, WM

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