Re: Cantor Confusion




mueckenh@xxxxxxxxxxxxxxxxx wrote:
Peter Webb schrieb:

You produce ANY list of all Reals, I can show you a missing real. Therefore
I can do it for ALL lists, and hence there is no complete list of Reals.

Obviously you have to tell me the list first, as there is no single Real
which is missing from every list,

No?

But the set of all lists is countable

Twaddle.

Let A ={ (x,0,0,0,...) | x is a real number }

Then A is a set of lists and A has as many elements
as there are real numbers.


(as is any quantized or
discontinuous set), so is the set of all list entries. Nevertheless,
there is no real number missing in every list? So every real number is
in at least one of the list? So every real number is one element of a
countable set of entries? And there is nothing real really outside of
this countable set?


Note any countable set of entries is a list (possibly
with a different ordering).

This is staight quantifier dyslexia

For every real number x, there exists a list, L_x such
that x is a member of L_x

There exists a list L, such that every real number x is
a member of L.

The first is true, the second is false. There is no way to put
all the L_x together to get a "countable set of entries"
(the list L).

- William Hughes

.



Relevant Pages

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  • Re: diagonal argument on ordered array of reals
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  • Re: diagonal argument on ordered array of reals
    ... You appear to be trying to show that the reals are countable. ... it is a wonderful insight by Cantor. ... nature, is evident in the infinite reals, the ones which have an infinite ... the proof would only apply to lists ...
    (sci.math)
  • Re: Cantors fatal mistake a la Zenkin
    ... > Any list of reals forms a countable set. ... Nothing forces lists to be mapped that way. ... The mapping between the ...
    (sci.logic)
  • Re: diagonal argument on ordered array of reals
    ... You appear to be trying to show that the reals are countable. ... it is a wonderful insight by Cantor. ... nature, is evident in the infinite reals, the ones which have an infinite ... the proof would only apply to lists ...
    (sci.math)