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Topic: Paradox with cardinality
Replies: 7   Last Post: Jan 17, 2012 10:01 PM

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fishfry

Posts: 1,368
Registered: 12/6/04
Re: Paradox with cardinality
Posted: Jan 17, 2012 8:08 PM
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In article
<60a44b45-0e06-4c73-91b1-2825f8a8c1ea@o13g2000vbf.googlegroups.com>,
Gustavo Broos <gpubgm@gmail.com> wrote:

> On 17 jan, 14:11, Gustavo Broos <gpu...@gmail.com> wrote:
> > Hello,
> >
> > I would like to propose this paradox with cardinality:
> >
> > Let T be the set of all infinite binary strings, so we have a 2D
> > infinity of 0s and 1s:
> >
> >    1   2   3  ... x
> >
> > 1 (0) (1) (1) ...
> >
> > 2 (1) (1) (1) ...
> >
> > 3 (0) (0) (1) ...
> >
> > .................
> >
> > y
> >
> > Let N be the set of natural numbers. Now we try to count with N not
> > the rows or columns, but the individual bits in the infinite 2D space
> > (this could be one of several ways to do so):
> >
> >    1   2   3  ... x
> >
> > 1 (1) (4) (9) ...
> >
> > 2 (2) (3) (8) ...
> >
> > 3 (5) (6) (7) ...
> >
> > .................
> >
> > y
> >
> > It seems that the diagonal argument can't be applied to this
> > individual bits, and we can count them all. Now each row or column not
> > only has one natural number assigned to it, but an infinite number of
> > different naturals asigned to it. However, if the diagonal argument
> > (or Cantor's theorem) is true then we have a contradiction, because we
> > don't have enough naturals to assign a different one to each row in T.

>
> Sorry, the argument was flawled. Discussion pointless.


It would be helpful to see clearly WHY it's flawed.

Initially you said that T is the *set* of all binary strings. But then
you implicitly assumed that you had a *list* of all such strings. Is
that your understanding of the flaw in your proof?



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