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



fishfry
Posts:
1,368
Registered:
12/6/04


Re: Paradox with cardinality
Posted:
Jan 17, 2012 8:08 PM


In article <60a44b450e064c7391b12825f8a8c1ea@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?



