Re: countability of reals
From: ken quirici (kquirici_at_yahoo.com)
Date: 01/07/05
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Date: 7 Jan 2005 04:47:07 -0800
1. I'm wondering about something. Is it possible to define a set of
rationals for which there is NO halting bijection to the natural
numbers? If so, that would be why the bwahaha folks can get away with
what they're getting away with - because we can't be sure our original
list isn't just one of those. So I think we need to prove there is NO
set for rationals for which is NO halting bijection to the natural
numbers.
2. Consider the following 'proof' diagonalizing against the rationals,
not the reals. It's at least amusing, maybe.
The proof would go like this (but I think it needs a little help, so
any would be appreciated)
it is easy to construct a function r(n,m)->p that maps any rational
n/m from 0 to 1 to an integer p and is bijective.
now diagonalize the rationals from 0 to 1 using the above listing
function r(n,m), and our standard diagonalization procedure - the
kth digit in the diagonal number is different from the kth digit of
the kth number on the list. We come up with a number x which is
clearly not in the list: if it were, it would be at some position p2.
But we define x's p2'th digit to be different from the one of the
rational at position p2. Contradiction.
We have just constructed (or we will when we finish the diagonalization
:))
(spoooky voice) a number that isn't rational. This is our first real.
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