Re: The natural numbers are uncountable
From: Arturo Magidin (magidin_at_math.berkeley.edu)
Date: 07/05/04
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Date: Mon, 5 Jul 2004 23:37:24 +0000 (UTC)
In article <7zgGc.2341$Rt4.1117@newsfe5-win.ntli.net>,
Andrew <stan370@btinternet.com> wrote:
>Given any simply infinite sequence of unique natural numbers, regardless of
>the specific identity of the natural number at each position in the
>sequence, there is first element in the sequence, a second element in the
>sequence, and so on, and so on, hence any simply infinite sequence of unique
>natural numbers may be substituted for an infinite sequence of the natural
>numbers in their natural order (i.e., 1, 2, 3, . . .). Which must mean that
>any such sequence is countable.
>
>On the other hand, if L is the set of all simply infinite sequences formed
>from the two symbols '0' and '1'. Let M be the set of all elements of L
>which are terminated in nothing but a string of zeros. Reading from left to
>right, each unique element of M corresponds with a unique element of the set
>of natural numbers. To see this, simply strip off the terminating string of
>zeros and each element becomes a natural number in base two - e.g.;-
>
>1000000000. . . = 1 = one
>0100000000. . . = 01 = two
>1100000000. . . = 11 = three
>etc.,
>
>Next, substitute this set M for the set M used by Cantor in his 1891 proof
>that the power set of natural numbers are uncountable (which originally used
>the set L whereas now we are using a countable subset of L). Cantor's
>diagonal method will now create a member of this set M which is guaranteed
>to not already be in the sequence.
This is where your argument breaks down. When you started talking
about trailing zeroes I figured this is where you were going, but I
thought I would wait and see.
Note that a sequence of 0's and 1's is in your "list" M IF AND ONLY IF
almost all entries are zeros; that is, it contains at most a finite
number of 1's.
When we use the diagonalization argument on this list, we end up with
a sequence of 0's and 1's, call it S.
Now, certainly, S is not any of the numbers on the list (you have to
be careful about dual representation in binary, but let us set it
aside for a moment).
However, the claim that S SHOULD be on the list is unwarranted: S will
correspond to a natural number if and only if it contains only a
finite number of 1's. That is, you would have to show that in your
original listing, only finitely many of the diagonal entries are 0's.
In fact, this will never occur; the sequence you produce is certainly
not on the original list, but as opposed to the situation with real
numbers (where ANY sequence corresponds to a number), the sequence you
produce DOES NOT correspond to a number that belongs on that list
anyway.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
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