Re: Cantor's wrong! (doubt it)

From: Russell Easterly (logiclab_at_comcast.net)
Date: 06/21/04


Date: Sun, 20 Jun 2004 22:26:32 -0700


"Mike Terry" <news.dead.person.stones@darjeeling.plus.com> wrote in message
news:2jjk04F12j25hU1@uni-berlin.de...
> "Russell Easterly" <logiclab@comcast.net> wrote in message
> news:25KdnWyv0s2i5U_dRVn-gQ@comcast.com...

> Here is what I think you really meant to say:
>
> Definition: a natural number is "oney" if it's decimal representation
> consists entirely of a string of 1s.
>
> (e.g. 1, 1111, 111111 etc. are oney numbers)
>
> Definition: given the sequence {N_i}, we will construct a new *sequence*
> {x_i} as follows:
>
> x_i = (smallest oney number which is greater than every oney number
> appearing in the set {N_j: j<=i})
>
> (e.g. if N_i is the sequence 7, 2, 111, 5, 1111, ...
> then x_i is the sequence 1, 1, 1111, 1111, 11111, ...)
>
> This is the construction you were getting at, right?

Right.

> This is a properly defined sequence {x_j}, and we can show it is
> monotonically increasing. For each natural number j, you have derived a
new
> natural number x_j. Note you have not produced a number x_oo which is
> "right at the end of the list", as the list has no end...
>
> Now, depending on the original sequence {N_i}, the corresponding sequence
> {x_i} may converge to a final value (x), or may be unbounded in which case
> there is no final (x) to which it converges.

"Converges" is not the right word. Let's just say that the list, x, contains
a
natural number that is not in list N.

> So where do you want to go now? You seem to be suggesting that the
sequence
> {x_i} always converges to a number x, which is not in the original
sequence
> {N_i}. Well, sure, if {x_i} does converge to a number x, then x is
missing
> from the original list. BUT what if your {x_i} is unbounded?

My point is that these same objections can be applied to Cantor's diagonal
argument.

For example, let's use the following rule to create our diagonal number:
Let N be a list of real numbers and let x_i be ith digit of the diagonal
number.

If digit i of N_i = 1 then x_i = 2 else x_i = 1.
Let N be the following list of real numbers:

.000...
.1000...
.11000...
.111000.
.1111000...
...

x_i is "unbounded" and x never "converges" to a real number not in N.

My problem with the diagonal argument is that it requires an infinite
number of computations. There was a thread some time ago where
we proved that the Halting problem is solvable if we allow a
computer to perform an infinite number of calculations in finite time.
The diagonal argument requires such a computer.

Here is another example:
Consider a set of sets of natural numbers written in unary.

Unary numbering system:
One = 1
Two = 11
Three = 111
etc.

Sets of unary numbers:
N_1 = {1}
N_2 = {1,11}
N_3 = {1,11,111}
etc.

Notice that each of these sets contains a member whose length
is equal to the cardinality of the set. For example:

length of 1 = cardinality of N_1
length of 11 = cardinality of N_2
length of 111 = cardinality of N_3

Now consider a N that contains every natural number in unary.
Does this N contain a member whose length is equal to the cardinality of N?

Russell
- 2 many 2 count



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