Re: Cantor's diagonal proof wrong?
From: Dave Seaman (dseaman_at_no.such.host)
Date: 11/21/04
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Date: Sun, 21 Nov 2004 00:19:07 +0000 (UTC)
On 20 Nov 2004 21:19:51 GMT, Curt Welch wrote:
> Dave Seaman <dseaman@no.such.host> wrote:
>> On 20 Nov 2004 18:23:08 GMT, Curt Welch wrote:
>> > The point above is that in Cantor's diagonal argument, what happens
>> > when the unspecified mapping function from natural numbers to real, is
>> > the same algorithm used to construct the anti-diagonal? The N digit
>> > real constructed from rows 1 to N is always placed in row N+1.
>> That doesn't even make sense. The diagonal is given by a mapping from
>> the naturals to the set of decimal digits, while the original mapping is
>> from the naturals to the set of reals in [0,1]. They aren't even the
>> same kind of mapping.
> First, we can easily remove your issues in relation to natural numbers and
> reals from the debate by changing the argument to another domain. Lets do
> that.
I am happy to substitute the powerset form of the diagonal argument and
dispense with digits altogether.
> Pretend we have a infinite table made up of an infinite strings of 1s and
> 0s on each row. So every row in this table is just an infinite sequence of
> some combination of 1s and 0s.
Cantor assumes instead that there is a mapping f: X -> P(X) for some X.
The diagonal shows that f is not a surjection.
> With this table, we make the argument that it is impossible for every
> combination of 1s and 0s to be listed in the table. We prove this by using
> the diagonal argument.
Given f: X -> P(X), we simply observe that the set
D = { x in X : not(x = f(x)) } is not in the range of f.
> So we start with the assumption that the table which includes all
> combinations of 1s and 0s does exist.
Cantor makes no such assumption. If you think there is a flaw in
Cantor's argument, then you should be able to point out where it lies in
the direct form of the argument.
> Then, we build the anti-diagonal value by selecting the numbers from the
> diagonal, and inverting them.
There are no numbers involved. Instead we have the set D defined above.
> This anti-diagonal value can not be in the table, because every table entry
> will be different by at least one digit.
D is not equal to f(x) for any x in X, because x in D iff x not in f(x).
> Therefore, we have created a contradiction with the starting assumption,
> that such a table can exist. Therefore, we have proved that the staring
> assumption is wrong, the table can not contain every possible string of 0's
> and 1's.
There is no contradiction involved. It's a direct proof.
> This is the exact same logic used in Cantor's diagonal proof, but this
> time, we are not dragging anything about math into it. (change 0 1 to A B
> if you like just to remove numbers from the picture).
Instead of arguing about whether your logic is the same as Cantor's,
let's simply use Cantor's.
> The result of this proof is the belief that any mapping function which maps
> table rows to a string, can not include every possible string, because the
> proof has constructed a string (from ANY mapping function) that is not in
> the table.
The result is that if f: X -> P(X) is given, then f is not a surjection.
> But, I can specify one special mapping function that works like this. Fill
> the first row with any combination of 1's and 0's you like. Then, fill
> every following row, with a copy of the row above it, with the diagonal
> digit inverted.
Let's consider the case X = N and define f(0) = {}. Then according to
your rule I believe I am supposed to define f(1) = {0} because 0 is the
diagonal element in f(0), right? What then? Do we take f(2) = {0,1}
because 1 is the diagonal element in f(1), or is that not what you meant?
If we proceed by that scheme, then we are going to get
f(0) = {},
f(1) = { 0 },
f(2) = { 0, 1 },
.
.
.
f(k) = { 0, 1, ..., k-1 }
.
.
.
from which it follows that D = { n in N : not(n in f(n)) } = N, and sure
enough, N is not a in the range of your function f. If you disagree,
then tell me for which n is f(n) = N?
> Now, what I can show about your diagonal value, is that every diagonal you
> construct is in fact in the table, in row N+1.
N is a set, not a number. What do you mean by N+1? You need to produce an n
such that f(n) = N.
> So your argument that it is not in the table, is no stronger than my
> argument that it is in the table.
It's Cantor's argument, not mine, and I suppose they are equally strong
if you ignore the minor point that his actually works and yours is
nonsense.
> If the table was finite in size, my argument would fall apart.
And also if the table is infinite in size.
[ rest snipped ]
-- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. <http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
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