On Fri, Dec 21, 2007 at 01:08:38PM +0100, Günther Greindl wrote:
Hi Russell,
Russell Standish wrote:
In your first case, the number (1,1,1,1...) is not a natural number,
since it is infinite. In the second case, (0,0,0,...) is a natural
number, but is also on the list (at infinity).
Hi,
Because zero even repeated an infinity of time is zero and is a natural
number. (1,1,1,...) can't be a natural number because it is not finite and a
natural number is finite. If it was a natural number, then N would not have a
total ordering.
Ok agreed: I was caught up in viewing it
Hi,
Le Friday 21 December 2007 13:08:38 Günther Greindl, vous avez écrit :
Hi Russell,
Russell Standish wrote:
In your first case, the number (1,1,1,1...) is not a natural number,
since it is infinite. In the second case, (0,0,0,...) is a natural
number, but is also on the list (at
Hi Russell,
Russell Standish wrote:
In your first case, the number (1,1,1,1...) is not a natural number,
since it is infinite. In the second case, (0,0,0,...) is a natural
number, but is also on the list (at infinity).
Why is (1,1,1,...) not in the list but (0,0,0,...) in the list at
Le 19-déc.-07, à 21:09, Barry Brent a écrit :
Excellent, Bruno, Thanks!
Well thanks. I will send a next diagonalization post and some
references next week,
Best,
Bruno
http://iridia.ulb.ac.be/~marchal/
--~--~-~--~~~---~--~~
You received this message
Hi Barry,
Le 18-déc.-07, à 18:52, Barry Brent a écrit :
Bruno--
Ahh, my amateur status is nakedly exposed. I'm going to expose my
confusion even further now.
That is the courageous attitude of the authentic scientists.
I like amateur because they have less prejudices, they have inner
Excellent, Bruno, Thanks!
Barry
On Dec 19, 2007, at 7:57 AM, Bruno Marchal wrote:
Hi Barry,
Le 18-déc.-07, à 18:52, Barry Brent a écrit :
Bruno--
Ahh, my amateur status is nakedly exposed. I'm going to expose my
confusion even further now.
That is the courageous attitude of the
Le 17-déc.-07, à 19:04, meekerdb (Brent Meeker) wrote:
Bruno wrote:
Exercise:
What is wrong with the following argument. (I recall that by
definition
a function from N to N is defined on all natural numbers).
(false) theorem: the set of computable functions from N to N is not
Bruno--
Ahh, my amateur status is nakedly exposed. I'm going to expose my
confusion even further now.
Never heard of a universal language. I thought I was familiar with
Church's thesis, but apparently no. I thought it was the claim that
two or three or four concepts (including recursive
Hi Daniel,
I agree with Barry, but apaprently you have still a problem, so I
comment your posts.
Le 16-déc.-07, à 10:49, Daniel Grubbs a écrit :
Hi Folks,
I joined this list a while ago but I haven't really kept up. Anyway,
I saw the reference to Cantor's Diagonal and thought perhaps
you seem to be after, even if you
fill in the details I mentioned.
Barry
On Dec 16, 2007, at 3:49 AM, Daniel Grubbs wrote:
Hi Folks,
I joined this list a while ago but I haven't really kept up.
Anyway, I saw the reference to Cantor's Diagonal and thought
perhaps someone could help me
On Sun, Dec 16, 2007 at 04:49:34AM -0500, Daniel Grubbs wrote:
Cantor's argument only works by finding a number that satisfies the
criteria for inclusion in the list, yet is nowhere to be found in the
list.
In your first case, the number (1,1,1,1...) is not a natural number,
since it is
seem to be after, even if you
fill in the details I mentioned.
Barry
On Dec 16, 2007, at 3:49 AM, Daniel Grubbs wrote:
Hi Folks,
I joined this list a while ago but I haven't really kept up.
Anyway, I saw the reference to Cantor's Diagonal and thought
perhaps someone could help me
Hi Dan,
Let me take your statements a few at a time.
Let me see if I am clear about Cantor's method. Given a set S,
and some enumeration of that set (i.e., a no one-one onto map from
Z^+ to S) we can use the diagonalization method to find an D
which is a valid element of S but is
Le 03-déc.-07, à 16:56, David Nyman a écrit :
On Nov 20, 4:40 pm, Bruno Marchal [EMAIL PROTECTED] wrote:
Conclusion: 2^N, the set of infinite binary sequences, is not
enumerable.
All right?
OK. I have to try to catch up now, because I've had to be away longer
than I expected, but I'm
On Nov 20, 4:40 pm, Bruno Marchal [EMAIL PROTECTED] wrote:
Conclusion: 2^N, the set of infinite binary sequences, is not
enumerable.
All right?
OK. I have to try to catch up now, because I've had to be away longer
than I expected, but I'm clear on this diagonal argument.
David
Hi,
Le 22-nov.-07, à 07:19, Barry Brent a écrit :
The reason it isn't a bijection (of a denumerable set with the set of
binary sequences): the pre-image (the left side of your map) isn't
a set--you've imposed an ordering. Sets, qua sets, don't have
orderings. Orderings are extra. (I'm not
{1,0,1,0,0,0,0,...}
6 {0,1,1,0,0,0,0,...}
7 {1,1,1,0,0,0,0,...}
8 {0,0,0,1,0,0,0,...}
...
omega --- {1,1,1,1,1,1,1,...}
What do we get if we apply Cantor's Diagonal to this?
Note also that in general, we start from what we want to prove, and
then do the math
Le 20-nov.-07, à 17:47, David Nyman a écrit :
On 20/11/2007, Bruno Marchal [EMAIL PROTECTED] wrote:
David, are you still there? This is a key post, with respect to the
Church Thesis thread.
Sorry Bruno, do forgive me - we seem destined to be out of synch at
the moment. I'm afraid I'm
Le 20-nov.-07, à 23:39, Barry Brent wrote :
You're saying that, just because you can *write down* the missing
sequence (at the beginning, middle or anywhere else in the list), it
follows that there *is* no missing sequence. Looks pretty wrong to me.
Cantor's proof disqualifies any
Le 21-nov.-07, à 08:49, Torgny Tholerus a écrit :
meekerdb skrev:Torgny Tholerus wrote:
An ultrafinitist comment to this:
==
You can add this complementary sequence to the end of the list. That
will make you have a list with this complementary sequence included.
But then you
0,0,1,0,0,0,0,...}
5 {1,0,1,0,0,0,0,...}
6 {0,1,1,0,0,0,0,...}
7 {1,1,1,0,0,0,0,...}
8 {0,0,0,1,0,0,0,...}
...
omega --- {1,1,1,1,1,1,1,...}
What do we get if we apply Cantor's Diagonal to this?
--
Torgny
--~--~-~--~~~---~--~~
You received
{0,1,1,0,0,0,0,...}
7 {1,1,1,0,0,0,0,...}
8 {0,0,0,1,0,0,0,...}
...
omega --- {1,1,1,1,1,1,1,...}
What do we get if we apply Cantor's Diagonal to this?
--
Torgny
Dr. Barry Brent
[EMAIL PROTECTED]
http://home.earthlink.net/~barryb0
On 20/11/2007, Bruno Marchal [EMAIL PROTECTED] wrote:
David, are you still there? This is a key post, with respect to the
Church Thesis thread.
Sorry Bruno, do forgive me - we seem destined to be out of synch at
the moment. I'm afraid I'm too distracted this week to respond
adequately - back
Hi,
David, are you still there? This is a key post, with respect to the
Church Thesis thread.
So let us see that indeed there is no bijection between N and 2^N =
2X2X2X2X2X2X... = {0,1}X{0,1}X{0,1}X{0,1}X... = the set of infinite
binary sequences.
Suppose that there is a bijection between N
Bruno Marchal skrev:
But then the complementary sequence (with the 0 and 1
permuted) is
also well defined, in Platonia or in the mind of God(s)
0 1 1 0
1 1 ...
But this infinite sequence cannot be in the list, above.
The "God" in question has to ackonwledge that.
The
Torgny Tholerus wrote:
Bruno Marchal skrev:
But then the complementary sequence (with the 0 and 1 permuted) is
also well defined, in Platonia or in the mind of God(s)
*0* *1* *1* *0* *1* *1* ...
But *this* infinite sequence cannot be in the list, above. The God
in question has to
meekerdb skrev:
Torgny Tholerus wrote:
An ultrafinitist comment to this:
==
You can add this complementary sequence to the end of the list. That
will make you have a list with this complementary sequence included.
But then you can make a new complementary sequence, that is
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