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
Hi.
Bruno could do this better, but I like the practice.
I guess you're trying to demonstrate that the form of Cantor's
argument is invalid, by displaying an example in which it produces an
absurd result.
Start with a set S you want to show is not enumerable. (ie, there is
no one-one
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
Hi Barry,
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 different from any particular indexed element in
the
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
Le 21-nov.-07, à 17:33, Torgny Tholerus a écrit :
What do you think of this proof?:
Let us have the bijection:
0 {0,0,0,0,0,0,0,...}
1 {1,0,0,0,0,0,0,...}
2 {0,1,0,0,0,0,0,...}
3 {1,1,0,0,0,0,0,...}
4 {0,0,1,0,0,0,0,...}
5
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
Bruno Marchal skrev:
Le 20-nov.-07, 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.
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 a specialist on this stuff
but I think
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
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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