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
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
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
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
inf
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
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 courageou
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 in
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
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
ally kept up.
> Anyway, I saw the reference to Cantor's Diagonal and thought perhaps
> someone could help me.
>
> Consider the set of positive integers: {1,2,3,...}, but rather than
> write them in this standard notation we'll use what I'll call
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'
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 b
any and all candidate enumerations. So
you don't arrive at the absurdity 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
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 infinite
e enumerations. So
you don't arrive at the absurdity 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
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.
Consider the set of positive integers: {1,2,3,...}, but rather than
write them in this standard notation we'l
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 ex
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,
,...}
> 4 {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
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
ction:
>
> 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 ---- {1,0,1,0,0,0,0,...}
> 6 {0,1,1,0,0,0,0,...}
> 7 {1,1,1,
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
>
> >
Dr. Barry Brent
[EMAIL PROTECTE
,0,...}
4 {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
--~--~-~--~~--
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.
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 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
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
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 candidate enumeration. You respond
by saying, "we
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 questi
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 complemen
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 -
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
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