Re: infinity
- From: Virgil <ITSnetNOTcom#virgil@xxxxxxxxxxx>
- Date: Fri, 02 Sep 2005 15:34:58 -0600
In article <MPG.1d825677830e968898a1d3@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> Daryl McCullough said:
> > What is L?
> L is a string length. With an alphabet of size S, we can make S^L
> strings of length L.
But what about the set of allowable values of L? If set of allowable
values of L the set of finite naturals, then the set of L's is
Cantor-infinite, and so the set of strings will also be
Cantor-infinite!
> >
> > >> You claim that the above summation is finite. According to you
> > >>
> > >> F = sum S^k for all finite k
> > >>
> > >> is a finite number.
> > >That is correct.
> > >
> > >> However if F is a finite number, then there are strings of
> > >> length F, and there are S^F strings of length F, which is
> > >> greater than F, the supposed number of finite strings. That's a
> > >> contradiction. Therefore F cannot possibly be a finite number.
> >
> > >Darryl just gave a similar "proof". When you say "for all finite
> > >k" then you are saying there is some upper bound to k.
> >
> > No, there is no (finite) upper bound to the set of finite natural
> > numbers.
> No, but k cannot be infinite if it is finite.
But the set of all k can be Cantor-infinite even though each k in that
set is finite.
> >
> > >k cannot be infinite. Therefore sum (x=1->k: S^x) is finite.
> >
> That sum is the number or strings less than or equal to k in length.
> It is finite for all finite k, since you are adding k terms of S^x
> where x is finite.
But the sum over all finite k is the sum over a Cantor-infinite set of
values, the Cantor-infinite set of finite naturals.
Until TO faces up to the Cantor-infintieness of the set of finite
naturals, he is being willfully blind to mathematical reality.
> >
> > >Here, now, you say, "All the words are finite because if you take
> > >a finite string and add a finite number of symbols you still get a
> > >finite string."
> >
> > No, what we are saying is that if U is a finite set of finite
> > strings, then there is a finite string that is not in U. From that,
> > it follows that U did not contain all finite strings.
> The contradiction comes from assuming you have defined the full set
> of all finite strings. This is exactly the "largest finite" argument,
> which of course causes contradictions, because you can always add 1.
> So what?? I suppose that would seem to indicate infinity, but it
> indcates infinity for the values in the set as well and the lengths
> of the strings.
But the set of _finite_ naturals has, according to TO, no largest
members, and is, therefore, clearly Cantor-infinite, however TO-finite
TO claims it to be.
> If you declare your values and strings finite, though
> you give no upper bound, there may be no upper bound to the size of
> the set, but it is finite as well, for all the reasons I keep having
> to repeat.
And any non-empty ordered set without an upper bound is Cantor-infinite!
> >
> > Yes. The sum over all finite values of L of S^L is infinite.
> So, you can add finite sets of strings to finite sets of strings in
> succession, and get an infinite result?
If one has infinitely many such finite terms, yes! There are infinite
series of positive numbers that diverge!
> But you can't add a finite
> number to a number in succession and get an infinite number? Hmmmm...
>
> Proof: The set of all finite strings on a finite alphabet is finite.
>
> L=0: Given N=S^L, there is S^0=1 string, the null string.
>
> L->L+1: Given a finite set of strings of length L or less, we add the
> set of strings of length L+1, which has S^(L+1) elements. Since L is
> finite, L+1 is finite, and S^(L+1) is finite. So we add this finite
> number of strings to the finite number of strings of length L or
> less, to get the number of strings of length L+1 or less. A finite
> plus a finite is finite.
But we have here an infinite series, S^1 + S^2 + S^3 + ... of positive
terms, which cannot converge unless the terms, S^n, converge to zero.
Does TO wish to state that the sequence f(n) = S^n ha kimit zero?
> If you say you
> can add the sizes of finite sets, and somehow get to infinity, but
> you cannot add other finite numbers and get infinity, then you are
> playing a shell game.
The sum of any finite number of positive terms is finite. The sum of any
infinite sequence of positive naturals is a divergent series.
> > There is no finite upper bound on the lengths of finite strings.
> Then there is no reason to conclude that they are all finite.
Since each is only one character longer than some other, all the way
down to one character strings, there is no possible way that any of them
can be infinite any more than the infinite set of finite naturals
contains any infinite naturals.
TO remains invincibly ignorant on these issues, in part due to his
incurable quantifier dyslexia.
.
- Follow-Ups:
- Re: infinity
- From: aeo6
- Re: infinity
- References:
- Re: infinity
- From: Daryl McCullough
- Re: infinity
- From: aeo6
- Re: infinity
- Prev by Date: Re: Another Coxeter group question
- Next by Date: Re: infinity
- Previous by thread: Re: infinity
- Next by thread: Re: infinity
- Index(es):
Relevant Pages
|
