Re: infinity

Virgil said:
> 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!
Sure, Cantor-infinite, but not actually infinite. Cantor's definitions suck.
> > >
> > > >> 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.
Yes, even though you have a finite number of finite terms, the sum is Cantor-
infinite. Doesn't that indicate a problem with Cantor's definitions/
> > >
> > > >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.
Yes, based on the misconception that you can have an infinite set of unique
finite naturals you can draw the conclusion that you can have an infinite set
of unique finite strings ona finite alphabet. This is your "Cantor-infinite".
>
> Until TO faces up to the Cantor-infintieness of the set of finite
> naturals, he is being willfully blind to mathematical reality.
Cantor-infiniteness isn't at issue. I don't care what Cantor says at this
point. There are obviously problems with Cantor's concept of infinity. I am
talking about actual logic here, not hocus pocus.
>
> > >
> > > >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.
Cantor may base his claims on such shakey ground as "largest member", but there
are clearly not an infinite number of unique finite naturals, nor of unique
finite strings on any finite alphabet.
>
> > 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!
Think for yourself for a change. Cantor is long gone.
> > >
> > > 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!
First of all, the infinitude of those terms has been in question since the
beginning of this conversation, since you are restricting yourself to finite
naturals, of which there are only a finite number. More importantly, my whole
point here is that you are willing to say that you can add finite numbers of
elements at a time, and eventually get an infinite set, but you refuse to
consider that, incrementing a value one unit at a time, it can ever reach
infinity. The set size IS a number, and you are adding finite quantities to
that number at each iteration. If that number can reach infinity through an
infinite series of finite additions, then why can't the natural numbers also
reach infinity through an infinite series of finite additions? This is totally
inconsistent.
>
> > 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?
Not if n is allowed to go to infinity, and yet, you have restricted n to finite
values, and therefore, the sum is finite.
>
> > 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.
Correct. That is why I say that performing the successor operation, or
incrementing, an infinite number of times to get an infinite set of whole
numbers yields infinite numbers as members of that set, since each natural
number is the sum of all the increments up to that point. If you restrict your
numbers to the finite, then there are a finite number of them, and if you use
these numbers as the lengths of your strings, then you only have finite-length
strings, and only have a finite set of them. If you want to have an infinite
number of whole numbers, then you need to include infinite whole numbers.
>
> > > 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.
If adding one natural to the set at a time can produce an infinite set by
incrementing the set size repeatedly, then incrementing the values in it can
just as easily produce an infinite value. Consider each string to be a set of
characters. You say adding one character at a time will never produce an
infinite string, but addding one natural at a time can produce an infinite set?



--
Smiles,

Tony
.

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