Re: infinity

Daryl McCullough said:
> Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> >
> >stephen@xxxxxxxxxx said:
>
> >> I have no idea what L is in your S^L. You are aware that there
> >> is more than one string length, so picking a single L does not
> >> make any sense. It almost makes sense if you think that L is
> >> the maximum string length, i.e. the largest finite natural number.
> >> Of course you also deny that there is a maximum string
> >> length, so I have no idea what S^L is supposed to mean.
>
> >Given any string length and alphabet, that is the maximum number of unique
> >srings in the language.
>
> What is L?
L is a string length. With an alphabet of size S, we can make S^L strings of
length L.
>
> >> 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.
>
> >k cannot be infinite.
> >Therefore sum (x=1->k: S^x) is finite.
>
> But that's not the sum we are talking about. We're not
> talking about the sum over all x less than k, we're talking
> about the sum over all *finite* x.
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.

>
> >> > Well, for any language you can define on these
> >> > finite words, I can define a larger language simply by adding this L+1
> >> > length word, so in the very same sense, the language is finite as well.
> >> > If the language is infinite because there is no upper obund, then the
> >> > word length is infinite for the same reason.
> >>
> >> What was that supposed to mean? You seem to be saying that
> >> because you can create a larger language, the original language
> >> could not be infinite? You are really making less and less sense.
> >That was in response to your similar statement.
>
> >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. 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.
>
> >So, in response, I say, "All the
> >languages are finite because if you take a finite language and add a finite
> >number of strings you still get a finite language." This directly applies to
> >your summation of subsets with variable L, above. Can you keep adding finite
> >sets of strings to a finite set of strings, and get an infinite set of strings?
>
> 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? 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.

Therefore: The set of all finite strings on a finite alphabet is finite.
>
>
> >> The set of all finite length strings over the alphabet {0,1}
> >> is an infinite set. There is no longest string in this
> >> set, and there is no L to plug into your S^L formula.
> >
> >If the string length cannot be infinite, then the language cannot be
> >infinite.
>
> It can't be finite, because for any finite set of strings, there
> is a finite string not in that set.
For any finite set that you define, you can define a larger one, but you cannot
define the entire set of naturals this way. If you say the set size is infinite
because it goes on forever, then I can say the values in the set are infinite
for the same reason. 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.
>
> >> That set is finite because there there is no finite bound
> >> on the number of elements in the set.
> >
> >I assume you mean infinite. Is there a finite bound on the length
> >of strings in the set?
>
> No. There is no finite bound on the lengths of finite strings.
>
> >What is the upper bound for string length?
>
> There is no finite upper bound on the lengths of finite strings.
Then there is no reason to conclude that they are all finite.
>
> --
> Daryl McCullough
> Ithaca, NY
>
>

--
Smiles,

Tony
.

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