Re: infinity
- From: stevendaryl3016@xxxxxxxxx (Daryl McCullough)
- Date: 2 Sep 2005 12:14:50 -0700
Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
>
>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.
The length of *which* string? '0' has length 1. '00' has length 2.
'000' has length 3. '0000' has length 4. If U = the set of all
finite sequences of 0s and 1s, then what is L?
>With an alphabet of size S, we can make S^L strings of
>length L.
You haven't said what L is. Is L a constant? Is it a variable?
Is it a natural number? What is L?
>> 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.
Didn't you claim that the set of all finite naturals is a finite set?
>This is exactly the "largest finite" argument, which of course
>causes contradictions
That's right: If you assume that the set of all finite numbers
is a finite set, that leads to a contradiction:
Let A = any finite set of finite naturals that includes 1 and 2.
Let m = the sum of all elements of A.
Then m is larger than any element of A.
Then m is not an element of A.
Then there exists a finite natural number that is not in A.
What that proves is
If A is a finite set of finite naturals, then A does not
contain all finite naturals.
So the collection of all finite naturals cannot be a finite set. QED
>because you can always add 1. So what??
So the collection of all finite naturals is not a finite set,
contrary to your claims.
>> 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?
To say that a sum of positive terms t_j is infinite is to say that
given any finite real number R, there exists a finite natural number
n(R) (note: n is a *function* of R) such that
t_0 + t_1 + ... + t_n(R) > R
That's clearly the case for the sum
S^0 + S^1 + ...
Given any finite real number R, we can find a number n(R) such that
S^0 + S^1 + ... + S_n(R) > R
>But you can't add a finite number to a number in
>succession and get an infinite number? Hmmmm...
The numbers grow without bound, but each number is finite.
>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.
That doesn't follow. What follows from your inductive proof is this:
forall L, the set of all strings of length less than or equal to L
is finite
If we let f(L) be the number of strings of length L or less, then
you can prove by induction:
forall L, f(L) is finite.
To prove that the set of *all* finite strings is finite, you need to
show something different:
exists K, forall L, f(L) < K
>> 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,
It doesn't matter whether I can define it or not. If U is
a finite set of finite strings, then the result of concatenating
all the strings in U is a finite string that is not in U.
>but you cannot define the entire set of naturals this way.
That's because it's *not* a finite set.
>If you say the set size is infinite because it goes on forever
That's what *you* have said. You've said that infinite means "without end".
>then I can say the values in the set are infinite for the same reason.
In the case where A = the set of all *finite* naturals, then
by definition of A, no element of A is infinite. You aren't making any
sense.
>If you say you can add the sizes of finite sets, and
>somehow get to infinity,
I said that if you have a *finite* set of finite naturals,
then you can add all the elements of the set and get a *finite*
result.
>> There is no finite upper bound on the lengths of finite strings.
>
>Then there is no reason to conclude that they are all finite.
Are you saying that there might be a finite string that is infinite?
--
Daryl McCullough
Ithaca, NY
.
- References:
- Re: infinity
- From: Daryl McCullough
- Re: infinity
- From: aeo6
- Re: infinity
- Prev by Date: Re: infinity
- Next by Date: Re: what makes it true?
- Previous by thread: Re: infinity
- Next by thread: Re: infinity
- Index(es):
Relevant Pages
|
