Re: infinity
- From: stephen@xxxxxxxxxx
- Date: Wed, 31 Aug 2005 19:23:00 +0000 (UTC)
Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> stephen@xxxxxxxxxx said:
>> I am not assuming that there is a longest word. A longest word
>> implies a largest natural number. You always deny that there
>> is a largest natural number, but once again you are using an
>> argument that depends on there being a largest natural number.
> I said "longest WORDS", not "longest word".
There are no longest words either, so I am not assuming
anything about them. There is no maximum finite word length.
>>
>> 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.
I asked what L is. L is not the maximum number of unique
strings in the language.
>>
>> You should have a summation:
>> S^0 + S^1 + S^2 + S^3 + .... S^k + ...
>> for all finite k. That is the number of finite length strings
>> over an alphabet with size S.
> Yes, and you are summing a finite number of terms, each of which is finite in
> value. Is the sum infinite? No.
But I am not summing a finite number of terms. I am summing
all of the finite k. There are an infinite number of finite k.
You cannot assume there are only a finite number of finite
k when trying to prove that there are only a finite number
of finite k.
>>
>> 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, I am saying "for all finite k". There is no finite upper
bound for k. I noticed that you have conveniently ignored my proof
above. What about the strings of length F? Which of the following
do you disagree with:
1. F is a finite number
2. There are strings of length F
3. There are S^F strings of length F
4. S^F > F
5. There are more than F finite strings.
You apparently disagree with 5, because you insist
there are only F finite strings. You definitely agree
with 1, and you seem pretty certain about 3 and 4. So
I guess 2 is the one you must disagree with, but
I do not see how you can claim that the set of all
finite strings does not include strings of length F
if F is a finite number.
> k cannot be infinite. Therefore sum
> (x=1->k: S^x) is finite. If we use an S-based digital number system, then each
> digit represents a power of S, and we get sum(x=1->k:S^x) is a string of k 1's.
> Now, how many 1's must we concatenate to get an infinite value? We need an
> infinite number of 1's to get an infinite value, with a finite based digital
> number system. So, in order to make this sum infinite, k must be infinite,
> which means you have infinitely long strings. QED.
This is all irrelevant and based on the assumption that
there is a largest finite number. There is not, and
you apparently agree that there is not, so why you keep
using it in your proofs is beyond me. Also, there
are no infinite values in the normal natural numbers,
so why you are making arguments about infinite values
is also beyond me.
>> > 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." 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?
Nobody is adding finite sets of strings to finite sets of strings
to form infinite languages. Languages are not formed by adding
elements one at a time, or by adding elements any finite quantity
at a time. Time is not involved.
L = a*(ba+ab)*b
is a language containing an infinite number of strings
(all of which are finite). I do not create this language
a string at a time. I define that language with some
finite structure, such as a regular expression, and I am done.
>>
>> 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.
So you keep saying without proof. That may be your definition
of infinite, but your definition of infinite does not
include unending sets that any reasonable person would call
infinite.
Stephen
.
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