Re: Orlow cardinality question

On Wed, 1 Jun 2005 14:15:50 -0400, Tony Orlow (aeo6)
<aeo6@xxxxxxxxxxx> wrote:

>Martin Shobe said:
>> On Tue, 31 May 2005 12:52:34 -0400, Tony Orlow (aeo6)
>> <aeo6@xxxxxxxxxxx> wrote:
>>
>> >David Kastrup said:
>> >> Uh, no. You string length is unlimited over the set, meaning that
>> >> there can't be a fixed maximum for the complete set, but every single
>> >> one of those lengths is finite.
>> >Why?
>>
>> There can't be a fixed maximum for the complete set becuase if there
>> was, we could append another member of S to the end of such a string
>> and have a finite string with a greater length.
>>
>> Every single one of those lengths is finite because the set in
>> question is the set of all finite strings.
>And what is the upper bound on the length of those strings?

1) There isn't a unique upper bound. Any infinite ordinal is an
upper bound (Since all the strings are finite.) There are no finite
upper bounds for reasons given earlier. Therefore, the least upper
bound is the smallest infinite ordinal. I.e. omega.

>> Because you would be talking about a different set. You can certainly
>> talk about sets with infinite strings. But there are no infinite
>> strings in the set of all finite strings.
>Yes, only a finite amount of finite strings in your set, unless of course
>you're working with an infinite set of symbols.

Nope. Proof. Let S be the set of symbols. let a be in S. Let A be
all the finite strings that contain only the symbol a. (I.e. A = { a,
aa, aaa, aaaa, ...}. A is obviously a subset of the set of finite
strings for the all the symbols in S. Now definie as follows. f(x)
is the strings xa. f is clearly a bijection. and there is no x in A
such that f(x) = a. Therefore A is infinite. Therefore, S is
infinite.

Martin

.

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