Re: infinity

David Kastrup said:
> Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> writes:
>
> > *** T. Winter said:
> >> In article <MPG.1d88f149318a1a7298a210@xxxxxxxxxxxxxxxxxxxxxxxxx> Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> writes:
> >> > *** T. Winter said:
> >> ...
> >> > > > For each L, there are 2^L strings of that length, and
> >> > > > sum(x=0->L: 2^x) strings up to that length. If L is finite,
> >> > > > then sum(x=0->L: 2^x) is a sum of a finite number of terms,
> >> > > > each of which is finite, so there are a finite number of
> >> > > > strings in a language with strings of finite length.
> >> > >
> >> > > The "so" part does not follow, it only follows if there would
> >> > > be a maximal length.
> >> >
> >> > No, it follows that if the length is always finite, then this
> >> > fact holds.
> >>
> >> No, it does not follow. *If* there is a maximal length L, then
> >> there are finitely many strings with length less than or equal to
> >> L. There is however no L such that all strings are in length less
> >> than that L, so it does not follow.
>
> > You miss the point.
>
> Well, you would not see it if it poked you in the eye.
>
> > For any finite L, the set of all strings less than or equal to L is
> > finite.
>
> Sure.
>
> > To get an infinite set of all string less than or equal to L you
> > need infinite L.
>
> But the set of all strings is not a set of strings less than or equal
> to some L.
For ANY finite length L, there are a finite set of strings of length L or less.
For which L in N is the set of all strings up to that length an infinite set?
For NONE of them. There is not ONE finite string which, in the ordered set of
strings, has an infinite set of predecessors. Therefore the set is finite,
since any infinite ordered set MUST have some members (an infinite number)
which have an infinite number of predecessors.
>
> > But, you have NO L which is infinite in your set of finite strings,
> > so therefore you cannot have any infinite initial segment of this
> > set.
>
> The whole set of strings does not have the structure of an "initial
> segment" since it has no end.
It has no "end", but never reaches infinity. It is unbounded but finite.
>
> >> Also correct. This still does *not* show that in a set of finite strings
> >> where there is no longest string (there are always longer ones available)
> >> is also finite. So your assertion:
>
> > Yes it does!!! You can only achieve an infinite set of such strings
> > if you allow infinite strings, as shown.
>
> Whining does not make it so.
Logic does. Making stupid insulting remarks does not make you look too smart,
if you ask me.
>
> > But, you do NOT allow such infinite strings, so you can NOT achieve
> > an infinite set. It doesn't get much simpler than that. Maybe it is
> > time to accuse me of quantifier dyslexia again, but that would only
> > look stupid.
>
> Well, it _looks_ quite stupid that you still don't get the difference
> between "arbitrary large" and "infinite".
YOU are the ones missing that distinction. Abitrarily large but finite naturals
comprise an unbounded but finite set. This is what I am trying to explain here.
>
>

--
Smiles,

Tony
.

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