Re: Cantorian pseudomathematics

MoeBlee said:
> Tony Orlow wrote:
> > MoeBlee said:
> > > Virgil wrote:
> > > > In article <MPG.1e35dfb2f0fbd3ad98a990@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > > > Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
> > > > > N=S^L, where N is the size of the set of strings of length L and S is
> > > > > the size of the alphabet from which they are made.
> > > >
> > > > This says that when L and S are finite naturals then N is a finite
> > > > natural, but says nothing about one is dealing wih a set of sring of
> > > > either unbounded string length or unbounded alpha bet size.
> > >
> > > I don't understand why you say that.
> > >
> > > Let 'x-pre-y' stand for the set of functions from y into x.
> > >
> > > Let 'x^y' stand for cardinal exponentiation, defined: x^y =
> > > card(x-pre-y).
> >
> > I am not sure this is what I mean, exactly, but maybe it is equivalent to
> > normal exponentiation, at least for finites. I mean 1 multiplied by x y times.
> >
> > >
> > > I take 'string' to mean a function whose domain is an ordinal. So a
> > > finite string is a function on a natural number, an omega-string is a
> > > function on omega, a denumerable string is a function on omega or on
> > > another denumerable ordinal, a countable string is a function on a
> > > natural number or a denumberable ordinal, and an uncountable string is
> > > a function on an uncountable ordinal.
> >
> > It's interesting that you allow uncountably long strings, if that's what you're
> > saying, since that was a point of contenntion in several other discussions. I
> > define a string as an order set, or, one or more symbols distinguished by their
> > order in the string, not by value.
> >
> > >
> > > So card(S-pre-L) = card(S)^card(L).
> > >
> > > So Orlow's formulation is a theorem of set theory, whether regarding
> > > finite or infintite sets.
> > >
> > > Or, if I'm mistaken, then what is the error in the above?
> >
> > The only problem I see here is the use of cardinal numbers, since card(L) for
> > all finite lengths is going to be aleph_0.
>
> If L is finite, then card(L) is finite. But card({L: L is a finite
> length}) is infinite. If that addresses your concern.
>
> > For me, if S is finite, then for any
> > finite L, N is finite.
>
> Correct.
>
> > You would have 2^aleph_0 for the number of binary
> > strings of finite length.
>
> No. The set of all binary strings of all finite lengths is a countable
> union of finite sets. This is countable. For each length, there are a
> finite number of strings. But there are a countable number of finite
> lengths. A countable union of finite sets.

This is precisely the kind of incorrect correction I am sick of. Did I say
there were 2^aleph_0 LENGTHS? No, I said there were 2^aleph_0 STRINGS, given
aleph_0 lengths. Now, that was incorrect. The correct statement is there are
sum(n=0->aleph_0: 2^n), or 2^(n+1)-1 strings. But, close enough. I didn't say
there were 2^aleph_0 LENGTHS. But you probably still think you caught me in
another one of my logical or set-theoretical "errors". Why can't mathemticians
read?

>
> > I say 2^N it is an unboundedly large but finite
> > number of strings
>
> You say whatever you want. But I'm not talking about your system in
> which you say whatever you want. My post was about SET THEORY. Anyway,
> as you said N = S^L (for a finite S and a fixed finite L), then N is
> finite and so is 2^N.

Right, and all of your naturals are finite, so this is true for all of them.

>
> > since N is an unboundedly large but finite number of finite
> > natural lengths, given the equivalence between count and value in the naturals.
>
> N becomes larger but finite as L becomes larger but finite. The UNION,
> for all L, of the N's is infinite.

not once you step away from the intoxication of the von Neumann ordinals.

>
> I don't address your "equivalence between count and value in the
> naturals" since my point was about SET THEORY, not about whatever your
> DIFFERENT system is supposed to be.

If all you want to do is drive standard set theory through my thick skull, then
you can give up now. I don't care to hear the same repetitions of unfounded
faith that continually flow out of the fountain/urinal in Cantor's Garden. If
you are not at all willing to discuss any other perspective than the one you
alreay have, then there's nothing to talk about. Why should I play your game,
while you scoff at mine?

>
> MoeBlee
>
>

--
Smiles,

Tony
.

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