Re: Orlow cardinality question

Martin Shobe said:
> On Thu, 16 Jun 2005 11:41:01 -0400, Tony Orlow (aeo6)
> <aeo6@xxxxxxxxxxx> wrote:
>
> >Martin Shobe said:
> >> On Wed, 15 Jun 2005 10:44:33 -0400, Tony Orlow (aeo6)
> >> <aeo6@xxxxxxxxxxx> wrote:
> >>
> >> >Martin Shobe said:
> >> >> On Mon, 13 Jun 2005 09:51:04 -0400, Tony Orlow (aeo6)
> >> >> <aeo6@xxxxxxxxxxx> wrote:
> >> >>
> >> >> >Martin Shobe said:
> >> >> >> 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.
> >> >> >Oh wow, so omega-1 must be that magical greatest infinite.
> >> >>
> >> >> I'll assume you mean greatest finite here, otherwise I don't see any
> >> >> connection to what I said. And, no omega - 1 is not the "magical
> >> >> greatest infinite". omega - 1 doesn't exist.
> >> >>
> >> >> That's one for the
> >> >> >books. If you simply declare omega to be the smallest infinity, why not declare
> >> >> >alpha to be the largest finite?
> >> >>
> >> >> I don't simply declare omega to be the smallest infinity. In ZFC, it
> >> >> is a theorem that a smallest infinite ordinal exists. Once we have
> >> >> proven this, we name that smallest infinite ordinal, omega.
> >> >A theorem proven by what axioms?
> >>
> >> ZFC, NBG are both capable of proving it, and from my point of view as
> >> a dabbler, they seem to be far and away the most common axioms used
> >> when dealing with sets. (Especially ZFC).
> >
> >That didn't answer the question. I'd like to saa anyone derive this "fact", and
> >I'll be happy to show you the false assumption upon which it rests.
>
> There exists a set, A, such that
> 1) The empty set, 0, is in A.
> 2) For all x, x is in A -> (x u {x}) is in A.
> [Infinity]
>
> There power set of A, P(A) exists.
> [Power Set]
>
> There exists a set, B, such that x is in B <->
> 1) x is in P(A)
> 2) 0 is in x.
> 3) for all y in x, (y u {y}) is in x.
> [Comprehension]
>
> There exists a set, C, such that x is in C <->
> 1) x is in A
> 2) for all y in B, x is in B.
> [Comprehension]
>
> Define h:C -> C such that, for all x in C, h(x) = x u {x}.
> [Power Set, Unordered Pair, Sum, Comprehension]
>
> let h(x) = h(y)
> {x u {x}} = {y u {y}}.
> x = y.
> [Foundation]
>
> h is one to one.
>
> for all x, x is in h(x).
>
> for all x, 0 =/= h(x).
>
> h(C) is a proper subset of C.
>
> Therefore C is infinite.
>
> There exists a set, D, such that x is in D <->
> 1) x is in C
> 2a) x = 0
> or
> 2b) There exists a y in C such that x = y u {y}.
> [Comprehension]
>
> 0 is in C
>
> 0 is in D
>
> Let x be in D.
> x is in C.
> There exists a y in C, such that y = x u {x}
> y is in D.
> x u {x} is in D.
>
> Therefore C is a subset of D.
>
> Let E be an infinite set.
>
> There exists a one to one function f:D -> E such that E is a proper
> subset of D.
>
> There exists an a, such that a is in D and a is not in E.
>
> Define g:C -> D such that
> 1) g(0) = a.
> 2) for x in C, y in D, g(x) = y -> g(x u {x}) = f(y).
> [Power Set, Unordered Pair, Sum, Comprehension]
>
> There exists a set, G, such that x is in G <->
> 1) x is in C.
> 2) for all y in C, g(x) = g(y) -> x = y.
> [Comprehension]
>
> g(0) = a
> g(0) is not in E.
>
> Let g(0) = g(x).
> x is in D.
> x = 0 or there exists a y in C such that x = y u {y}.
> x = 0 or g(x) = g(y u {y}) = f(g(y)).
> x = 0 or g(x) is in E.
> g(x) is not in E.
> x = 0.
>
> Therefore, 0 is in G.
>
> Let x, y be in G.
> Let g(x u {x}) = g(y u {y})
> g(x u {x}) = f(g(x)) = g(y u {y}) = f(g(y)).
> g(x) = g(y).
> x = y.
>
> Therefore (x u {x}) is in G.
>
> Therefore C is a subset of G.
>
> Therefore g is one to one.
>
> Therefore |C| <= }E|.
>
> Therefore |C| is the smallest cardinal.
>
> Your turn.
>
> >> >There is no such thing,
> >> >and if ZFC claims there is, that's just another glob o' junk for the bucket.
> >> >What axioms prove this? They need review.
> >>
> >> Then don't use ZFC. But, you should keep two things in mind.
> >>
> >> 1) Just because you don't like a result, it doesn't make it
> >> inconsistant. You have claimed that ZFC is inconsistant, but so far,
> >> you have only shown that it proves things you wish it wouldn't.
> >
> >I have shown that it makes assumptions that are inconsistent with other
> >assumptions within set theory and outside of it.
>
> And anyone can show any theory to be inconsistent by making
> assumptions outside of that theory. On the other hand, you have never
> shown it to be inconsistent without making assumptions outside of set
> theory.
That's true, but the rules outside of set theory that I am using are ones which
apply to the elements in any given exmaple. You can't use quantities or strings
to prove your point, and then ignore the properties of quantities or strings.
>
> >>
> >> 2) You are going to have a hard time making what you want consistent
> >> without introducing even bigger problems (from the point of view of
> >> most mathematicians).
> >
> >I have heard that opinion numerous times, along with admonishments about how
> >it's not that easy, etc. Yet, no one has pointed out, despite their claims to
> >the contrary, any inconsistencies in my understanding, either internally or
> >externally, with anything except cardinality, which doesn't agree with anything
> >else anyway. There are no bigger problems this introduces. Only ones it
> >addresses.
>
> Let's see. You don't think having the size of a set be dependent on
> the characteristics of another set isn't a problem?
You mean like comparing to a standard set?

> You don't think
> that having sets for which no size exists isn't a problem?
I think there may be sets which are hard to compare to standard sets, but all
sets have a size. What set did I say has no size? The set of finite naturals? I
don't see that as any more of a problem than having no maximal element in your
finite range. Live with it.

> You don't
> think having sets with multiple sizes isn't a problem?
I don't have sets with multiple sizes. I have set expressions which can be
interpreted as different sets, depending on whther you are talking about
quantity or symbolic representation. That's a step toward clarification, when
you are using mixtures of these concepts and getting mixed results.

> Those are all
> problems that your system has introduced. All so you can avoid having
> a smallest cardinal.
All so I can achieve measures of infinite sets which are precise and intuitive.
>
> Martin
>
>
I'll have to find some time to go through your proof above. I can't do it now,
but I have printed it out for later perusal. Thanks.
--
Smiles,

Tony
.

Relevant Pages

  • Re: ZFC in another shape.
    ... So x is an ordinal iff x is transtive and every member in x is ... power set. ... Not unless ZFC is inconsistent, ...
    (sci.math)
  • Re: Would it matter if ZF was inconsistent?
    ... a _claim_ that ZF is inconsistent ... Platonists are concerned with what they _believe_ to be true. ... Formalists use ZFC because it hasn't been proved inconsistent. ... and mutually exhaustive, for mathematicians who currently ...
    (sci.logic)
  • Re: Rational numbers, irrational numbers: each dense in real numbers
    ... And your "arguments" as to your opinion that ZFC is inconsistent are ... specified transfinite recursion schema not defined for ordinals up to ... The property that there remains an uncountable number or irrationals ...
    (sci.math)
  • Re: Would it matter if ZF was inconsistent?
    ... a _claim_ that ZF is inconsistent ... Platonists are concerned with what they _believe_ to be true. ... Formalists use ZFC because it hasn't been proved inconsistent. ... and mutually exhaustive, for mathematicians who currently ...
    (sci.logic)
  • Re: The modern mathematical concept of infinity is ...
    ... were to prove PA inconsistent, ... Notice that since Gentzen proved if ZFC is consistent, ... I believe the answer is tetration. ... If Nelson succeeds, the so-called "cranks" will be vindicated. ...
    (sci.math)
Quantcast