Re: Cardinality of Real Numbers

Martin Shobe wrote:
> On 31 Aug 2005 18:10:12 -0700, "Ross A. Finlayson"
> <raf@xxxxxxxxxxxxxxx> wrote:
>
> >Jonathan Hoyle wrote:
> >> I think the issue is that without the Axiom of Choice, you cannot
> >> assume that all cardinals are orderable. That is to say, assuming the
> >> Axiom of Choice and given any two arbitrary sets S1 and S2, we know
> >> that exactly one of the following three holds:
> >>
> >> 1. |S1| = |S2| (there is a bijection between S1 and a S2)
> >> 2. |S1| < |S2| (there is a bijection between S1 and a proper subset of
> >> S2, but not between S1 & S2)
> >> 3. |S1| > |S2| (there is a bijection between a proper subset of S1 and
> >> S2, but not between S1 & S2)
> >>
> >> Without AC, you cannot assume that exactly one of the above three must
> >> hold. You could have a case in which S1 and S2 are incomparable; that
> >> is to say, none of the above conditions hold.
> >>
> >> The wikipedia entry may get you started:
> >> http://en.wikipedia.org/wiki/Cardinal_number
> >>
> >> As for the reals, the Axiom of Choice proves that there exists a
> >> well-ordering of them.
> >>
> >> Good Luck!
> >>
> >> Jonathan Hoyle
> >
> >There is a simple existence proof of an ordinal that bijects to that
> >set S1, correct?
>
> Not without using the axiom of choice.
>
> >Where that is so, there is guaranteed the existence of an ordinal
> >bijecting to any set, so any set is well-orderable.
> >
> >That is to say, for S1, regardless of what S2 is, there exists at least
> >one ordinal that bijects with S1: true?
>
> Not necessarily. There are models of ZF where this isn't true.
>
> >If a set has a cardinality, then there is an equivalent ordinal, and
> >the set is well-orderable. From what I understand of you saying so,
> >some sets have varying cardinality?
>
> No. They don't have any, um, ordinality.
>
> > Doesn't that violate a trichotomy
> >of cardinals, and thus infinite sets are equivalent?
>
> ZF can't prove the trichotomy of cardinals (that requires the axiom of
> choice). But regardless of whether it can or can't, it can still show
> that bijections form an equivalence relation, and that there is more
> than one equivalence class of infinite sets.
>
> >So anyways, besides the discussion about ramifications of well-ordering
> >the reals, which are a well-orderable set in ZFC for everybody, does
> >the existence of an ordinal for each cardinal, even if it is not simple
> >to say what it is, does the existence of said ordinal imply a
> >well-ordering exists for that set, any set? Similarly, there is the
> >existence proof of a well-ordering of the reals in ZFC, and if the
> >reals are a set, then they are well-orderable, and the well-ordering
> >implies the extension of Cantor's first to that there must be adjacent
> >points, regardless of whether it implies that there are more than
> >countably many disjoint intervals.
>
> NO. In order to reach the contradiction, Cantor's first requires that
> the well-ordering be order-equivalent to the natural numbers. Since
> any well-ordering of the reals will not be order-equivalent to the
> natural numbers, no contradiction is reached, and there are no
> adjacent points or uncountable collections of disjoint intervals.
>
> Martin

Heh heh heh heh.

I'm talking about an extension of Cantor's first, where for any
well-ordered set that bijects to the reals, there are that many
disjoint intervals generated.

I think it's accepted that there's an ordinal, in the cumulative
hierarchy that diverges to Ord, for each cardinal, Ord is large, and an
ordinal.

What do you mean: "Cantor's first requires the well-ordering be
order-equivalent to N?" Do you mean that to say that Cantor's first
applies to a bijection from N to R only, or what? I'm talking about
any set mapping to R, in extension. What do you mean by
"order-equivalent"?

Skolemize, your model is countable. The generic extension, a
nonstandard definition, contains no extra elements. A model is to a
theory as a class is to a set, to whit, no classes in set theory. V =
L. All infinite sets are equivalent.

With regularity in ZF, is the existence if not identity of an ordinal
for each cardinal certain? All there needs be is an ordinal for any
greater cardinal.

Regards,

Ross

--
Ross A. Finlayson

.

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