Re: Cardinality of Real Numbers

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
.

Relevant Pages

  • Re: Cardinality of Real Numbers
    ... >>> I think the issue is that without the Axiom of Choice, ... >>> assume that all cardinals are orderable. ... They don't have any, um, ordinality. ... besides the discussion about ramifications of well-ordering ...
    (sci.math)
  • Re: Cardinal arithmetic question
    ... had proved that the Axiom of Choice implied the ... Well-Ordering Theorem, and after Frankel showed ... See H. Rubin and J. Rubin, ... if maxis always defined for infinite cardinals k and u ...
    (sci.math)
  • Re: Skolems Paradox
    ... inversion axiom. ... cannot be a set, in that supposed "set theory", and cannot be a class. ... that the rationals and irrationals alternate in the well-ordered reals, ... characteristics as NCD sets for the reals in the well-ordering of the ...
    (sci.logic)
  • Re: Amateur continuum hypothesis question
    ... There are two common kinds of transfinite "number": ordinals and cardinals. ... The well-ordering theorem I stated above implies ... correspondence with an initial segement of Y. The basic property I ...
    (sci.math)
  • Re: True = [ proven | provable ]
    ... > constant symbols, function symbols, relation symbols, and variables. ... > had also thought that a well-ordering of the sentences would be sufficient ... > to prove that every consistent set of sentences has a consistent complete ... I had thought that the axiom of choice would only be ...
    (comp.theory)