Re: Cardinality of Real Numbers

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?

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?

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? Doesn't that violate a trichotomy
of cardinals, and thus infinite sets are equivalent? I guess from
scanning that AC article that AC is there described to imply dichotomy
of cardinals, I generally interpret AC to mean that there's a choice
function, which is the same thing as that the set is well-orderable.
If each set has a cardinality, then there is at least one ordinal for
that cardinality, the guaranteed existence of an ordinal, and it's
well-orderable.

Well-order the reals. That's one of Hilbert's problems and a solution
will bring to you accolades.

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.

Infinite sets are equivalent.

Regards,

Ross

--
Ross A. Finlayson

.

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