CH yet again.
- From: "Bill Taylor" <w.taylor@xxxxxxxxxxxxxxxxxxxxx>
- Date: 30 Jan 2006 20:17:43 -0800
MoeBlee writes:
>> ...i.e. there exists a function F witnessing the fact that
>> every infinite subset of R is if cardinality N or R.
>> (I haven't bothered to define "#" precisely.)
>
> Let me see if I understand correctly:
>
> F is a function on the set of infinite subsets of R such that for every
> X that is an infinite subset of R,
>
> F(X) is a bijection between X and R
> or
> F(X) is a bijection between X and N.
>
> And a version of CH is to say that such an F exists.
Yep, that's it.
> So, the second version implies the first version.
Yep, so without AC it is stronger, or "more definite" as you put it.
In the presence of AC it is merely equally strong.
> But for the first version to imply the second version, we would have to
> choose for each infinite subset X a bijection to be the value of F(X).
> Right? So that's where AC comes in, right?
Yes, that's a nice way of putting it.
> But the second version is not CH as we commonly know it.
Well, it's not *worded* the way we commonly hear it, but if
"commonly" means using ZFC, as seems reasonable, then it *is* CH.
> You're kind of upping the stakes, trying to make CH more "definite" and
> showing that that definiteness comes with a price, which is AC, which
> is notoriously "indefinite" in its way. Am I at least getting the basic
> idea here?
Yes, that's an interesting way to put it, and yes, you are.
I leave aside any further comment on GCH, which is a rather abstruse
product of "surreal" math, and not my real interest.
>> Even in the above, the result is not exactly the same if the
>> (1-1) functions are weakened to onto functions in reverse order.
> Maybe I should save that for an exercise for when I'm more up to speed.
Yes it's an intriguing thing. Every injection can be trivially
reversed to a surjection; but the converse cannot be assured
without AC. This leads to all sorts of intriguing things in ZF(no C).
e.g. it is easy to prove the Schroder-Bernstein theorem that
if A and B inject to each other, then they biject. But the "dual"
version, that if they surject to each other, then they must biject,
cannot be proved without some form of AC. I have been informed
that it is strictly weaker than AC, that there are models where AC
is false, but dual-S-B is true, but I have no idea what they look like.
There are many other intriguing results along these lines, but OT here.
-------------------------------------------------------------------------
Bill Taylor W.Taylor@xxxxxxxxxxxxxxxxxxxxx
-------------------------------------------------------------------------
Mathematics runs under set theory just as Freecell runs under windows.
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