[FOM] Cardinality in the absence of AC
Floris van Doorn
fpvdoorn at gmail.com
Fri Jun 7 03:47:39 EDT 2019
You can take
A = { X in powerset(R) | X is not Lebesgue measurable }
B = powerset(R)
- Floris
On Thu, 6 Jun 2019 at 21:29, Timothy Y. Chow <tchow at math.princeton.edu>
wrote:
> On Wed, 5 Jun 2019, Noah Schweber wrote:
> > Unless I'm missing something the answer is no, that situation is not
> > possible - and this is provable in (much less than) ZF alone.
> > Let f: B->C be injective and g: A->C be surjective, fix b in B and
> > consider the map h from A to B given by:
> > - If g(a) is in the range of f, then h(a) is f^{-1}(g(a)) (which is
> > well-defined since f is injective).
> > - Otherwise h(a)=b.
> > This seems to be a surjection from A to B (since f is defined on all of
> > B).
>
> This argument looks correct to me (although I'd phrase the last
> parenthetical comment differently---the reason every b' in B has a
> preimage is that the surjectivity of g guarantees the existence of a in A
> such that g(a) = f(b')). Thanks!
>
> I have a followup question. With my proposed definition of "B outnumbers
> A" as "there is an injection from A to B but no surjection from A to B,"
> can we construct sets A and B such that
>
> 1. in ZFC we can prove that |A| = |B|, but
> 2. in ZF+LM we can prove that B outnumbers A?
>
> The way I've defined "B outnumbers A" blocks the example of A = R and B =
> R/Q that I mentioned previously.
>
> Tim_______________________________________________
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