# [FOM] Cardinality in the absence of AC

Timothy Y. Chow tchow at math.princeton.edu
Thu Jun 6 10:39:20 EDT 2019

```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
```