[FOM] Cardinality in the absence of AC
Timothy Y. Chow
tchow at math.princeton.edu
Wed Jun 5 11:06:23 EDT 2019
Let us work in ZF + "all subsets of reals are Lebesgue measurable" (or
ZF+LM for short).
Is the following situation possible?
We have three sets, A, B, and C. There exists an injection from A to B,
and an injection from B to C. There *does not exist* a surjection from A
to B, and there *does not exist* a surjection from B to C. Nevertheless,
there exists a surjection from A to C.
---
I was led to ask this question after reading an expository article by Alan
Taylor and Stan Wagon in the April 2019 issue of the American Mathematical
Monthly, about the fact that in ZF one can prove the existence of an
injection from R to R/Q and in ZF+LM one can prove the nonexistence of an
injection from R/Q to R. They interpret this fact as saying that in ZF+LM
one can prove that there are "more" elements in R/Q than in R or that R/Q
"outnumbers" R.
It occurred to me that another interpretation could be that what is wrong
is the definition of "|A| < |B|" as "there is an injection from A to B but
no bijection from A to B." Perhaps it is better to define "|A| < |B|" as
"there is an injection from A to B but no surjection from A to B." I
believe that in the absence of choice, these two definitions are not
equivalent.
What is not immediately clear to me is whether the "<" relation as I have
proposed it is transitive. Hence my question.
Tim
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