Cardinality of sets of reals without choice

101 Views Asked by At

Assuming just ZF (no axiom of choice): Does $\aleph_n\leq|\mathbb{R}|$ for all $n<\omega$ imply $\aleph_\omega\leq|\mathbb{R}|$?

(with $\kappa\leq|\mathbb{R}|$ meaning that there is a set of reals of cardinality $\kappa$)

It is clear that this would follow from countable choice.

1

There are 1 best solutions below

0
On BEST ANSWER

Nope. It does not imply that.

This is a result due to Derrick and Drake. The idea, in its essence, is to add $\aleph_n$ Cohen reals, for each $n$, and take only the sets which are definable by a bounded part of the forcing.