## 580 -Some choiceless results (5)

February 2, 2009

[This lecture was covered by Marion Scheepers. Many thanks! The notes below also cover lecture 6.]

We want to prove the following result and a few related facts.

Theorem. (Specker). ${\rm CH}({\mathfrak m})$ and ${\rm CH}(2^{\mathfrak m})$ imply $2^{\mathfrak m}=\aleph({\mathfrak m})$.

It follows immediately from the theorem that ${\sf GCH}$ implies ${\sf AC}$, since the result gives that any (infinite) ${\mathfrak m}$ embeds into $\aleph({\mathfrak m})$.