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}).

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