How is it possible to reconcile the following...
In 1970, Solovay constructed Solovay's model, which shows that it is consistent with standard set theory, excluding uncountable choice, that all subsets of the reals are measurable.
A not too well known application of the Boolean prime ideal theorem is the existence of a non-measurable set[2] (the example usually given is the Vitali set, which requires the Axiom of Choice). From this and the fact that the BPI is strictly weaker than the Axiom of Choice, it follows that the existence of non-measurable sets is strictly weaker than the axiom of choice.
Also, a similar question regarding...
The ultrafilter lemma is equivalent to the Boolean prime ideal theorem, with the equivalence provable in ZF set theory without the axiom of choice. Many other theorems of general topology that are often said to rely on the axiom of choice are in fact equivalent to BPI. For example, the theorem that a product of compact Hausdorff spaces is compact is equivalent to it. If we leave out "Hausdorff" we get a theorem equivalent to the full axiom of choice.
Now, Janich's Topology gives a proof of the full blown Tychonoff theorem from the Ultrafilter Lemma (after showing Zorn's Lemma implies the Ultrafilter Lemma). But this should not be possible since Tychonoff is equivalent to AC and UL is weaker than AC. So do I need to read his proof more carefully...is he still using the full power of AC elsewhere?