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$\begingroup$ Your statement that the existence of non-Lebesgue measurable sets is equivalent over ZF to the Hahn-Banach theorem is surprising to me. Could you give a reference (or a proof) for the implication from non-measurable sets to HB? $\endgroup$

Andreas Blass
– Andreas Blass

2017-01-27 19:05:20 +00:00
Commented Jan 27, 2017 at 19:05
$\begingroup$ @AndreasBlass To my knowledge the original result is due to Foreman and Wehrung in the following paper from Fundamenta Mathematicae (1991): matwbn.icm.edu.pl/ksiazki/fm/fm138/fm13812.pdf $\endgroup$

Dawid K
– Dawid K

2017-01-27 23:12:53 +00:00
Commented Jan 27, 2017 at 23:12
1

$\begingroup$ Ah, I see what you did there. The word equivalent is misleading, since the proved implication is only from HB to non-measurable sets, and not the other way. Amending the original answer. $\endgroup$

Dawid K
– Dawid K

2017-01-27 23:25:59 +00:00
Commented Jan 27, 2017 at 23:25
$\begingroup$ It may also be worth mentioning that HB is related to the ultrafilter lemma rather than "full" choice. But of course the ultrafilter lemma is enough to prove non-measurable sets and is arguably the part of choice most important for that. IIRC, the BPIT equivalent to Tychonoffs theorem for hausdorff spaces which is sufficient to prove HB. $\endgroup$

saolof
– saolof

2022-09-20 07:30:39 +00:00
Commented Sep 20, 2022 at 7:30

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