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MathOverflow
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2009/12/18 — But why should there be a definable Vitali set? Of course, some models of set theory have definable Vitali sets, because sometimes there is ...
回答 2 件
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(Edit.) With a closer reading of your question, I see that you asked for a very specific notion ...
What is the least α such that Lα contains a non-measurable set
2024年9月10日
Measurable and definable sets - MathOverflow
2013年10月20日
Is definability of a basis for - R - N - independent of ZFC?

3）
Springer
link.springer.com › article
VG Kanovei 著 · 2017 · 被引用数: 33 — The consistency of the existence of a countable definable set of reals, containing no definable elements, is established.
What does the Axiom of Choice know about when creating ...

4）
Quora
www.quora.com › What-does-the-Axiom-of-Ch...
Vitali sets were one of the first examples of a non-measurable set. ... For other reasons it's fairly plausible that no Vitali set is definable.
回答 5 件
As others answers have already remarked, the Axiom of Choice is a mathematical statement - and ...
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この最後のQuoraの回答より
Samuel Gomes da Silva
Ph.D. in Mathematics & Set Theory, University of São Paulo (USP) (Graduated 2004)Author has 660 answers and 242.8K answer views3y
As others answers have already remarked, the Axiom of Choice is a mathematical statement - and so it does not “know” about anything.

What possibly the OP was asking about is what do we “know” or “see” when we invoke the Axiom of Choice in order to get the Vitali set. Well, the Axiom of Choice is regarded as a non-constructive statement, so we do not have an algorithm or something like that in order to properly “construct” the Vitali set. We only use the Axiom of Choice to assert its existence, and that’s all.

The existence of free ultrafilters over the naturals is also a non-constructive statement: it follows from the Axiom of Choice, but it is a weaker assumption. Sierpinski has proved in the 30’s that a free ultrafilter itself (identifying its members with their characteristic functions and then with the corresponding real numbers of the unit interval written in the binary base) is a non-measurable subset of the real line. So, if you wanna “see” something, after assuming the existence of a free ultrafilter over the naturals (and this remains to be a non-constructive process, do not forget about that) then the verification of its non-measurability is somehow more intuitive than the procedure with Vitali sets, IMO.
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