Answer by gowers for Non-Borel sets without axiom of choice
There are some very nice examples of non-Borel sets. Two that I particularly like are the differentiable functions (as a subset of the space of continuous functions on [0,1], say) and the set of all...
View ArticleAnswer by George Lowther for Non-Borel sets without axiom of choice
Measure theory without the Axiom of Choice (not even countable choice) is discussed in Fremlin, Measure Theory, Volume 5, Chapter 56. This is freely available online. Thanks to MO and...
View ArticleAnswer by Andrés E. Caicedo for Non-Borel sets without axiom of choice
No, it is not possible. It is consistent with ZF without choice that the reals are the countable union of countable sets. (*)From this it follows that all sets of reals are Borel. Of course, the...
View ArticleAnswer by Joel David Hamkins for Non-Borel sets without axiom of choice
If you assume the countable axiom of choice, then most setsof reals are not Borel. Under AC, what you get is thatthere are continuum many Borel sets, that is,$2^{\aleph_0}$ many, but $2^{2^{\aleph_0}}$...
View ArticleNon-Borel sets without axiom of choice
This is a simple doubt of mine about the basics of measure theory, which should be easy for the logicians to answer. The example I know of non Borel sets would be a Hamel basis, which needs axiom of...
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