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75 votes
4 answers
24k views

Non-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 ...
Anweshi's user avatar
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4 votes
1 answer
718 views

Is every element of $\omega_1$ the rank of some Borel set?

It is well known that we can obtain the $\sigma$-algebra of Borel subsets of $2^{\omega}$ in the following way: Let $B_0$ be the collection of all open subsets of $2^{\omega}$. For $\alpha=\beta+1$, ...
Hannes Jakob's user avatar
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