# Tag Info

## Hot answers tagged borel-sets

Accepted

### Proof that the Cardinality of Borel Sets on $\mathbb R$ is $c$ without using the ordinals .

Your idea is sound, but it requires more work. As pointed out, you have only described so far a very small subcollection of the Borel sets. Instead, show that you can associate to each Borel set a ...
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Accepted

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### A set in the Borel $\sigma$-algebra over $[0,1]$ that isn't in the algebra generated by open sets

Claim: For any set $X$ obtainable the boundary is nowhere dense. Proof: Boundary of open set is nowhere dense. Now for the operations: 1) complement - since boundary is the same for a set and its ...
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### Is every Borel set a countable union of intervals?

The Cantor set $C$ is an uncountable Borel set and it does not contain any non-trivial interval (i.e. a singleton). Hence $C$ can not be written an as countable union of intervals. However, its ...
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### Example of limit of a net of Borel functions not Borel

Let $A$ be a subset of $\mathbb R$ that is not Borel. I'll produce a net of Borel functions on $\mathbb R$ whose limit is the characteristic function of $A$. The index set $I$ for my net is the set ...
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### Can we obtain the Borel $\sigma$-algebra on $[0;1]$ as a limit of finite algebra?

Yes there is a natural and useful construction of $\sigma$-algebras as a "limit". But not as the limit of a sequence, rather the limit of an uncountable collection, indexed by the countable ordinals (...
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### What sets are obtained by adding $\aleph_1$ unions and intersections to the Borel algebra?

This is an excellent question! Note first that, just from cardinal arithmetic considerations, whether we obtain all sets is independent (and therefore so is whether we obtain all Lebesgue measurable ...
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### What is the Sigma Algebra generated by Jordan measurable sets?

I found the answer in this journal paper. The Sigma algebra generated by the Jordan measurable sets is the collection of all sets which can be written as a union of a Borel set and a subset of a ...
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### $[0,1)$ is in both $G_{\delta}$ and $F_{\sigma}$

$G_\delta$ is not the complement of $F_\sigma$. The complement of an $F_\sigma$ set is a $G_\delta$ set.
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