# All Borel $\sigma$-algebras are separating and countably generated?

I am studying the book Large Networks and Graph Limits by L. Lovasz. However, because of my CS background, I am not very knowledgable in topology and measure theory (but I try to catch up). In particular, I am stuck on the following paragraph, taken from http://www.cs.elte.hu/~lovasz/bookxx/hombook-almost.final.pdf, page 341:

We start with a quite general notion. Let $(\Omega, \mathcal{B})$ be a Borel sigma-algebra; then $(\Omega,\mathcal{B})$ is separating and generated by a countable family $\mathcal{J} = \{J_1 , J_2 \ldots\}$ of subsets of $\Omega$. It will be convenient to assume that the generator set $\mathcal{J}$ is closed under complementation and finite intersections, which implies that it is a Boolean algebra (not a sigma-algebra!). We call the sets in $\mathcal{B}$ and also in $\mathcal{B}\times\mathcal{B}$ etc. Borel sets. (The reader who likes more concrete structures can think of this as the sigma-algebra of Borel sets in $[0, 1]$, with $\mathcal{J}$ consisting of finite unions of open intervals with rational endpoints.)

I have three questions regarding the second sentence of the paragraph.

1. I would like to clarify if my understanding of the definition of Borel $\sigma$-algebra is correct. If $(\Omega,\mathcal{B})$ is a Borel $\sigma$-algebra, it means that there is in a fact an underlying topology $\mathcal{X}$ on $\Omega$ (set of open sets) and $\mathcal{B}$ is then the smallest $\sigma$-algebra such that $\mathcal{X}\subseteq\mathcal{B}$. Is this correct?

2. By "separating", does the author mean "separable"? Or "separated"? This is not clear, as the the example $\Omega=[0,1]$ is both separable ($[0,1]\cap\mathbb{Q}$ is a dense subset) and separated (clearly).

3. The author claims that every Borel $\sigma$-algebra $(\Omega,\mathcal{B})$ is generated by a countable family of subsets of $\Omega$. Should this be obvious? After some google-search, I have not found a proof of this proposition. I do see how the open intervals with rational endpoints generate Borel sets in $[0,1]$, but not the general case. The proposition seems suspicious to me, since $(\Omega,\mathcal{X})$ can be arbitrarily weird topological space. To get that the Borel $\sigma$-algebra is countably generated, we would need to use some intrinsic property of Borel sets, independent of the space itself?

• Have you looked at Appendix A.3? – Meta-мета-μετα-meta-мета-μετα Aug 30 '16 at 13:30
• Oh, I have not. And everything is clear now, the author uses some non-standard terminology. Thank you. I am kind of embarrassed for my ignorance now. Should I make an answer clarifying matters, or delete the question, or what? – JS_ Aug 30 '16 at 13:37
• It's probably pretty much up to you. (Speaking strictly for myself, I'm not certain how useful a question which was resolved by simply reading more from the source is.) – Meta-мета-μετα-meta-мета-μετα Aug 30 '16 at 14:37
• The correct terminology is "standard Borel space", which means "Borel space of a Polish space". – justt Aug 30 '16 at 16:36