Confusion regarding the definition and generators of Borel $\sigma$-algebra We have defined the Borel $\sigma$-algebra as following:
Let $\Omega$ be a topological space. The $\sigma$-algebra generated by open sets (i.e. the smallest $\sigma$-algebra contains all open sets) is called the Borel $\sigma$-algebra, and noted $\mathcal{B}(\Omega)$. An element of $\mathcal{B}(\Omega)$ is called a Borel(measurable) set.
And then we were given the following example:
On $\mathbb{R}$, the Borel $\sigma$-algebra is also generated by intervals $]a,b[,-\infty \leq a < b \leq \infty$. It is also generated by intervals $[a,b], ]a,b]$ or $[a,b[$.
What does it mean here, when we say that Borel $\sigma$-algebra is generated by the interval $]a,b[$? In my understanding a $\sigma$-algebra generated by for ex. $\varepsilon$ is the intersection of all $\sigma$-algebras containing  $\varepsilon$.
If I apply this to our example doesn't that mean that on $\mathbb{R}$ the Borel $\sigma$-algebra can be generated by any set, or any number. Fox example:
The Borel $\sigma$-algebra generated by $\{1\}$ is the intersection of all $\sigma$-algebras containing $\{1\}$. This can be done for any number, for any set. Is that right?
 A: If $\mathcal{A}$ is a family of subsets of $X$ then $\sigma(\mathcal{A})$ is defined as the intersection of all $\sigma$-algebras on $X$ that contain $\mathcal{A}$. It is the same as the smallest $\sigma$-algebra containing $\mathcal{A}$ and could also be constructed conceptually by starting with $\mathcal{A}$ and then adding only those subsets that we have to add because of the properties of a $\sigma$-algebra, iteratively. So $\emptyset,X$ have to be in it and we need to add all countable unions and intersections, and complements of sets that we already added, until we get nothing new anymore. This can be formalised as a transfinite recursion over ordinals $< \omega_1$, but this is usually not what's done in measure theory. The definition I gave at the beginning is (with some cleverness ) enough to work with the $\sigma$-algebra with generators $\mathcal{A}$.
Read the remark correctly: the Borel $\sigma$-algebra is $\sigma(\mathcal{A})$ where $\mathcal{A} = \{]a,b[: a < b, a, b \in \Bbb R\}$, so not a single interval, but the set of all open intervals. This $\sigma$ algebra contains all open sets, because open sets are all countable unions of such intervals. So it will also contain all Borel sets.
The same applies to all other types of intervals: we need all (or lots, really) of them to generate all open sets and hence all Borel sets.
If a $\sigma$-algebra contains all sets of the form $[a,b)$ it will also contain all open intervals as $(a,b) = \bigcup_n [a+\frac{1}{n}, b)$ and hence all open sets etc.
