How can we work with the Borel $\sigma$-algebra given that we can't exactly list what the smallest $\sigma$-algebra set is? For the Borel $\sigma$-algebra, it is defined to be the $\sigma$-algebra generated by all open intervals $(a,b)$ with $a,b \in \mathbb{R}$. By generated, it means the smallest such $\sigma$-algebra. However, and this might be a bad question, how do we know that it is the smallest? In my book, they start with an open interval $(a,b)$ and then take unions, complements, so forth to get larger and larger sets of the Borel $\sigma$-algebra. This process continues without bound so how do we know that the Borel $\sigma$-algebra is indeed the smallest if we cannot physically list out what all possibilities are? 
 A: You define the Borel $\sigma$ algebra as the smallest $\sigma$-algebra containing all open intervals. From theorems we have proven, we know that this definition is sound (because there actually exists a unique smallest element among the set of all $\sigma$-algebras with that property).
So we start with the fact that the Borel $\sigma$-algebra exists, and by definition, we know it is the smallest one that contains open intervals. 
Now, we can start thinking about what sets it contains. You can, for a given set, determine whether that set is in the Borel $\sigma$-algebra from only that fact.
For example, is $(0,1)\cup (5,6)$ in the Borel $\sigma$-algebra $\mathcal B$?
Well:


*

*$(0,1)$ is open, so it is in $\mathcal B$.

*$(5,6)$ is open, so it is in $\mathcal B$.

*We know that $\mathcal B$ is a $\sigma$-algebra. So we know that if $A\in \mathcal B$ and $B\in \mathcal B$, then $A\cup B\in\mathcal B$.


From those $3$ points, we conclude that $(0,1)\cup (5,6)\in \mathcal B$
A: Suppose $\{\mathcal{A}_{\alpha}\}_{\alpha \in I}$ are $\sigma$-algebras on a set $X$, for some indexing set $I$. Then $\bigcap_{\alpha \in I} \mathcal{A}_{\alpha}$ is itself a $\sigma$-algebra. Showing this is a matter of checking properties against the definition of a $\sigma$-algebra.
This means that if $\mathcal{S}$ is a collection of subsets of some set $X$, we can define the $\sigma$-algebra generated by $\mathcal{S}$ to be the intersection of all $\sigma$-algebras containing $\mathcal{S}$. We denote this as $\sigma(\mathcal{S})$. The intersection here is nontrivial because every collection $\mathcal{S}$ is contained in at least one $\sigma$-algebra on $X$, namely the power set $2^{X}$.
In the sense of inclusion being a partial order on $\sigma$-algebras, this $\sigma(\mathcal{S})$ will be the "smallest" $\sigma$-algebra containing $\mathcal{S}$. If we specialize to the case $X$ is a topological space and $\mathcal{S}$ is the collection of open sets, then $\sigma(\mathcal{S})$ gives the Borel $\sigma$-algebra, which is sometimes written $\mathcal{B}(X)$.
