Borel $\sigma$-algebra definition question So I am studying measure theory and I have found myself struggling to fully understand the concept of the Borel $\sigma$-algebra in depth. We know that the Borel $\sigma$-algebra is the smallest $\sigma$-algebra containing all open sets. The part that I cannot clearly grasp is the word smallest. The way we can generate a Borel $\sigma$-algebra is take all the open sets and take all possible set operations between them. Won't this always produce a unique $\sigma$-algebra of sets? How can a $\sigma$-algebra be larger? Can anyone provide an intuitive example? Thank you in advance!   
 A: Lets consider $\Omega=\{1,2,3,4\}$
$\sigma(\{1\})$ is the smallest $\sigma$-algebra which contains $1$.
So we must take any other elements of $P(\Omega)$ such that the conditions for being a $\sigma$-algebra are fulfilled.
It does clearly contains $1$. Also $1^C=\{2,3,4\}$. And $\Omega,\emptyset$.
So we have $\sigma(\{1\})=\{\Omega,\emptyset,\{1\},\{2,3,4\} \}$. Indeed this is a $\sigma$-algebra. Also we got this set for taking as much elements of the power until our conditions are fulfilled for the "first time".
An example for a $\sigma$-algebra with is not the smallest one but is containing $\{1\}$ is:
B:=$\{\Omega,\emptyset,\{1\},\{2\},\{1,2\},\{3,4\},\{2,3,4\},\{1,3,4\} \}$.
Clearly $\sigma(\{1\})\subset B$
A: Say we are doing this on the real line $\mathbb R$.  Let $\mathcal G$ be the collection of all open sets.  We are interested in $\sigma$-algebras $\mathcal F$ such that $\mathcal F \supseteq \mathcal G$.  There may be many such $\sigma$-algebras.  For example, the power set $\mathcal P$, consisting of all subsets of $\mathbb R$ is one.  But that is the largest one, we want the smallest one.  
As a comment noted, this largest and smallest are in the sense of set inclusion.  The Borel $\sigma$-algebra $\mathcal B$ satisfies $\mathcal B \subseteq \mathcal F$ for any $\sigma$-algebra $\mathcal F$ such that $\mathcal F \supseteq \mathcal G$.  That is the sense in which $\mathcal B$ is smallest.
A: if you agree that the borel $\sigma$-algebra isn't neccessarily all the subsets of the topological space, then you might also agree that $P(X)$ (or $2^X$ in a different notation, the power set) is a larger $\sigma$-algebra. Larger means more sets in the $\sigma$-algebra.
Also note that the generated $\sigma$-algebra by the open sets, which is the borel $\sigma$-algebra, is the intersection of all $\sigma$-algebras containing the open sets - which is smaller than any other $\sigma$-algebra by the sense of being a subset.
