smallest sigma algebra Let $A$ be any collection of subsets of the real numbers. How would I show that there is always a smallest sigma algebra $B$ containing $A$.
I was thinking if I could show that the intersection of sigma algebras was once again a sigma algebra then maybe I could show that the different intervals in the real numbers could be made up of various intervals' unions and intersections.
 A: There is a general technique for this sort of problem which I call intersection enclosure. 
1) show there is a sigma algebra containing the subsets: in this case the power-set of R will do.
2) observe that if you take an arbitrary intersection of sigma algebras you get a sigma algebra
3) take the intersection of all sigma algebras that contain your collection of subsets. By 1) the intersection is non-empty. By 2, it is a sigma algebra. and it clearly contains the required subsets. 
Done.
(see my book "proof patterns" for further discussion.)
A: Method 1.(Advanced).Let $A$ be a non-empty family of subsets of $R.$ Let $A'=A\cup \{R$ \ $a: a\in A\}. $ Let $A''$  be the set of all $\cap B$ over all finite $B\subset A'. $  Let $ A''' $  be the set of all $\cup C$ over all finite $C\subset A''.$ Exercise: $A'''$ is a Boolean algebra of subsets of $R.$
Let $\omega_1$ be the set of countable ordinals. Let $B(0)=A'''.$ For $x\in \omega_1$ let $B(x+1)=(B(x)\cup C(x))'''$ where $C(x)$ is the set of all $\cap C$ over all countable $C\subset B(x).$ For non-zero limit ordinal $x\in \omega_1$ let $B(x)=(\cup_{y<x} B(y))'''.$ And let $A^*=\cup_{x\in \omega_1}B_x.$
Any $\sigma$-algebra $S$ on $R$ such that $A\subset S$ must satisfy $S\supset A^*.$ The regularity of $\omega_1$ (A countable union of members of $\omega_1$ is a member of $\omega_1$) implies that $A^*$ is a $\sigma$-algebra.  
For completeness' sake, if $A$ is empty let $A^*=\{R,\phi\}.$
Method 2.(Elementary). Let $A^s$ be the set of all $\sigma$-algebras on $R$ that have $A$ as a subset. Then $A^s$ is not empty because $P(R),$ the set of all subsets of $R,$ belongs to $A^s.$ Let $A^*=\cap A^s=\{x:\forall y\in A^s\;(x\in y)\}.$ 
Lemma.If $F$ is a non-empty family of $\sigma$-algebras on a set $G$ then $\cap F$ is a $\sigma$-algebra on $G.$ Proof:$$ (i).\forall f\in F \;(G\in x)\implies G\in \cap F.$$ $$ (ii). x\in \cap F \implies \forall f\in F \;(\;G\backslash x \in f\;)\implies G\backslash x \in \cap F. $$   (iii). When $T=\{x_n:n\in N\}\subset \cap F$ then $\forall f\in F (\cap T\in f \land \cup T\in f)$ which implies $(\cap T\in \cap F \land \cup T\in \cap F).$
So $A^*=\cap A^s$ is a $\sigma$-algebra on $R$ with $A$ as a subset, and if $B$ is a $\sigma$-algebra on $R$ with $B\supset A,$ then $B\in A^s$ so $B\supset \cap A^s=A^*.$
