Proving this set is an algebra. Let $J = \{$all intervals contained in $[0,1]\}$  and $B_0 = \{$all finite unions of elements of $J\}$.
Prove that $B_o$ contains $[0,1], \emptyset$, and is closed under formation of complements and of finite unions and intersections, i.e., prove that $B_0$ is an algebra.
How do I prove that $B_0$ is closed under complements?
 A: Thank you for letting me work through this. Below are my attempts at proofs, corrections are ,as always, warmly welcome.
Finite intersection of intervals are finite unions of intervals Proof:
An interval is a set of real numbers with the property that any number that lies between two numbers in the set is also included in the set. 
Take any two intervals $A = [i_1, i_2]$ and $B=[i_3,i_4]$ where $i_1,i_2,i_3,i_4 \in R$.
$$A \cap B = \{x \in R : i_1\le x \le i_2 \text{ and } i_3 \le x \le i_4 \}$$


*

*In the case $i_2 < i_3 \ A \cap B = \emptyset = \emptyset \cup \emptyset $ 

*In the case $i_3 < i_2 < i_4  \ A \cap B = \{x \in R : i_3\le x \le i_2 \}$ 

*...


We can see that in all the cases we obtain a single interval and because $A \cap B \cap C  = (A \cap B) \cap C$, where $C$ is an other interval $\in R$ we can extend this to any finite intersection.
Complement of an interval is a union of two intervals Proof:
Take any interval $A = [i_1, i_2]$ and $i_1,i_2 \in R$.
$$R \backslash A = \{ x \in R : x \notin A\}=\{x \in R : x < i_1 \text{ or } x > i_2 \} = (- \infty, i_1) \cup (i_2, +\infty)   $$
Then by Demorgans law the result I was trying to prove follows.
