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?
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 $i_1,i_2,i_3,i_4 \in R$.
$$A \cap B = \{x \in R : i_1\le x \le i_2 \text{ and } i_3 \le x \le i_4 \}$$
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.