How do I prove that the complement of the closed interval $[a,b]$ is an open set. How do I prove that the complement of the closed interval $[a,b]$ is an open set.  
I have a theorem that says an open set is a union of open intervals.
Can I simply say the complement of the closed interval $[a,b]$ is $(-\infty, a) \cup (b, \infty)$?
 A: I denote $A=[a,b]^c$ the complement of $[a,b]$.
$$x\in A\implies x\notin[a,b]\implies \exists \delta>0: ]x-\delta,x+\delta[\cap [a,b]=\emptyset$$
and thus $]x-\delta,x+\delta[\subset A.$
Therfore, $A$ is open.
A: Yes, that works. You would want to say more explicitly that $[a,b]$ is closed (in $\mathbb R$) because its complement (in $\mathbb R$) is open.
You can also prove it strictly by definition.We want to show that if a sequence $\{x_n\} \in [a,b] \subset \mathbb R$ converges to some limit $x \in \mathbb R$, then $x \in [a,b]$. 
Suppose to the contrary that a closed interval $[a,b]$ is not closed. Then there exists some sequence $\{x_n\} \in [a,b]$ which converges to $x \in \mathbb R$ but $x \not \in [a,b]$. Then either $a \in (-\infty, a)$ or $a \in (b, \infty)$. 
Then the distance between $x$ and either of the endpoints $a$ or $b$ must be nonzero. Let $s = min \{|x-a|, |x-b|\} > 0$ by construction. Then $B_s(x)$, the open ball around $x$ with radius $s$, contains zero points in $[a,b]$. 
This contradicts the definition of a limit point. $\textit Any$ open ball around a limit point of a set must contain infinitely points of that set. (Recall the definition of a limit of a sequence in $\mathbb R$. If $\{x_n\}$ converges to $x$, then for any $\epsilon > 0$, there exists an $N$ such that $\forall n \ge N$, $|x_n - x| < \epsilon$. But here, if you set $\epsilon$ equal to $\frac s 2$, no such $N$ exists. In other words, you cannot pick a number in the interval $[a,b]$ close enough to $x$ to "bridge the distance.")
