Forgive my poor LaTeX, I'm very new to it (as in, reading guides as I go just to write this).
In my Elementary Real Analysis course, we're asked to prove a finite union of closed sets is itself closed. I've already done it using de Morgan's law with the complement of a finite intersection of open sets; however, I got to thinking about how I would go about proving it in the specific case of pairwise disjoint closed intervals without de Morgans. I know that if the unions themselves form a new interval, then we can just create a 'ball':
$$
\ A \subset \mathbb R^n\,.
\forall r>0\,, \alpha=\inf{A}\,, \gamma=\sup{A}\,,\,\, B(\alpha,r) \not\subset{A}\ and\, B(\gamma,r) \not\subset{A}
$$
Where $\alpha$ is the minimum of the infima of all the intervals, and $\gamma$ is the max of the suprema of all the intervals.
And so I was wondering how the argument would look if the intervals were all disjoint? I tagged it as recreational since, while it is related to my studies, is more for me than anything else.
Also, forgive me if this has been answered; I looked around at many different questions that seemed similar enough, but didn't find one that had asked this precisely.
Suppose the intervals are $[x_1,y_1],[x_2,y_2],\dots,[x_n,y_n]$, ordered so that $x_1\le y_1<x_2\le y_2<\dots<x_n\le y_n$. Then the complement of $\bigcup_{k=1}^n[x_k,y_k]$ is
$$(-\infty,x_1)\cup(y_1,x_2)\cup(y_2,x_3)\cup\dots\cup(y_{n-1},x_n)\cup(y_n,\infty),$$
and each open interval is open: the bounded ones can be written as $(a,b)=B(\frac{a+b}2,\frac{a-b}2)$ and the unbounded ones can be written in terms of bounded intervals as $(-\infty,x_1)=\bigcup_{z<x_1}(z,x_1)$ and $(y_n,\infty)=\bigcup_{z>y_n}(y_n,z)$. Thus a finite disjoint union of closed intervals has a complement which is a union of open balls in $\Bbb R$, and so is closed by definition.