Give an example of subsets $A$ and $B$ of $\mathbb{R}$ such that all three of the following conditions hold.

Question

(a) Neither A nor B is open;
(b) $A \cap B = \emptyset$;
(c) $A \cup B$ is open.

I've considered $A = (0,1] $ and $B=(1,2]$, and $A = (0,1] $ and $B=[1,2)$, but neither possible subset pairs satisfy all three conditions. Is this even possible?

Maybe it is possible if we consider infinite union subsets for $A$ and $B$. As in,

$A=\bigcup\limits_{n=0}^{\infty}(2n, 2n+1]$ and $B=\bigcup\limits_{n=0}^{\infty}(2n+1, 2n+2]$

so that $A \cup B = (0, \infty)$ which is open.

Answer 1

A very basic strategy could take two disjoint open intervals, and have them each "trade ownership" of a point from their interiors.

So, for example,

$$A = (0,1){\setminus}\{1/2\} \cup \{5/2\}$$

$$B = (2,3){\setminus}\{5/2\} \cup \{1/2\}$$

Answer 2

Take any $A \subset \mathbb R$ which is neither open nor closed and define $B = \mathbb R \setminus A$.

Let us observe that we cannot find any two intervals $A, B$ with the desired properties.

Assume $A, B$ are intervals with these properties. Then each of $A, B$ must be either a (bounded) closed interval or a half-open interval.

Let $A$ be an interval boundary point of $A$. W.l.o.g. we may assume that it is the right interval boundary point (the other case can be treated similarly). In order that $a$ becomes an interior point of $A \cup B$, we must have $B = (a,b]$ with $b > a$. Note that $B$ cannot contain $a$ and cannot be open. But then $b$ is the right boundary point of the interval $A \cup B$, thus $A \cup B$ is not open.