Is every set that's a union of two sets disconnected? I'm curious about the definition of disconnected and I'm sure this can't be right. But let's say that $S= A \cup B$, then isn't $S$ automatically disconnected? Since I can re-write It as $(S \cap C) \cup (S \cap D)$ with $C=A$ and $D=B$? This is a circular definition since it basically uses an identity about unions and intersection, but is my understanding of disconnected totally off? Or is this just a trivial case we toss aside? I'm a tad confused here.
 A: Let $A=[0,1]$ and $B=[1,2]$; both sets are connected, and $A\cup B=[0,2]$, which is also connected. Let $A$ be the set of irrational numbers and $B$ the set of rational numbers; then both $A$ and $B$ are totally disconnected, but $A\cup B=\Bbb R$ is connected.
In order for $S$ not to be connected, you must be able to write it as a disjoint union of non-empty sets that are both relatively open in $S$: $S=U\cup V$, say, where $U\ne\varnothing\ne V$, $U\cap V=\varnothing$, and $U\cap\operatorname{cl}V=V\cap\operatorname{cl}U=\varnothing$. Notice the requirements: $U$ and $V$ are non-empty, they’re disjoint, their union is $S$, and each is a relatively open subset of $S$. All of these conditions must be met in order for $U$ and $V$ to be a separation of $S$.
A: Disconnectedness requires also that $A,B$ are disjoint.
A: Assuming the topic is topology, I'm interpreting the question is follows: If $X$ is a topological space and $A,B\subseteq X$, is $A\cup B$ always disconnected?
The answer is no, for two reasons. The first is trivial, simply if $A$ and $B$ are not disjoint, then their union may be connected (e.g., take $X=\mathbb R$, A=B=[0,1]). More interestingly, there are famous examples of two disjoint sets whose union is connected, for instance take $X=\mathbb R$, $A=(0,1]$, and $B=\{0\}$. A more interesting example is for instance the topologist's since curve.
A: $S$ is disconnected if it is the union of two disjoint, open, non-empty subsets $A$ and $B$.
So if $S = A \cup B$ for some $A$ and $B$, it could still be connected because $A$ and $B$ are not disjoint or one of $A$ or $B$ is not open.
Example 1: If $S = (0, 1)$, we can write $S = A \cup B$ where $A = (0, 3/4)$ and $B = (1/4, 1)$. $A$ and $B$ are both open but not disjoint, so this doesn't contradict the fact that $(0,1)$ is connected.
Example 2: Again take $S = (0,1)$, but this time take $A = (0,1/2)$ and $B = [1/2, 1)$. Then $S = A \cup B$ and $A$ and $B$ are disjoint, but the problem here is that $B$ is not open in $(0,1)$.
