Separated sets, disconnection, and connected subset of $\mathbb{R}$

Let $A,B,E\subseteq \mathbb{R}$, with closure as in Interior, closure, boundary and exterior of a subset of $\mathbb{R}$.

• $A$ and $B$ are separated when $\bar{A}\cap B=\mathrm{\varnothing }\text{and}A\cap \bar{B}=\mathrm{\varnothing }\mathrm{.}$
• A disconnection of $E$ is a pair $(A,B)$ of nonempty separated sets with $E=A\cup B$.
• $E$ is disconnected when it admits a disconnection, and connected when it does not.

Separated is strictly stronger than disjoint. Since $A\subseteq \bar{A}$ (Interior, closure, boundary and exterior of a subset of $\mathbb{R}$), the first displayed condition already gives $A\cap B\subseteq \bar{A}\cap B=\mathrm{\varnothing }$, so separated sets are disjoint. The converse fails: $(0,1)$ and $[1,2)$ are disjoint, yet every neighbourhood of $1$ meets $(0,1)$, so $1$ is an adherent point of $(0,1)$ and lies in $\overline{(0,1)}$ (The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points), while $1\in [1,2)$; hence $\overline{(0,1)}\cap [1,2)\mathrm{\ne }\mathrm{\varnothing }$ and the pair is not separated. What separation adds to disjointness is exactly this: neither set of a separated pair may contain a point adherent to the other, which is what makes a disconnection a genuine splitting rather than a bookkeeping partition.

Separation does not ask the two closures to be disjoint. Each condition tests one closure against the other set, never closure against closure. The pair $(0,1)$, $(1,2)$ illustrates the difference and is separated: $[0,1]$ is a closed set containing $(0,1)$, so $\overline{(0,1)}\subseteq [0,1]$ (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Interior, closure, boundary and exterior of a subset of $\mathbb{R}$) and $\overline{(0,1)}\cap (1,2)=\mathrm{\varnothing }$; symmetrically $\overline{(1,2)}\subseteq [1,2]$ and $(0,1)\cap \overline{(1,2)}=\mathrm{\varnothing }$. The two closures nevertheless share the point $1$, so a definition demanding $\bar{A}\cap \bar{B}=\mathrm{\varnothing }$ would be a different and strictly stronger condition, and it is not the one used here.

• Why separation and not "both pieces open". For a subset $E$ of $\mathbb{R}$ the pieces of a splitting are rarely open as subsets of $\mathbb{R}$: in the disconnection of $\mathbb{Q}\cap [0,2]$ used by $\mathbb{Q}\cap [0,2]$ is bounded and disconnected, so being an interval of $\mathbb{Q}$ is not enough ↗ neither piece is open in $\mathbb{R}$. Rudin's separated-sets formulation avoids introducing a second topology relative to $E$, and it is the only formulation this page uses. Nothing below refers to sets open "in $E$".

• Every one-point set and the empty set are connected. A disconnection requires two nonempty pieces with union $E$, and if $E$ has at most one point no two nonempty disjoint sets have union $E$.

• Connectedness of a subset of $\mathbb{R}$ turns out to be an order property: $E$ is connected exactly when it is order-convex (A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval). That is a theorem about $\mathbb{R}$ and uses its completeness; the definition above mentions no order at all.