Separated Sets

What It Means for Sets to Be Separated

Connectedness is defined by the absence of a split into two nonempty, relatively open pieces. To recognize such a split in practice, it helps to have a condition that describes how the pieces sit near one another. Separated sets provide that condition: each set must avoid not only the other set, but also every point of the other set’s closure.

All closures below are taken in the ambient metric space $X$, unless a subspace is specified. This matters because a closure can change when the ambient space changes. The definition is stronger than disjointness, but it does not require a positive minimum distance between the sets.

Definition: Let $A$ and $B$ be subsets of a metric space $X$. They are separated if $$ A\cap\overline{B}=\varnothing \qquad\text{and}\qquad \overline{A}\cap B=\varnothing. $$ In words, neither set contains a point of the other set’s closure. Separated sets are necessarily disjoint, since $A\cap B$ is contained in both $A\cap\overline B$ and $\overline A\cap B$.

The definition is symmetric: exchanging $A$ and $B$ leaves its two conditions unchanged. It also handles empty sets without any special convention: if either set is empty, the pair is separated. In applications to a partition, both sets will be required to be nonempty.

Worked Example: Two Intervals Separated at a Missing Point

In $\mathbb{R}$, let $A=(0,1)$ and $B=(1,2)$. These sets are disjoint. Their closures are $\overline A=[0,1]$ and $\overline B=[1,2]$. Although the closures meet at $1$, that point belongs to neither $A$ nor $B$. Consequently,

A\cap\overline B=(0,1)\cap[1,2]=\varnothing, \qquad \overline A\cap B=[0,1]\cap(1,2)=\varnothing.

Thus $A$ and $B$ are separated. This example also shows why separated sets need not have disjoint closures: their closures may meet at a point belonging to neither set.

Relative Openness Characterizes Separation

The closure condition has a useful equivalent form. Consider the union $A\cup B$ as a metric space with its restricted metric. If $A$ and $B$ are separated, each point of $A$ has a neighborhood that avoids $B$, and each point of $B$ has one that avoids $A$. Thus each set is open within the union. Conversely, if both are relatively open in their union, neither can approach a point of the other set.

Theorem (Characterization of Separated Sets): Let $A,B\subseteq X$ be disjoint. The following are equivalent:

$A$ and $B$ are separated in $X$.
$A$ and $B$ are both open in the subspace $A\cup B$.
$A$ and $B$ are both closed in the subspace $A\cup B$.

Proof. Suppose first that $A$ and $B$ are separated. Fix $a\in A$. Since $a\notin\overline B$, the definition of closure gives an $r>0$ such that $B_r(a)\cap B=\varnothing$. Therefore $B_r(a)\cap(A\cup B)\subseteq A$, so $A$ is open in $A\cup B$. Applying the same argument to each $b\in B$, using $b\notin\overline A$, shows that $B$ is open there as well.

Since $A$ and $B$ are disjoint and have union $A\cup B$, each is the complement of the other in that subspace. A set is closed precisely when its complement is open, so both are also closed in $A\cup B$.

Now suppose $A$ and $B$ are both open in $A\cup B$. For any $a\in A$, relative openness supplies $r>0$ such that $B_r(a)\cap(A\cup B)\subseteq A$. In particular, $B_r(a)\cap B=\varnothing$, which means $a\notin\overline B$. This holds for every $a\in A$, so $A\cap\overline B=\varnothing$. The corresponding argument for each $b\in B$ gives $\overline A\cap B=\varnothing$. Hence $A$ and $B$ are separated. This proves the equivalence of the first two conditions; the complement argument proves the third as well. $\square$

Disjointness alone does not give relative openness. A point in one set may be arbitrarily close to points of the other. The closure conditions rule out precisely this possibility at points that belong to either set; they do not rule out a common limit point that belongs to neither.

Worked Example: Disjoint Sets That Are Not Separated

Let $A=(0,1)$ and $B=[1,2)$ in $\mathbb{R}$. They are disjoint: $A$ contains only numbers strictly less than $1$, while every point of $B$ is at least $1$. However, $1\in\overline A$ and $1\in B$. Thus

\overline A\cap B\supseteq\{1\}\ne\varnothing.

The sets are not separated. In the subspace $A\cup B=(0,2)$, the set $B=[1,2)$ is not open: every ball around $1$, after intersection with $(0,2)$, contains points less than $1$ that belong to $A$. The characterization theorem therefore predicts that the pair is not separated.

Testing Separation with Distances

In a metric space, closure membership can be tested by distance to a set. For a nonempty subset $E\subseteq X$, recall that $d(x,E)=0$ exactly when $x\in\overline E$. This gives a pointwise distance test for separated sets. It does not, however, turn separation into a requirement that the distance between the two entire sets be positive.

Proposition (Pointwise Distance Test for Separation): Let $A$ and $B$ be nonempty subsets of a metric space $X$. They are separated if and only if $$ d(a,B)>0\text{ for every }a\in A, \qquad d(b,A)>0\text{ for every }b\in B. $$

Proof. By the Distance Characterization of Closure, for each $a\in A$, the condition $a\notin\overline B$ is equivalent to $d(a,B)>0$. Likewise, for each $b\in B$, $b\notin\overline A$ is equivalent to $d(b,A)>0$. The definition of separated sets requires exactly that $A\cap\overline B$ and $\overline A\cap B$ are empty. These are equivalent to the two pointwise conditions in the proposition. $\square$

Worked Example: Separation Does Not Require Positive Set Distance

Return to $A=(0,1)$ and $B=(1,2)$. For every $a\in A$, the distance from $a$ to $B$ is $1-a$. Indeed, every $b\in B$ satisfies $b-a>1-a$, and values of $b$ can be taken arbitrarily close to $1$, so the infimum is $1-a>0$. Similarly, for every $b\in B$, $d(b,A)=b-1>0$. The pointwise distance test confirms that $A$ and $B$ are separated.

Nevertheless, the infimum of all distances between a point of $A$ and a point of $B$ is zero. For every positive integer $n\geq 2$, choose $a_n=1-1/n\in A$ and $b_n=1+1/n\in B$. Then

d(a_n,b_n)=|a_n-b_n| =\left|1-\frac1n-\left(1+\frac1n\right)\right| =\frac2n\longrightarrow0.

Thus each individual point has positive distance from the opposite set, even though there is no single positive lower bound that works for every pair of points.

Worked Example: Separated Sets with a Common Closure Point

In $\mathbb{R}$, let $A=\{1/n:n\text{ is a positive integer}\}$ and $B=\{-1/n:n\text{ is a positive integer}\}$. Both sets have $0$ as a limit point, so $\overline A=A\cup\{0\}$ and $\overline B=B\cup\{0\}$. The only common point of the closures is $0$, and $0$ belongs to neither $A$ nor $B$. Hence

A\cap\overline B=\varnothing, \qquad \overline A\cap B=\varnothing.

The sets are separated, even though $\overline A\cap\overline B=\{0\}$. For any fixed $a=1/n\in A$, its distance to $B$ is $1/n>0$: the points of $B$ approach $0$, and no point of $B$ is closer to $1/n$. For a fixed $b=-1/n\in B$, the distance to $A$ is also $1/n>0$. The distances can become small as $n$ grows, but each pointwise distance is positive.

Separated Sets and Connected Subsets

The definition of connectedness from the previous tutorial asks whether a space can be written as a union of two disjoint, nonempty, relatively open sets. The characterization above translates that condition into the language of separated sets. This translation is useful because separation can be checked using closures, even when a proposed partition is not presented as open sets at first.

Theorem (Connectedness and Separated Pairs): A metric space $E$ is disconnected if and only if there are nonempty separated subsets $A,B\subseteq E$ such that $E=A\cup B$. Equivalently, $E$ is connected if and only if it has no such representation.

Proof. Suppose $E$ is disconnected. By the definition of connectedness, there are disjoint, nonempty sets $A,B$ open in $E$, with $E=A\cup B$. The Characterization of Separated Sets, applied in the metric space $E$, shows that $A$ and $B$ are separated in $E$.

Conversely, suppose $E=A\cup B$, where $A$ and $B$ are nonempty separated subsets of $E$. The same characterization shows that both sets are open in $A\cup B=E$. They are disjoint by the definition of separation, and their union is $E$. Thus they form a separation of $E$, so $E$ is disconnected. $\square$

Here the closures used in the separation test are closures in $E$. If $E$ is itself a subset of a larger space $X$, it is important not to substitute closures in $X$ without justification. The closure-in-a-subspace theorem relates these two closures; in particular, a point outside $E$ that is a common ambient limit does not itself belong to the subspace $E$.

Theorem (A Connected Subset Lies on One Side): Let $A$ and $B$ be separated subsets of a metric space $X$, and let $C\subseteq A\cup B$ be connected. Then $C\subseteq A$ or $C\subseteq B$.

Proof. Since $C\subseteq A\cup B$ and $A,B$ are disjoint, the sets $C\cap A$ and $C\cap B$ are disjoint and have union $C$. Suppose both are nonempty. For any $c\in C\cap A$, separation gives $c\notin\overline B$, so some ball about $c$ misses $B$; its intersection with $C$ lies in $C\cap A$. Thus $C\cap A$ is open in $C$. The same reasoning shows that $C\cap B$ is open in $C$. They would form a separation of $C$, contradicting its connectedness. Therefore at least one of $C\cap A$ and $C\cap B$ is empty. Since their union is $C$, this means $C\subseteq B$ or $C\subseteq A$, as claimed. $\square$

This result applies even if $C$ is not all of $A\cup B$. It says that a connected set contained in two separated regions cannot meet both regions. A common use is to locate a connected subset once a larger space has been divided into separated pieces.

Worked Example: A Discrete Metric Space

Let $X=\{p,q,r\}$ with the discrete metric, and take $A=\{p\}$ and $B=\{q,r\}$. Every subset of a discrete metric space is closed: if $x\notin E$, the ball $B_{1/2}(x)=\{x\}$ misses $E$, so the complement of $E$ is open. Hence $\overline A=A$ and $\overline B=B$. The sets are disjoint, and therefore

A\cap\overline B=\{p\}\cap\{q,r\}=\varnothing, \qquad \overline A\cap B=\{p\}\cap\{q,r\}=\varnothing.

Thus $A$ and $B$ are separated, and their union $X$ is disconnected. More generally, any two disjoint subsets of a discrete metric space are separated because every subset is closed. This differs from the behavior of intervals in $\mathbb{R}$, where sets may be disjoint but fail to be separated because one contains a limit point of the other.

Why the Closure Conditions Matter

Separatedness is a precise way to say that two sets do not meet at a point belonging to either set. The distinction between this and disjointness is essential: $(0,1)$ and $[1,2)$ are disjoint but not separated, whereas $(0,1)$ and $(1,2)$ are separated. Conversely, separated sets need not have disjoint closures, as the sequences approaching $0$ from opposite sides demonstrate.

A second common pitfall is to replace pointwise positive distances by a uniform positive distance. The pointwise distance test requires $d(a,B)>0$ for each fixed $a\in A$, and $d(b,A)>0$ for each fixed $b\in B$. It does not require a constant $\delta>0$ such that $d(a,b)\geq\delta$ for every $a\in A$ and $b\in B$. The two intervals $(0,1)$ and $(1,2)$ are the basic example where that stronger condition fails.

When a connectedness argument involves a proposed partition, the practical sequence is therefore: verify that the pieces are disjoint and cover the space, then test whether either piece contains a point of the other’s closure. If both closure intersections are empty, the pieces are relatively open and give a separation. If one intersection is nonempty, the proposed partition is not a separation.

Check Your Understanding

Use the definition and results above to answer the following questions.

Can two separated sets have intersecting closures? Give an example and identify the common closure point.
Why does $a\notin\overline B$ imply that some open ball about $a$ misses $B$?
For nonempty sets $A,B$, how does the pointwise distance test express the two closure conditions?
Why do $(0,1)$ and $(1,2)$ have set-to-set distance zero but remain separated?
If $C$ is connected and contained in the union of two separated sets, why can $C$ not meet both sets?

What You'll Learn