Separated Sets

When Are Two Sets Separated?

In the previous tutorial, a separation of a set was a partition into two nonempty, disjoint, relatively open pieces. The word “separated” also describes a relationship between two sets: neither set contains a point of the other set’s closure. This definition lets us test whether two pieces are separated even before deciding whether they partition a particular set.

Throughout, closure means closure in $\mathbb{R}$. Recall that a point belongs to the closure of a set if every open interval around that point meets the set. Thus, to check whether $A$ and $B$ are separated, it is not enough to check that they are disjoint. We must also check that points of either set are not limit points of the other.

Definition: Two subsets $A,B\subseteq\mathbb{R}$ are separated if $$ A\cap\overline{B}=\varnothing \quad\text{and}\quad \overline{A}\cap B=\varnothing. $$ Here, $\overline{A}$ and $\overline{B}$ denote their closures in $\mathbb{R}$. In particular, separated sets are disjoint.

The definition is symmetric: exchanging $A$ and $B$ leaves the two conditions unchanged. It also distinguishes separated sets from sets whose closures are disjoint. Separated sets may have a common closure point, provided that point belongs to neither set. This distinction matters when two sets approach the same boundary point from opposite sides.

Separated Pieces and Relative Openness

The definition becomes especially useful when the two sets together form a set $E$. If $E=A\cup B$ and $A$ and $B$ are disjoint, each is the complement of the other within $E$. The next result connects separation, closures, and relative openness.

Theorem (Separated Pieces and Relative Closedness): Suppose $A,B\subseteq\mathbb{R}$ are disjoint and $E=A\cup B$. Then $A$ and $B$ are separated if and only if both are relatively closed in $E$. In this case, both are also relatively open in $E$.

Proof. The closure of $A$ relative to $E$ is $E\cap\overline{A}$. Since $E=A\cup B$, $$ E\cap\overline{A} =(A\cap\overline{A})\cup(B\cap\overline{A}) =A\cup(B\cap\overline{A}). $$ Therefore $A$ is relatively closed in $E$, meaning $E\cap\overline{A}=A$, exactly when $B\cap\overline{A}=\varnothing$. By the same reasoning, $B$ is relatively closed in $E$ exactly when $A\cap\overline{B}=\varnothing$. These are precisely the two conditions in the definition of separated sets. This proves the equivalence.

If both pieces are relatively closed in $E$, then $A=E\setminus B$ is relatively open in $E$, because $B$ is relatively closed. Likewise, $B=E\setminus A$ is relatively open. $\square$

When $A$ and $B$ are nonempty as well, disjoint, separated, and cover $E$, they form a separation of $E$ in the sense of the previous tutorial. The nonempty condition is important: separated sets can include an empty set, but a separation of $E$ requires two nonempty pieces.

Worked Example: Two Finite Sets

Let $A=\{-3,2\}$ and $B=\{0,5\}$. Each finite subset of $\mathbb{R}$ is closed, so $\overline{A}=A$ and $\overline{B}=B$. Their intersection is empty: $$ A\cap\overline{B}=\{-3,2\}\cap\{0,5\}=\varnothing, \qquad \overline{A}\cap B=\{-3,2\}\cap\{0,5\}=\varnothing. $$ Thus $A$ and $B$ are separated. Their union $E=\{-3,0,2,5\}$ is partitioned into the two nonempty relatively open pieces $A$ and $B$, so this partition is a separation of $E$.

A Shared Closure Point Does Not Prevent Separation

It is tempting to strengthen the definition and require $\overline{A}\cap\overline{B}=\varnothing$. That condition is sufficient for separation, but it is not required. If $A$ and $B$ approach the same point without containing it, their closures may meet even though neither set meets the other’s closure.

Worked Example: Two Intervals Approaching the Same Point

Take $A=(0,1)$ and $B=(1,3)$. Their closures are $\overline{A}=[0,1]$ and $\overline{B}=[1,3]$, so the closures meet at $1$. But $1$ belongs to neither $A$ nor $B$. Consequently, $$ A\cap\overline{B}=(0,1)\cap[1,3]=\varnothing, \qquad \overline{A}\cap B=[0,1]\cap(1,3)=\varnothing. $$ Thus $A$ and $B$ are separated, despite having a common closure point. In their union, each interval is relatively open and relatively closed. The missing point $1$ is what keeps either piece from containing a point in the closure of the other.

The endpoint cases show why both closure conditions must be checked. If one of these intervals includes $1$, the pair is no longer separated: that endpoint belongs to one set and to the closure of the other. Disjointness alone does not rule out this failure.

Worked Example: Disjoint Sets That Are Not Separated

Let $A=(0,1)$ and $B=\{1\}$. These sets are disjoint, but $\overline{A}=[0,1]$, and hence $$ \overline{A}\cap B=[0,1]\cap\{1\}=\{1\}\ne\varnothing. $$ Therefore $A$ and $B$ are not separated. The point $1$ belongs to $B$ and is also approached by points of $A$. In the union $E=(0,1]$, the piece $B$ is not relatively open: every open interval about $1$, when intersected with $E$, contains points of $A$. This agrees with the theorem on separated pieces.

Separation Is Weaker Than Positive Distance

A positive gap between two sets guarantees separation, but separated sets need not have such a gap. To make this precise, for nonempty sets $A,B\subseteq\mathbb{R}$, define their distance by $$ \operatorname{dist}(A,B)=\inf\{|a-b|:a\in A,\ b\in B\}. $$ The infimum can be zero even when the sets do not intersect. In particular, it can be zero for separated sets.

Theorem (Positive Distance Implies Separation): If $A,B\subseteq\mathbb{R}$ are nonempty and $\operatorname{dist}(A,B)>0$, then $A$ and $B$ are separated.

Proof. Write $d=\operatorname{dist}(A,B)>0$. Fix $b\in B$. For every $a\in A$, the definition of infimum gives $|a-b|\geq d$. Thus the open interval $(b-d/2,b+d/2)$ contains no point of $A$. This means $b\notin\overline{A}$. Since $b$ was arbitrary, $B\cap\overline{A}=\varnothing$. The same argument, fixing any $a\in A$, shows $a\notin\overline{B}$, so $A\cap\overline{B}=\varnothing$. Both conditions in the definition hold, and $A$ and $B$ are separated. $\square$

Worked Example: Sets with a Positive Gap

Let $A=(-4,-2)$ and $B=(1,3)$. For $a\in A$ and $b\in B$, we have $a<-2$ and $b>1$, so $$ b-a>1-(-2)=3. $$ Values of $b-a$ can be made arbitrarily close to $3$ by choosing $a$ close to $-2$ and $b$ close to $1$. Therefore $\operatorname{dist}(A,B)=3>0$, and the theorem shows that $A$ and $B$ are separated. Directly, $\overline{A}=[-4,-2]$ and $\overline{B}=[1,3]$; neither set intersects the other’s closure.

The converse to the positive-distance theorem is false. For the intervals $A=(0,1)$ and $B=(1,3)$, the sets are separated, but points of the two intervals can be chosen arbitrarily close to $1$. For example, $a_n=1-1/n\in A$ and $b_n=1+1/n\in B$ for every integer $n\geq2$, and $$ |a_n-b_n|=\left|1-\frac{1}{n}-1-\frac{1}{n}\right|=\frac{2}{n}\longrightarrow 0. $$ Thus their distance is zero. Separation rules out points of either set lying in the other’s closure; it does not require a uniform positive gap between them.

Connected Sets Cannot Cross Between Separated Pieces

Separated pieces provide a useful test for connectedness. If a connected set is contained in the union of two separated sets, it cannot meet both. Otherwise, its intersections with the two pieces would divide it into disjoint, nonempty, relatively open subsets.

Theorem (Connected Sets Lie on One Side of a Separation): Suppose $A$ and $B$ are separated, and $C\subseteq A\cup B$ is connected. Then either $C\subseteq A$ or $C\subseteq B$.

Proof. Set $E=A\cup B$. If $C$ met both $A$ and $B$, then $C\cap A$ and $C\cap B$ would be nonempty and disjoint, and their union would be $C$. By the Separated Pieces and Relative Closedness Theorem, $A$ and $B$ are relatively open in $E$. Hence $C\cap A$ and $C\cap B$ are relatively open in $C$: each is the intersection of $C$ with a relatively open subset of $E$. They would form a separation of $C$, contradicting its connectedness. Therefore $C$ cannot meet both pieces, which proves the claim. $\square$

This result includes the case $C=\varnothing$, since the empty set is a subset of both $A$ and $B$. If $C$ is nonempty, the conclusion says all of its points lie in just one of the two separated pieces. The theorem does not say that every subset of $A\cup B$ must lie on one side; connectedness is essential.

Worked Example: A Connected Interval in a Union of Separated Sets

Let $A=(-2,0)$, $B=(0,4)$, and $C=[-1,0)$. The intervals $A$ and $B$ are separated: their closures are $[-2,0]$ and $[0,4]$, and $$ A\cap\overline{B}=(-2,0)\cap[0,4]=\varnothing, \qquad \overline{A}\cap B=[-2,0]\cap(0,4)=\varnothing. $$ The set $C$ is an interval, so it is connected by the Intervals Are Connected Theorem. Also $C\subseteq A\cup B$, and in fact $C\subseteq A$. The connected-set theorem explains why a connected subset of this union cannot include points from both sides of $0$. The point $0$, which would separate the two intervals, belongs to neither.

What to Check When Using the Definition

For two proposed sets $A$ and $B$, check both intersections with closures:

Check disjointness.
If either set contains a point of the other, the sets are not separated.

Check each closure condition.
Verify $A\cap\overline{B}=\varnothing$ and $\overline{A}\cap B=\varnothing$. A point in either intersection rules out separation.

Distinguish separation from a gap.
A positive distance is sufficient for separation, but a shared closure point or distance zero does not by itself prevent separation.

When $A$ and $B$ partition $E$, the relative-closedness theorem is often the more convenient test: show that both pieces are relatively closed in $E$, or equivalently that both are relatively open there. When they do not partition a set, return to the definition and check the two closure intersections directly. Keeping these situations distinct avoids confusing a relationship between two sets with a separation of an entire set.

Check Your Understanding

Use the definition and results in this tutorial to answer the following questions.

What two conditions must $A$ and $B$ satisfy to be separated?
Can separated sets have intersecting closures? If so, what must be true of their common closure points?
Why does positive distance imply separation, and why does separation not imply positive distance?
If $A$ and $B$ are disjoint and cover $E$, how is their separation related to relative openness in $E$?
Why can a connected subset of the union of two separated sets not meet both sets?

What You'll Learn