Recognizing a Separation by Its Closures
In a metric space, a separation is a way to divide a set into two nonempty pieces that are open when viewed from within the set. The definition describes those pieces using relative openness. An equivalent test uses relative closures: each piece must avoid the relative closure of the other. This test is useful when it is easier to locate limit points than to find suitable open balls directly.
Let \((X,d)\) be a metric space, let \(E\subseteq X\), and suppose \(E=A\cup B\), where \(A\) and \(B\) are disjoint. Write \(\overline{A}^{\,E}\) for the closure of \(A\) in the subspace \(E\). The relative closure is \(\overline{A}^{\,E}=E\cap\overline{A}^{\,X}\), where \(\overline{A}^{\,X}\) is the closure in \(X\). The distinction matters: a point outside \(E\) can be in the ambient closure without being in the relative closure.
Proof. Suppose first that \(A\) and \(B\) form a separation, so both are relatively open in \(E\). Since \(E\setminus A=B\) is relatively open, \(A\) is relatively closed in \(E\). Thus \(\overline{A}^{\,E}=A\), and \(\overline{A}^{\,E}\cap B=A\cap B=\varnothing\). Likewise, \(B\) is relatively closed, so \(\overline{B}^{\,E}=B\) and \(A\cap\overline{B}^{\,E}=\varnothing\).
Conversely, suppose the two closure intersections are empty. Since \(\overline{A}^{\,E}\subseteq E=A\cup B\), and it contains \(A\), the condition \(\overline{A}^{\,E}\cap B=\varnothing\) implies \(\overline{A}^{\,E}=A\). Thus \(A\) is relatively closed in \(E\), so its complement \(B\) is relatively open. The other condition gives \(\overline{B}^{\,E}=B\), so \(B\) is relatively closed and its complement \(A\) is relatively open. The pieces are disjoint, nonempty, relatively open, and cover \(E\). They form a separation. \(\square\)
Nonemptiness is a necessary hypothesis in this statement. The closure conditions alone do not ensure a separation: if \(E=A=\{0\}\) and \(B=\varnothing\), both intersections are empty, but one piece is empty. In a separation argument, check the nonempty pieces as well as the closure conditions.
Using Ambient Closures
The criterion can also be applied by working in the whole space \(X\). For a partition \(E=A\cup B\), the identities \(\overline{A}^{\,E}=E\cap\overline{A}^{\,X}\) and \(\overline{B}^{\,E}=E\cap\overline{B}^{\,X}\) show that the relative-closure conditions are equivalent to \(\overline{A}^{\,X}\cap B=\varnothing\) and \(A\cap\overline{B}^{\,X}=\varnothing\). The ambient closures themselves need not be disjoint: they may meet at points outside \(E\). What matters for the partition is that neither piece contains a point of the other piece’s closure.
Worked Example: A Separation on Two Sides of a Missing Point
Let \(E=\mathbb{R}\setminus\{0\}\), \(A=(-\infty,0)\), and \(B=(0,\infty)\), with the usual metric. These sets are disjoint, nonempty, and cover \(E\). The ambient closure of \(A\) is \((-\infty,0]\), which does not meet \(B\); the ambient closure of \(B\) is \([0,\infty)\), which does not meet \(A\). Therefore the relative-closure criterion shows that \(A\) and \(B\) form a separation of \(E\).
The ambient closures do intersect: both contain \(0\). But \(0\notin E\), so it is in neither relative closure in \(E\). In fact, \(\overline{A}^{\,E}=A\) and \(\overline{B}^{\,E}=B\). This example shows why one must distinguish closure in the whole space from closure in the set being separated.
Worked Example: Why the Nonempty-Pieces Condition Cannot Be Dropped
Set \(E=\{3,5\}\), \(A=E\), and \(B=\varnothing\), with the usual metric on \(\mathbb{R}\). Then \(\overline{A}^{\,E}=E\) and \(\overline{B}^{\,E}=\varnothing\), so \(\overline{A}^{\,E}\cap B=\varnothing\) and \(A\cap\overline{B}^{\,E}=\varnothing\). Nevertheless, \(A\) and \(B\) do not form a separation because \(B\) is empty. The example confirms that the closure test is an equivalence only when the partition pieces are required to be nonempty.
This is a general proof-checking habit: when a theorem concerns a definition with several requirements, an equivalent test must retain every requirement not supplied by the test itself. Here the closure conditions capture relative openness, but they do not capture nonemptiness.
Positive Distance as a Separation Test
A distance estimate can often verify the closure conditions without computing either closure. For nonempty subsets \(A,B\subseteq X\), define their distance by \(\operatorname{dist}(A,B)=\inf\{d(a,b):a\in A,\ b\in B\}\). If this infimum is positive, every point in one set has a ball that misses the other set. This is a useful sufficient condition; it is stronger than the closure criterion.
Proof. Put \(\delta=\operatorname{dist}(A,B)>0\). Fix \(b\in B\). For every \(a\in A\), the definition of infimum gives \(d(a,b)\geq\delta\). Hence the open ball \(B(b,\delta)\) does not meet \(A\): if \(a\in B(b,\delta)\), then \(d(a,b)<\delta\), a contradiction. Thus \(b\notin\overline{A}^{\,X}\). Since \(b\in B\) was arbitrary, \(\overline{A}^{\,X}\cap B=\varnothing\). The same argument with the roles reversed gives \(A\cap\overline{B}^{\,X}=\varnothing\).
For \(E=A\cup B\), these ambient conditions imply the relative-closure conditions, so the Relative-Closure Criterion gives a separation of \(E\). Both pieces are nonempty by hypothesis. Therefore \(E\) is disconnected. \(\square\)
Worked Example: Positive Distance Separates Two Planar Sets
In the Euclidean plane, let \(A=\{(x,0):-2\leq x\leq-1\}\) and \(B=\{(x,0):1\leq x\leq2\}\). Both sets are nonempty. For \(a=(x,0)\in A\) and \(b=(y,0)\in B\), we have \(x\leq-1\) and \(y\geq1\), so \(d(a,b)=|y-x|=y-x\geq2\). Equality holds for \(a=(-1,0)\) and \(b=(1,0)\), and therefore \(\operatorname{dist}(A,B)=2\).
The theorem implies that \(A\) and \(B\) are separated, and their union is disconnected. Directly, the open ball of radius \(2\) centered at any point of either piece misses the other piece. The distance calculation supplies a uniform buffer between the two parts.
Positive Distance Is Not Necessary
Separated sets need not be a positive distance apart. A separation rules out points of one piece lying in the closure of the other, but it does not require a single positive lower bound for all distances between the pieces. This distinction is especially important in unbounded spaces, where points can get arbitrarily close while escaping every bounded region.
Worked Example: Separated Sets at Distance Zero
In \(\mathbb{R}\), let \(A=\{n:n\geq2\}\) and \(B=\{n+1/n:n\geq2\}\), where \(n\) ranges over the integers. The sets are disjoint: every element of \(A\) is an integer, whereas \(n+1/n\) is not an integer for \(n\geq2\). Both sets are closed in \(\mathbb{R}\). Indeed, each has only finitely many points in any bounded interval, so it has no finite limit point outside the set. Consequently their closures are themselves and neither meets the other. They are separated.
For each integer \(n\geq2\), the points \(n\in A\) and \(n+1/n\in B\) have distance \(1/n\). Hence \(\operatorname{dist}(A,B)\leq1/n\) for every \(n\geq2\). Distances are nonnegative, so \(\operatorname{dist}(A,B)=0\). Thus positive distance is a sufficient test for separation, not a necessary one.
The finite-in-every-bounded-interval observation in this example is specific to these sets. A sequence of their points that converges in \(\mathbb{R}\) must be bounded; only finitely many points of either set lie in the bounded region containing that sequence. Any convergent sequence from one of the sets must therefore have a constant subsequence, whose limit belongs to the set. This verifies that neither set has a missing finite limit point.
Connected Sets Cannot Cross a Separation
A separation is useful not only for proving that a set is disconnected. It also constrains any connected subset lying inside the separated set. This principle lets us locate connected pieces without having to characterize all connected subsets of the ambient space.
Proof. Suppose instead that \(C\) meets both \(A\) and \(B\). The sets \(C\cap A\) and \(C\cap B\) are then nonempty, disjoint, and cover \(C\). We show that they are relatively open in \(C\).
Take \(x\in C\cap A\). Since \(A\) and \(B\) are separated, \(x\notin\overline{B}^{\,X}\). By the definition of closure, there is an \(r>0\) such that \(B(x,r)\cap B=\varnothing\). Because \(C\subseteq A\cup B\), every point of \(B(x,r)\cap C\) must then lie in \(A\). Thus \(B(x,r)\cap C\subseteq C\cap A\), proving that \(C\cap A\) is relatively open in \(C\). For any \(y\in C\cap B\), the other separation condition gives \(y\notin\overline{A}^{\,X}\), so a ball around \(y\) misses \(A\); its intersection with \(C\) lies in \(C\cap B\). Hence \(C\cap B\) is relatively open in \(C\) as well.
The two nonempty relatively open sets \(C\cap A\) and \(C\cap B\) form a separation of \(C\), contradicting its connectedness. Therefore \(C\) cannot meet both pieces, and it lies entirely in one of them. \(\square\)
Worked Example: Connected Subsets of a Punctured Line
For \(E=\mathbb{R}\setminus\{0\}\), the sets \(A=(-\infty,0)\) and \(B=(0,\infty)\) form a separation, as verified earlier. If \(C\subseteq E\) is connected, the theorem shows that \(C\subseteq(-\infty,0)\) or \(C\subseteq(0,\infty)\). In particular, no connected subset of the punctured line can contain both a negative and a positive number.
The conclusion uses the separation of \(E\), not a claim that every subset of either side is connected. A subset of one side may itself be disconnected. The theorem only says that a connected subset cannot occupy both sides of this particular separation.
Choosing the Right Separation Argument
When testing whether a set can be split, first identify a proposed partition and verify that its pieces are disjoint, cover the set, and are nonempty. Then use the most convenient method to establish relative openness: compute relative closures, or find neighborhoods that stay on one side. A positive distance estimate is often quick, but the example with distance zero shows why it cannot be treated as a necessary condition.
The Relative-Closure Criterion converts a separation problem into a limit-point question. The connected-subset theorem then turns a separation of a larger set into a restriction on every connected subset inside it. These arguments work in metric spaces without relying on an order or an interval description, extending the separation methods developed earlier in this course.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why must nonemptiness be stated separately in the Relative-Closure Criterion for a Separation?
- For a partition \(E=A\cup B\), how do the relative-closure conditions imply that both pieces are relatively open?
- Why does positive distance imply that no point of either piece lies in the other piece’s closure?
- How can two separated sets have distance zero?
- Which part of the proof shows that a connected subset cannot meet both pieces of a separation?