From Separated Sets to a Separation of a Set
The previous tutorial studied when two sets are separated: neither contains a point of the other’s closure. A separation of a set uses two such pieces to partition an entire set. The distinction is important. Two separated sets need not cover anything in particular, while a separation must cover the set under consideration and have two nonempty pieces.
For a subset \(E\subseteq\mathbb{R}\), openness and closedness of its pieces are understood relative to \(E\). Thus, a piece can be relatively open in \(E\) even when it is not open in \(\mathbb{R}\). The result from the previous tutorial, Separated Pieces and Relative Closedness, gives an equivalent way to recognize a separation: the two pieces are separated sets whose union is \(E\).
The nonempty requirement prevents the empty piece from making every set look separated. Since \(A\) and \(B\) are disjoint and cover \(E\), each is the relative complement of the other. Consequently, if both are relatively open, both are also relatively closed in \(E\). This is the Clopen Criterion for Connectedness from earlier in the course, not a separate test that needs to be established again here.
Missing Intermediate Points Detect Separations
On the real line, there is a direct way to decide whether a set admits a separation: look for a gap between two of its points. A gap here means an omitted real number strictly between points of \(E\). Any such omission creates a separation by dividing the set to the left and right of the gap. Conversely, if a set has a separation, points in opposite pieces must have an omitted point between them.
Proof. First suppose \(E=A\cup B\) is a separation. Choose \(a\in A\) and \(b\in B\); these points exist because both pieces are nonempty. If necessary, exchange their names so that \(a<b\). We claim that some point strictly between \(a\) and \(b\) is not in \(E\).
Suppose instead that every point between \(a\) and \(b\) belonged to \(E\). Then the whole closed interval \([a,b]\) would be a subset of \(E=A\cup B\). The interval \([a,b]\) is connected by the Intervals Are Connected Theorem. The result Connected Sets Lie on One Side of a Separation, established in the previous tutorial, would then imply \([a,b]\subseteq A\) or \([a,b]\subseteq B\). Neither is possible: \(a\in A\) and \(b\in B\). This contradiction proves that there is some \(c\in(a,b)\) with \(c\notin E\).
Conversely, suppose \(x,y\in E\) and \(c\notin E\) lies strictly between them. After exchanging \(x\) and \(y\) if needed, assume \(x<c<y\). Define $$ A=E\cap(-\infty,c) \quad\text{and}\quad B=E\cap(c,\infty). $$ The points \(x\) and \(y\) show that \(A\) and \(B\) are both nonempty. They are disjoint. Every point of \(E\) lies either below or above \(c\), because \(c\notin E\), so \(E=A\cup B\). Finally, \((-\infty,c)\) and \((c,\infty)\) are open in \(\mathbb{R}\), making \(A\) and \(B\) relatively open in \(E\). Thus they form a separation. \(\square\)
This criterion says that connected subsets of the real line cannot omit an intermediate point between two of their points. It is closely related to the Characterization of Connected Subsets of \(\mathbb{R}\) from earlier in the course: connected subsets of \(\mathbb{R}\) are exactly the intervals. The gap criterion is especially useful when the set is given by a description that makes its missing points easy to identify.
Worked Example: A Gap Between Two Intervals
Let \(E=[-5,-3]\cup[2,6]\). The point \(c=0\) is not in \(E\), and \(-4,4\in E\) satisfy \(-4<0<4\). The gap criterion therefore gives a separation. The corresponding pieces are $$ A=E\cap(-\infty,0)=[-5,-3], \qquad B=E\cap(0,\infty)=[2,6]. $$ They are nonempty and disjoint, and their union is \(E\). They are relatively open in \(E\): for example, \(A=E\cap(-\infty,0)\), where \((-\infty,0)\) is open in \(\mathbb{R}\); the same reasoning applies to \(B\). Thus the missing point \(0\) gives an explicit separation.
Worked Example: Separating the Rational Numbers at an Irrational Point
Let \(E=\mathbb{Q}\), the set of rational numbers, and take \(c=\sqrt{2}\). Since \(\sqrt{2}\notin\mathbb{Q}\), it is a missing point of \(E\). For example, \(1,\ 2\in E\) and \(1<\sqrt{2}<2\). The gap criterion shows that \(\mathbb{Q}\) has a separation: $$ A=\mathbb{Q}\cap(-\infty,\sqrt{2}), \qquad B=\mathbb{Q}\cap(\sqrt{2},\infty). $$ Both pieces are nonempty, as \(1\in A\) and \(2\in B\). No rational number equals \(\sqrt{2}\), so they cover \(\mathbb{Q}\); they are disjoint and relatively open there because each is the intersection of \(\mathbb{Q}\) with an open ray. Notice that a gap in \(E\) need not be an interval free of points of \(E\): it is enough that one intermediate point is missing to form this particular left-right partition.
Every Separation Forces a Gap, but Need Not Follow One Cut
The converse part of the gap criterion constructs a particular separation: all points below \(c\) go in one piece and all points above it go in the other. It does not say that every separation has this form. A set can have a separation whose pieces alternate in their positions on the line.
Worked Example: A Separation That Does Not Come from One Cut
Take \(E=\{0,1,2\}\), with \(A=\{0,2\}\) and \(B=\{1\}\). These are nonempty, disjoint pieces and \(E=A\cup B\). To check relative openness directly, each point of \(E\) can be isolated by an open interval: \(E\cap(-1/3,1/3)=\{0\}\), \(E\cap(2/3,4/3)=\{1\}\), and \(E\cap(5/3,7/3)=\{2\}\). Therefore every subset of \(E\), including \(A\) and \(B\), is relatively open, and this is a separation.
But this partition cannot be made by putting all points below a single omitted cut on one side and all points above it on the other. The points \(0\) and \(2\) are in the same piece, while \(1\), which lies between them, is in the other. This example is consistent with the gap criterion: \(E\) has omitted points between its elements, such as \(1/2\). The criterion guarantees that a set has at least one separation arising from a gap; it does not classify every possible separation of that set.
A practical consequence of the first half of the gap criterion is that every separation has a witness: choose one point from each piece, and there is a missing point strictly between them. The missing point need not determine the original pieces by a left-right division. In the three-point example, \(0\) and \(1\) lie in opposite pieces, as do \(1\) and \(2\); a gap between either pair witnesses the existence of a gap, but the original partition still groups the two outer points together.
Restricting a Separation to a Subset
A separation can also be used on part of the original set. The restriction is useful because a subset may inherit a separation from a larger set, provided it meets both pieces. If it meets only one piece, the restricted partition has an empty side and is not a separation.
Proof. The sets \(F\cap A\) and \(F\cap B\) are nonempty by hypothesis. They are disjoint because \(A\cap B=\varnothing\), and their union is \(F\), since every point of \(F\subseteq E=A\cup B\) belongs to \(A\) or \(B\). Because \(A\) is relatively open in \(E\), there is an open set \(U\subseteq\mathbb{R}\) such that \(A=E\cap U\). Then $$ F\cap A=F\cap(E\cap U)=F\cap U, $$ using \(F\subseteq E\). Hence \(F\cap A\) is relatively open in \(F\). The same argument, using an open set \(V\subseteq\mathbb{R}\) with \(B=E\cap V\), shows that \(F\cap B=F\cap V\) is relatively open in \(F\). The two nonempty pieces therefore form a separation of \(F\). \(\square\)
The hypothesis that \(F\) meets both sides cannot be dropped. If \(F\subseteq A\), then \(F\cap B=\varnothing\), so the intersections do not give two nonempty pieces. Conversely, this restriction theorem gives a quick way to prove that a subset is disconnected: find a known separation of a larger set and verify that the subset has points on both sides.
Worked Example: Restricting to a Smaller Set
Use the separation of \(E=[-5,-3]\cup[2,6]\) at \(0\), and let \(F=\{-4,3,5\}\subseteq E\). Then $$ F\cap A=\{-4\}, \qquad F\cap B=\{3,5\}. $$ Both intersections are nonempty, so the Restriction of a Separation Theorem shows that they separate \(F\). Directly, they are disjoint and cover \(F\). Since \(F\) is finite, each of its points can be isolated by an open interval, so both pieces are relatively open in \(F\). If instead we took \(F=\{-4\}\), the intersection with \(B\) would be empty, and this restriction would not be a separation.
How to Apply the Gap Criterion
When deciding whether a subset of \(\mathbb{R}\) is disconnected, it is often simplest to use the following procedure. This avoids guessing a partition first and then checking relative openness from scratch.
To produce a separation, start with \(x,y\in E\) and look between them. To rule one out using the criterion, check whether every point between every such pair belongs to \(E\).
If \(x<c<y\) and \(c\notin E\), the left and right parts of \(E\) at \(c\) are both nonempty.
Use \(E\cap(-\infty,c)\) and \(E\cap(c,\infty)\). Verify that they are disjoint, cover \(E\), and are relatively open.
A gap proves that at least one separation exists. It does not imply that every separation of \(E\) is a left-right partition at a single point.
The converse test is equally useful: if \(E\) has a separation, choose points in opposite pieces. The Gap Criterion for Separation guarantees an omitted intermediate point. Thus an interval, which contains every point between any two of its points, cannot have a separation. This conclusion also follows from the earlier Intervals Are Connected Theorem; the gap argument explains exactly what obstruction to connectedness would have to appear on the real line.
Check Your Understanding
Use the definition and results in this tutorial to answer the following questions.
- What conditions must two pieces satisfy to form a separation of \(E\)?
- How does a missing point strictly between two points of \(E\) produce a separation?
- Why must every separation of a subset of \(\mathbb{R}\) have a missing intermediate point between points in opposite pieces?
- Does the gap criterion say that every separation is determined by one left-right cut? Explain using \(E=\{0,1,2\}\).
- Under what condition does a separation of \(E\) restrict to a separation of a subset \(F\subseteq E\)?