Order Gaps as a Test for Disconnectedness
The order on the real line gives a direct way to detect some disconnected subsets. If a point \(c\) is absent from a subset \(A\subseteq\mathbb{R}\), and \(A\) has points both below and above \(c\), then the points of \(A\) on the two sides form a separation. This gives a useful obstruction to connectedness: a connected subset of \(\mathbb{R}\) cannot have such an order gap.
Throughout, \(\mathbb{R}\) has its standard topology, and every subset \(A\subseteq\mathbb{R}\) has the subspace topology. Thus a set \(U\subseteq A\) is open in \(A\) precisely when \(U=A\cap W\) for some open set \(W\subseteq\mathbb{R}\).
Openness here is relative to \(A\), not necessarily openness in \(\mathbb{R}\). For example, a subset of the integers can be open in the subspace topology even if it is not open as a subset of the real line. The construction below uses intersections with open half-lines, so it directly verifies the required relative openness.
The Gap-Point Separation
Proof. Since \(c\notin A\), every point of \(A\) is either less than \(c\) or greater than \(c\). Therefore \(U\cup V=A\), and the two sets are disjoint because no real number is both less than and greater than \(c\). The inequalities \(a<c<b\), with \(a,b\in A\), show that \(a\in U\) and \(b\in V\), so both sets are nonempty. The half-lines \((-\infty,c)\) and \((c,\infty)\) are open in \(\mathbb{R}\); consequently \(U\) and \(V\) are open in \(A\), since each is the intersection of \(A\) with an open subset of \(\mathbb{R}\). Thus \(U,V\) form a separation of \(A\), so \(A\) is disconnected. \(\square\)
The point \(c\) is a gap point for \(A\) when \(c\notin A\) and \(A\) contains at least one point below \(c\) and at least one point above \(c\). The theorem shows that having a gap point is sufficient for disconnectedness. The two nonemptiness requirements matter: a point missing from \(A\) does not by itself give a separation.
Worked Example: A Finite Subset
Let \(A=\{-4,-1,3,7\}\subseteq\mathbb{R}\), and test the missing point \(c=0\). The set \(A\) is given the subspace topology inherited from the standard topology on \(\mathbb{R}\).
First, \(0\notin A\). The elements of \(A\) less than \(0\) are \(-4\) and \(-1\), while those greater than \(0\) are \(3\) and \(7\). Thus the proposed pieces are \(U=A\cap(-\infty,0)=\{-4,-1\}\) and \(V=A\cap(0,\infty)=\{3,7\}\). They are disjoint, both nonempty, and their union is \(\{-4,-1,3,7\}=A\). Since \((-\infty,0)\) and \((0,\infty)\) are open in \(\mathbb{R}\), these pieces are open in \(A\). They form a separation, so \(A\) is disconnected.
Check: the proposed split assigns every listed element exactly once: \(-4,-1<0\), and \(3,7>0\). In particular, there is no element of \(A\) left out at \(0\), because \(0\notin A\).
Worked Example: A Gap in the Integers
Take \(A=\mathbb{Z}\subseteq\mathbb{R}\) and \(c=\frac{3}{2}\), with \(\mathbb{Z}\) carrying the subspace topology. The number \(\frac{3}{2}\) is not an integer: it lies strictly between the consecutive integers \(1\) and \(2\), since \(1<\frac{3}{2}<2\).
Splitting at \(c\) gives \(U=\mathbb{Z}\cap(-\infty,\frac{3}{2})=\{\ldots,-2,-1,0,1\}\) and \(V=\mathbb{Z}\cap(\frac{3}{2},\infty)=\{2,3,4,\ldots\}\). The first set contains \(1\), and the second contains \(2\), so neither is empty. Every integer is either at most \(1\) or at least \(2\), so the two sets are disjoint and their union is \(\mathbb{Z}\). They are relatively open because each is the intersection of \(\mathbb{Z}\) with one of the open half-lines determined by \(\frac{3}{2}\). Hence they separate \(\mathbb{Z}\).
Check: the strict inequalities place \(1\) in \(U\), because \(1<\frac{3}{2}\), and \(2\) in \(V\), because \(2>\frac{3}{2}\). No integer equals \(\frac{3}{2}\), so no integer is omitted from the union.
The Order Condition for Connected Subsets
The gap-point test has a useful contrapositive. A connected subset of the real line cannot skip a point that lies strictly between two of its elements. This order property is often called order-convexity.
Proof. Suppose \(A\) is not order-convex. By the definition of order-convexity, there are \(a,b\in A\) and \(c\in\mathbb{R}\) such that \(a<c<b\) but \(c\notin A\). The gap-point criterion then gives a separation of \(A\), so \(A\) is disconnected. This proves the contrapositive: if \(A\) is connected, it must be order-convex. \(\square\)
This theorem is a necessary condition for connectedness: failure of order-convexity proves disconnectedness. It does not, by itself, establish that every order-convex subset is connected. In particular, the theorem is a gap test, not a complete classification of connected subsets of \(\mathbb{R}\).
Worked Example: A Gap in the Rationals
Let \(A=\mathbb{Q}\subseteq\mathbb{R}\), with the subspace topology, and consider \(c=\sqrt{2}\). First, \(c\notin\mathbb{Q}\). To verify this, suppose \(\sqrt{2}=m/n\) for integers \(m,n\) with \(n\ne0\), written in lowest terms. Then \(m^2=2n^2\), so \(m^2\) is even and \(m\) is even. Write \(m=2k\). Substitution gives \(4k^2=2n^2\), hence \(n^2=2k^2\), so \(n\) is even as well. This contradicts that \(m/n\) was in lowest terms.
The rational numbers \(1\) and \(2\) lie on opposite sides of \(c\): \(1<\sqrt{2}\) follows from \(1^2<2\), and \(\sqrt{2}<2\) follows from \(2<2^2\). Thus \(c\) is a gap point for \(A\). The resulting pieces are \(U=\mathbb{Q}\cap(-\infty,\sqrt{2})\) and \(V=\mathbb{Q}\cap(\sqrt{2},\infty)\). They are disjoint, relatively open, and cover \(\mathbb{Q}\) because \(\sqrt{2}\notin\mathbb{Q}\). The rational numbers \(1\) and \(2\) show respectively that \(U\) and \(V\) are nonempty. Therefore \(\mathbb{Q}\) is disconnected in its subspace topology.
Check: \(1\in U\) because \(1<\sqrt{2}\), and \(2\in V\) because \(\sqrt{2}<2\). The split uses strict inequalities, while the proof above verifies that the dividing point itself is not rational.
A Common Pitfall: A Missing Point Is Not Enough
To produce a separation by cutting at \(c\), both sides of the cut must contain points of \(A\). Omitting this condition can lead to a pair of sets that covers \(A\) but is not a separation, because one piece is empty.
Common mistake. For \(A=\{-4,-1,3,7\}\), one might choose the missing point \(c=8\) and claim that the two parts form a separation. The calculation is \(A\cap(-\infty,8)=\{-4,-1,3,7\}=A\) and \(A\cap(8,\infty)=\varnothing\). The second set is empty, so the pair fails the definition of a separation. The correction is to choose a missing point between elements of \(A\), such as \(c=0\): then \(A\cap(-\infty,0)=\{-4,-1\}\) and \(A\cap(0,\infty)=\{3,7\}\), and both pieces are nonempty.
Likewise, the order-convexity theorem can be used in one direction with no extra hypotheses: if a subset skips a point between two of its elements, it is disconnected. If no such gap is found, this test alone does not prove connectedness. The distinction is important whenever a criterion gives an obstruction rather than a full equivalence.
| What to check | Consequence |
|---|---|
| \(c\notin A\), with elements of \(A\) on both sides of \(c\) | The two half-line intersections separate \(A\) |
| A point between two elements of \(A\) is missing | \(A\) is not order-convex and is disconnected |
| A point is missing, but all elements of \(A\) lie on one side | This cut does not give a separation |
| No gap between elements of \(A\) has been found | The gap test alone gives no conclusion about connectedness |
The order on \(\mathbb{R}\) makes a proposed separation especially concrete: once a gap point is identified, the two pieces are specified by inequalities, and their relative openness follows from open half-lines. This converts an order calculation into a topological proof.
Check Your Understanding
For each question, use the subspace topology inherited from the standard topology on \(\mathbb{R}\) whenever the set is a subset of \(\mathbb{R}\).
- For \(A=\{-5,0,4\}\), choose a gap point and write down a separation, verifying that both pieces are nonempty.
- Show that \(\mathbb{Z}\) has a gap point at \(c=\frac{1}{2}\), and describe the two sets in the resulting separation.
- Can a missing point \(c\) give the gap-point separation if \(A\) has no elements greater than \(c\)? Explain using the definition of separation.
- Suppose \(A\subseteq\mathbb{R}\) is connected, \(a,b\in A\), and \(a<c<b\). What must be true of \(c\), and which theorem gives the conclusion?