What a Separation Must Do
A space is disconnected when its points can be divided into two nonempty pieces that are open from the point of view of that space. The phrase “from the point of view of that space” matters: for a subset of the real line, openness is usually relative openness, not openness in the whole real line. We will use this distinction to build explicit separations of the rationals, the irrationals, and punctured intervals.
Throughout, \(\mathbb{R}\) has its standard topology. Every subset \(A\subseteq\mathbb{R}\) has the subspace topology: its open sets are the intersections \(A\cap U\), where \(U\) is open in \(\mathbb{R}\). A subset of \(A\) that is open in this topology is called relatively open in \(A\).
Thus, to prove that a particular subset \(A\) of \(\mathbb{R}\) is disconnected, it is enough to exhibit two sets \(U,V\subseteq A\) and verify four things: they are nonempty, they are disjoint, they cover \(A\), and they are relatively open in \(A\). A set can be open in \(A\) even when it is not open in \(\mathbb{R}\).
Proof. Suppose \(X\) is disconnected, and let \(U,V\) be a separation. Since \(X=U\cup V\) and \(U\cap V=\varnothing\), we have \(X\setminus U=V\). The set \(V\) is open, so \(U\) is closed; by the definition of separation, \(U\) is also open, nonempty, and not all of \(X\). Conversely, suppose \(U\) is a nonempty proper subset of \(X\) that is both open and closed. Then \(V=X\setminus U\) is nonempty because \(U\ne X\), and open because \(U\) is closed. The sets \(U,V\) are disjoint, open, nonempty, and have union \(X\), so they form a separation. \(\square\)
A Cut That Gives a Separation
In the real line, a missing point can serve as a cut: points of the set on its two sides give the two pieces. The next result records precisely what needs to be checked.
Proof. Set \(U=A\cap(-\infty,c)\) and \(V=A\cap(c,\infty)\). The hypothesis gives that both are nonempty. No real number is both less than and greater than \(c\), so \(U\cap V=\varnothing\). Since \(c\notin A\), each point of \(A\) is either less than \(c\) or greater than \(c\); therefore \(A=U\cup V\). Finally, \((-\infty,c)\) and \((c,\infty)\) are open in \(\mathbb{R}\), so their intersections with \(A\), namely \(U\) and \(V\), are open in the subspace topology on \(A\). Thus \(U,V\) are a separation. \(\square\)
The missing-cut condition is useful because it reduces the construction to finding one point absent from \(A\), with points of \(A\) on both sides. The strict inequalities matter: the cut itself belongs to neither piece, so it must not be a point of \(A\).
Worked Example: Separating the Rationals
Consider \(A=\mathbb{Q}\subseteq\mathbb{R}\), with the subspace topology. We separate it using \(c=\sqrt{2}\). First, \(\sqrt{2}\notin\mathbb{Q}\). Indeed, if \(\sqrt{2}=m/n\) for relatively prime positive integers \(m,n\), then \(m^2=2n^2\). Thus \(m^2\) is even, so \(m\) is even; writing \(m=2k\) gives \(n^2=2k^2\), so \(n\) is even as well, contradicting that \(m,n\) are relatively prime.
Define \(U=\mathbb{Q}\cap(-\infty,\sqrt{2})\) and \(V=\mathbb{Q}\cap(\sqrt{2},\infty)\). The rational number \(1\) belongs to \(U\), since \(1<\sqrt{2}\), and the rational number \(2\) belongs to \(V\), since \(2>\sqrt{2}\). Hence neither set is empty. No number can be both less than and greater than \(\sqrt{2}\), so the sets are disjoint. Every rational number differs from the irrational number \(\sqrt{2}\), and hence is either less than or greater than \(\sqrt{2}\); therefore \(U\cup V=\mathbb{Q}\). Each is relatively open because it is the intersection of \(\mathbb{Q}\) with an open real interval. These checks prove that \(U,V\) separate \(\mathbb{Q}\), so \(\mathbb{Q}\) is disconnected.
Check: The complement of \(U\) within \(\mathbb{Q}\) is \(V\), not a set containing \(\sqrt{2}\), because \(\sqrt{2}\notin\mathbb{Q}\). Thus \(U\) is also relatively closed, in agreement with the proposition.
Separating the Irrationals
The same cut method works for a different subset, but the cut should now be rational. Write \(\mathbb{R}\setminus\mathbb{Q}\) for the set of irrationals. At a rational cut, no irrational point is left out, so the two strict sides cover the entire set.
Worked Example: Separating the Irrationals
Let \(A=\mathbb{R}\setminus\mathbb{Q}\), with the subspace topology, and choose the rational cut \(c=0\). Set \(U=A\cap(-\infty,0)\) and \(V=A\cap(0,\infty)\). The number \(-\sqrt{2}\) is irrational and negative, so it belongs to \(U\); the number \(\sqrt{2}\) is irrational and positive, so it belongs to \(V\). Both pieces are therefore nonempty. They are disjoint because a real number cannot be simultaneously negative and positive. Since \(0\) is rational, it does not belong to \(A\); every irrational number is consequently either negative or positive. This proves \(A=U\cup V\). The two pieces are relatively open because they are intersections of \(A\) with open intervals in \(\mathbb{R}\). Hence they form a separation, and the irrationals are disconnected.
Check: The strict inequalities exclude \(0\), but that causes no gap in the partition: \(0\notin A\). If the set being split were all of \(\mathbb{R}\), these same two pieces would fail to cover it, since they would omit \(0\).
Punctured Intervals
Removing an interior point from an interval creates a gap. The pieces on either side of the missing point are relatively open in the remaining set. This works whether the original interval includes its endpoints or not; the relative topology handles endpoint membership automatically.
Worked Example: An Open Interval with Its Midpoint Removed
Take \(A=(0,2)\setminus\{1\}\), with the subspace topology inherited from \(\mathbb{R}\). Let \(U=(0,1)\) and \(V=(1,2)\). Both are subsets of \(A\). They are nonempty, since \(1/2\in U\) and \(3/2\in V\). They are disjoint because no real number is simultaneously less than and greater than \(1\). Every point of \(A\) lies in \((0,2)\) and differs from \(1\), so it is either less than \(1\) or greater than \(1\); thus \(A=U\cup V\). Also \(U=A\cap(-\infty,1)\) and \(V=A\cap(1,\infty)\). Both are therefore relatively open in \(A\), and they separate \(A\).
Check: The point \(1\) is not part of the set \(A\), so neither piece needs to contain it. The union is exactly \((0,1)\cup(1,2)=(0,2)\setminus\{1\}\), as required.
Worked Example: A Closed Interval with an Interior Point Removed
Let \(A=[0,4]\setminus\{2\}\). The two pieces are \(U=[0,2)\) and \(V=(2,4]\). Each is nonempty: \(0\in U\) and \(4\in V\). They are disjoint, and every point of \(A\) is either less than \(2\) or greater than \(2\), so \(U\cup V=A\). To verify relative openness rather than relying on the endpoint notation, write \(U=A\cap(-1,2)\) and \(V=A\cap(2,5)\). The intervals \((-1,2)\) and \((2,5)\) are open in \(\mathbb{R}\), so \(U,V\) are open in \(A\). They form a separation, proving that the punctured closed interval is disconnected.
Check: The endpoint \(0\) belongs to \(U\), and \(4\) belongs to \(V\); neither is lost when the pieces are described as intersections with open real intervals. The cut point \(2\) is absent from \(A\), so the two pieces cover all of \(A\).
A Common Mistake: A Split That Misses a Point
A proposed pair of sets is not a separation unless its union is the whole space. For example, in \(\mathbb{Q}\), the two sets \(\mathbb{Q}\cap(-\infty,0)\) and \(\mathbb{Q}\cap(0,\infty)\) are disjoint and relatively open, but their union is \(\mathbb{Q}\setminus\{0\}\), not \(\mathbb{Q}\). The rational number \(0\) has been omitted, so this is not a separation. Assigning \(0\) to one side does not fix the problem while keeping both pieces open: for instance, \(\mathbb{Q}\cap(-\infty,0]\) is not relatively open in \(\mathbb{Q}\), since every open real interval around \(0\) contains positive rationals as well as \(0\).
The correction is to choose a cut that is not in the set being separated. For \(\mathbb{Q}\), \(\sqrt{2}\) is such a cut; for the irrationals, \(0\) is such a cut. In each case the two strict sides cover the set because the cut itself is absent. Before concluding that a space is disconnected, check nonemptiness, disjointness, coverage, and relative openness separately.
Check Your Understanding
For each proposed separation, identify the set being split and verify the partition conditions.
- Use the cut \(c=1\) to construct a separation of \(\mathbb{Q}\setminus\{1\}\).
- For \(A=[-2,3]\setminus\{0\}\), write the two pieces of a separation and express each as an intersection with an open subset of \(\mathbb{R}\).
- Explain exactly why \(\mathbb{Q}\cap(-\infty,0)\) and \(\mathbb{Q}\cap(0,\infty)\) do not separate \(\mathbb{Q}\).
- Show that if \(c\notin A\subseteq\mathbb{R}\) and \(A\) has points on both sides of \(c\), then \(A\) is disconnected.
- In the subspace \(A=(0,1)\setminus\{1/3\}\), identify a point in each side of the separation determined by \(1/3\), and state why the pieces are relatively open.