What Must a Separation Do?
A separation divides a space into two pieces that are open from the point of view of the space itself. Four requirements work together: the pieces must be open, nonempty, disjoint, and cover the whole space. Missing even one condition means that the proposed pair is not a separation. This makes the definition useful both for constructing examples and for ruling them out.
Let \((X,\mathcal{T})\) be a topological space. A set is open in \(X\) if it belongs to \(\mathcal{T}\). When a set \(Y\subseteq X\) is given the subspace topology, a set open in \(Y\) has the form \(Y\cap O\), where \(O\) is open in \(X\). Thus, “relatively open” always means open in the space currently under consideration.
The sets in a separation partition the space: every point belongs to exactly one of them. The order of the pair does not matter; if \((U,V)\) is a separation, then \((V,U)\) is one as well. A pair of disjoint open sets that leaves out even one point is not a separation. Nor is a pair that covers the space but has an empty member.
Verify that each proposed piece is open in the topology of \(X\), not merely open in some larger ambient space.
Each piece must contain at least one point.
Verify that the two pieces have no point in common.
Verify that every point of \(X\) belongs to one of the two pieces.
Building a Separation in a Subspace
Worked Example: The Punctured Real Line
Let \(X=\mathbb{R}\setminus\{0\}\), with the subspace topology inherited from the standard topology on \(\mathbb{R}\). We will verify that \(U=(-\infty,0)\) and \(V=(0,\infty)\) form a separation of \(X\).
Openness. The interval \((-\infty,0)\) is open in \(\mathbb{R}\), and \(U=X\cap(-\infty,0)=(-\infty,0)\). Therefore \(U\) is open in the subspace \(X\). Likewise, \((0,\infty)\) is open in \(\mathbb{R}\) and \(V=X\cap(0,\infty)=(0,\infty)\), so \(V\) is open in \(X\).
Nonemptiness and disjointness. We have \(-1\in U\) and \(1\in V\), so both sets are nonempty. A real number cannot be both negative and positive, so \(U\cap V=\varnothing\).
Coverage. Each \(x\in X\) is a nonzero real number. It is therefore either negative or positive, so \(x\in U\cup V\). Conversely, both \(U\) and \(V\) are subsets of \(X\). Thus \(U\cup V=X\), and all four requirements hold.
This example also shows why the space under consideration must be named. The same two sets do not cover \(\mathbb{R}\): their union is \(\mathbb{R}\setminus\{0\}\), which omits \(0\). They do cover \(X\), because \(0\) is not a point of \(X\).
A Simple Way to Rule Out Separations
Proof. Assume, to the contrary, that \(U,V\) form a separation of \(X\). Since both are nonempty and open, the hypothesis gives \(p\in U\) and \(p\in V\). Hence \(p\in U\cap V\), contradicting the requirement that \(U\) and \(V\) be disjoint. Therefore no separation exists. \(\square\)
The theorem gives a direct test: if all nonempty open sets must pass through one common point, two such sets cannot be disjoint. The following example checks this test in a small space by listing its open sets.
Worked Example: The Sierpinski Space
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). The nonempty open sets are exactly \(\{a\}\) and \(X\), and both contain \(a\). The theorem therefore shows that \(X\) has no separation.
We can also check this directly. A proposed pair of nonempty open sets must choose each member from \(\{a\}\) and \(X\). If both members are \(X\), their intersection is \(X\), which is nonempty. If one member is \(\{a\}\) and the other is \(X\), their intersection is \(\{a\}\), which is nonempty. If both are \(\{a\}\), their intersection is again \(\{a\}\). Thus no pair of nonempty open sets is disjoint, so in particular none can be a separation.
Worked Example: An Indiscrete Space
Let \(X=\{p,q,r\}\) and give \(X\) the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\). The only nonempty open set is \(X\), which contains \(p\). By the theorem, \(X\) has no separation.
Checking the definition directly gives the same result. Any two nonempty open sets must both equal \(X\). Their intersection is \(X\), so they are not disjoint. The fact that \(X\) has three points does not itself produce a separation: the available open sets, not just the underlying points, determine whether one exists.
Separations in Discrete Spaces
At the opposite extreme, in a discrete topology every subset is open. This makes it possible to choose pieces freely, subject only to the requirements that both be nonempty, disjoint, and cover the space.
Proof. First suppose \(X\) has at least two points. Choose \(p\in X\) and let \(U=\{p\}\), \(V=X\setminus\{p\}\). The set \(U\) is nonempty. Since \(X\) has at least two points, there is a point other than \(p\), so \(V\) is nonempty. The sets are disjoint and \(U\cup V=X\). Because the topology is discrete, both sets are open. Hence they form a separation.
Conversely, suppose \(X\) has fewer than two points. If \(X=\varnothing\), it has no nonempty subsets. If \(X\) has exactly one point, any two nonempty subsets both contain that point and so cannot be disjoint. In either case a separation is impossible. This proves both directions. \(\square\)
Worked Example: A Three-Point Discrete Space
Take \(X=\{p,q,r\}\) with the discrete topology. Set \(U=\{p\}\) and \(V=\{q,r\}\). Each is open because every subset of a discrete space is open. Neither is empty, and \(U\cap V=\varnothing\), since \(p\) is not \(q\) or \(r\). Their union is \(\{p,q,r\}=X\). Thus \((U,V)\) is a separation.
As a check, each point of \(X\) appears in exactly one member: \(p\in U\), while \(q,r\in V\). The pair therefore covers all three points without overlap.
An Infinite Space with No Separation
Having many open sets does not guarantee that a separation exists. For example, in the finite-complement topology on an infinite set, any two nonempty open sets must intersect.
Proof. Suppose \(U\) and \(V\) are nonempty open sets. By the definition of the topology, \(X\setminus U\) and \(X\setminus V\) are finite. If \(U\cap V=\varnothing\), then every point of \(X\) is outside \(U\) or outside \(V\), so \(X=(X\setminus U)\cup(X\setminus V)\). The right-hand side is a union of two finite sets and is therefore finite. This contradicts the assumption that \(X\) is infinite. Thus any two nonempty open sets intersect, and no pair can meet the disjointness requirement for a separation. \(\square\)
Worked Example: The Finite-Complement Topology on the Integers
Give \(X=\mathbb{Z}\) the finite-complement topology. The sets \(U=\mathbb{Z}\setminus\{0\}\) and \(V=\mathbb{Z}\setminus\{1\}\) are open because their complements are finite. They are nonempty: \(1\in U\) and \(0\in V\). But they are not disjoint. In fact, \(2\in U\) and \(2\in V\), so \(2\in U\cap V\).
This pair is not a separation because it fails the disjointness condition. More generally, the theorem shows that no different choice of two nonempty open subsets of \(\mathbb{Z}\) can be disjoint. Thus \(\mathbb{Z}\) with this topology has no separation.
A Common Mistake: Checking Only Disjointness
A frequent error is to find two disjoint open sets and conclude that they form a separation without checking whether they cover the space. For instance, in \(\mathbb{R}\) with the standard topology, \(U=(-\infty,0)\) and \(V=(0,\infty)\) are open and disjoint. But the proposed calculation \(U\cup V=\mathbb{R}\) is wrong: in fact \(U\cup V=\mathbb{R}\setminus\{0\}\), so \(0\) is missing.
The correction depends on the space being considered. For the whole real line, this pair fails the coverage condition. For the subspace \(X=\mathbb{R}\setminus\{0\}\), it does cover the space, and the earlier verification shows that it is a separation. A valid check must use the actual underlying set \(X\) at every step.
| Requirement | Question to ask | Failure in the example above on \(\mathbb{R}\) |
|---|---|---|
| Open | Are both sets open in the stated topology? | No failure: both intervals are open. |
| Nonempty | Does each set contain a point? | No failure: \(-1\in U\) and \(1\in V\). |
| Disjoint | Is \(U\cap V=\varnothing\)? | No failure: a number cannot be both negative and positive. |
| Covering | Is \(U\cup V=X\)? | Failure: \(0\in\mathbb{R}\) but \(0\notin U\cup V\). |
Separations are the precise set-and-topology structure used in the study of connected spaces. At this stage, the key practice is to test all four defining conditions and to keep the topology and underlying set explicit. The examples show why neither the number of points alone nor disjointness alone decides whether a separation exists.
Check Your Understanding
For each question, state the topology and check openness, nonemptiness, disjointness and coverage wherever a separation is proposed.
- In the discrete topology on \(\{a,b,c,d\}\), give a separation with one piece containing exactly \(a\) and \(b\).
- In the indiscrete topology on a two-point set, explain why no pair of nonempty open sets can be disjoint.
- For \(X=\mathbb{R}\setminus\{2\}\) with the subspace topology, do \((-\infty,2)\) and \((2,\infty)\) form a separation of \(X\)? Verify each condition.
- In the finite-complement topology on an infinite set \(X\), explain why two nonempty open sets cannot be disjoint.
- Can a pair of disjoint, nonempty, open subsets of \(X\) fail to be a separation? Give an example and identify the failed condition.