Separating a Subset Inside a Space
A subset inherits its own topology from the space containing it. Consequently, a separation of that subset is not required to divide the whole ambient space: it divides only the points of the subset, using sets that are open relative to it. This distinction gives a practical test for connectedness. Instead of first asking whether the subset is a topological space in its own right, we can look for ambient open sets whose intersections with the subset form a separation.
Let \((X,\mathcal{T})\) be a topological space and \(A\subseteq X\). The subspace topology on \(A\) consists of the sets \(A\cap U\) for \(U\in\mathcal{T}\). A set open in this topology is called relatively open in \(A\). A set is relatively closed if its complement in \(A\) is relatively open.
In this definition, “relatively open” is essential: \(P\) and \(Q\) need not be open in \(X\). They are open in the topology \(A\) inherits. The empty set and every singleton are connected, since neither can be expressed as a union of two disjoint nonempty subsets.
Separations and Clopen Subsets
Proof. Suppose \(A=P\cup Q\) is a separation. Since \(P\) and \(Q\) are relatively open, \(P\) is relatively open. Also \(A\setminus P=Q\) is relatively open, so \(P\) is relatively closed. It is nonempty, and it is proper because \(Q\ne\varnothing\).
Conversely, let \(C\) be a nonempty proper subset of \(A\) that is relatively open and relatively closed. Then \(A\setminus C\) is relatively open because \(C\) is relatively closed. The sets \(C\) and \(A\setminus C\) are both nonempty, disjoint, and have union \(A\). They form a separation. \(\square\)
A set that is both open and closed is often called clopen. The theorem says that to prove a subset connected, it is enough to rule out every nonempty proper clopen subset in its subspace topology. This test does not require the subset itself to be open or closed in \(X\).
Using Ambient Open Sets
Proof. Suppose \(A\) is disconnected, with separation \(P,Q\). By the definition of the subspace topology, there are open sets \(U,V\in\mathcal{T}\) such that \(P=A\cap U\) and \(Q=A\cap V\). Since \(P,Q\) are nonempty, disjoint, and cover \(A\), the intersections \(A\cap U,A\cap V\) have the required properties.
Conversely, suppose open sets \(U,V\) have the stated properties. Their intersections with \(A\) are relatively open by the definition of the subspace topology. They are nonempty, disjoint, and cover \(A\), so they are a separation of \(A\). \(\square\)
The theorem requires the intersections to be disjoint, but it does not require \(U\) and \(V\) themselves to be disjoint. Any overlap \(U\cap V\) must simply contain no points of \(A\). Requiring ambient opens to be disjoint would impose an extra condition that need not be possible in a general topological space.
Worked Example: Separating Two Points in the Real Line
Give a separation of \(A=\{0,1,3\}\) as a subset of \(\mathbb{R}\) with its standard topology, and verify it using ambient open sets.
Set \(U=(-1,1.5)\) and \(V=(2,4)\). Both are open in \(\mathbb{R}\). For \(U\), the point \(0\) satisfies \(-1<0<1.5\), the point \(1\) satisfies \(-1<1<1.5\), and \(3\) does not belong to \(U\). Thus \(A\cap U=\{0,1\}\). For \(V\), \(3\) satisfies \(2<3<4\), while \(0\) and \(1\) do not belong to \(V\). Thus \(A\cap V=\{3\}\).
The intersections are nonempty and disjoint, and \(\{0,1\}\cup\{3\}=\{0,1,3\}=A\). They are relatively open because they are intersections of \(A\) with ambient open sets. Therefore they give a separation of \(A\), so \(A\) is disconnected.
Check: The complementary pieces in \(A\) are exactly \(\{0,1\}\) and \(\{3\}\). Their intersection is empty and their union is all three points of \(A\), as required.
Worked Example: Relative Pieces from Overlapping Ambient Opens
Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}=\{\varnothing,\{c\},\{a,c\},\{b,c\},X\}\), and let \(A=\{a,b\}\). Show that the ambient opens \(\{a,c\}\) and \(\{b,c\}\) give a separation of \(A\), even though they overlap.
First, this collection is a topology: it contains \(\varnothing\) and \(X\), and unions of its sets are again in the collection. The only intersection of the two distinct two-point sets that needs checking is \(\{a,c\}\cap\{b,c\}=\{c\}\), which is also in the collection; the other finite intersections are in the collection as well.
Take \(U=\{a,c\}\) and \(V=\{b,c\}\). Intersecting with \(A\) gives \(A\cap U=\{a\}\) and \(A\cap V=\{b\}\). These are nonempty, disjoint, and their union is \(A\). Hence they form a separation in the subspace topology on \(A\). Yet \(U\cap V=\{c\}\ne\varnothing\).
In fact, there are no disjoint ambient open sets containing \(a\) and \(b\), respectively. The open sets containing \(a\) are \(\{a,c\}\) and \(X\); those containing \(b\) are \(\{b,c\}\) and \(X\). Every choice of one from each list has nonempty intersection. The relative separation still works because the shared point \(c\) is outside \(A\).
When an Ambient Separation Restricts
An ambient separation can sometimes provide a quick way to detect that a subset is disconnected. If \(X\) is divided into two disjoint open sets, intersecting those sets with \(A\) produces disjoint relatively open pieces. They form a separation of \(A\) precisely when both intersections are nonempty.
Proof. Because \(X=U\cup V\) and \(U\cap V=\varnothing\), the sets \(A\cap U\) and \(A\cap V\) are disjoint and have union \(A\). They are relatively open because \(U\) and \(V\) are open in \(X\). The assumed nonempty intersections therefore form a separation of \(A\). \(\square\)
The nonempty-intersection condition matters. If \(A\subseteq U\), for example, then \(A\cap V=\varnothing\), and the ambient separation gives no separation of \(A\). More generally, a subset can be connected even when the ambient space is disconnected.
Worked Example: A Connected Subset of a Disconnected Space
Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}=\{\varnothing,\{a,b\},\{c\},X\}\). Let \(A=\{a,b\}\). Determine whether \(X\) and \(A\) are connected.
The sets \(\{a,b\}\) and \(\{c\}\) are nonempty, disjoint, open, and have union \(X\), so they separate \(X\). Thus \(X\) is disconnected. For the subset \(A\), intersect each open set in \(\mathcal{T}\) with \(A\): the resulting subspace topology is \(\{\varnothing,A\}\). Its only nonempty relatively open set is \(A\) itself.
A separation of \(A\) would need two disjoint nonempty relatively open sets. Since both would have to equal \(A\), they could not be disjoint. Therefore \(A\) has no separation and is connected. The ambient split does not separate \(A\), because \(A\cap\{c\}=\varnothing\).
Check: The only candidates for relatively open pieces are \(\varnothing\) and \(\{a,b\}\). No pair of nonempty candidates has empty intersection, confirming that no separation exists.
A Common Pitfall
A common error is to treat a separation of \(A\) as if it had to be a separation of \(X\). In the preceding overlapping-open example, one might claim that \(U=\{a,c\}\) and \(V=\{b,c\}\) cannot witness a separation because \(U\cap V=\{c\}\ne\varnothing\). That calculation is correct for the ambient sets, but it tests the wrong condition: the definition requires \((A\cap U)\cap(A\cap V)=\varnothing\), not \(U\cap V=\varnothing\). Here the relevant calculation is \(\{a\}\cap\{b\}=\varnothing\), and \(\{a\}\cup\{b\}=A\). The relative pieces do form a separation.
Conversely, finding two ambient open sets whose intersections with \(A\) are disjoint is not by itself enough. Both intersections must be nonempty, and together they must cover all of \(A\). These checks prevent an empty piece or an uncovered point from being mistaken for a separation.
Check Your Understanding
Use the definitions of relative openness and separation to answer each question.
- For \(A=\{1,2\}\subseteq\mathbb{R}\), find ambient open sets whose intersections with \(A\) form a separation.
- Explain why disjoint ambient open sets are not required in the ambient-open test for a separation of a subset.
- Show that a nonempty proper relatively clopen subset of \(A\) gives a separation of \(A\).
- If \(X=U\cup V\) is a separation of \(X\), what additional condition on \(A\) ensures that intersecting \(U,V\) with \(A\) separates \(A\)?
- Can a subset be connected when its ambient space is disconnected? Give an example and verify the subspace topology.