Open Covers, Subcovers, and Refinements

Covering a Space with Open Sets

An open cover packages a local description of a space into a family of open sets that reaches every point. The family may have finitely many members or infinitely many. It is important to check both parts of the definition: every member must be open in the topology being used, and their union must contain the whole space (or the specified subset). We will also distinguish a subcover, which selects members from a given cover, from a refinement, which may use different open sets.

Definition: Let \((X,\mathcal{T})\) be a topological space and \(A\subseteq X\). A cover of \(A\) is a family \(\mathcal{U}\) of subsets of \(X\) such that \(A\subseteq\bigcup_{U\in\mathcal{U}}U\). An open cover of \(A\) in \(X\) is a cover of \(A\) whose members all belong to \(\mathcal{T}\). When \(A=X\), we call it an open cover of \(X\). A subcover of \(\mathcal{U}\) is a subfamily \(\mathcal{V}\subseteq\mathcal{U}\) that is still a cover of \(A\); it is a finite subcover if \(\mathcal{V}\) has finitely many members.

For \(A=\varnothing\), the empty family is a cover, since \(\varnothing\subseteq\bigcup_{U\in\varnothing}U=\varnothing\). If \(A\ne\varnothing\), an empty family cannot cover \(A\). A cover of \(A\) need not cover all of \(X\). Also, when a family is indexed, a subcover can be described by selecting indices: repeated sets do not prevent the selected subfamily from being finite.

If \(A\subseteq X\), the subspace topology on \(A\) consists of the sets \(A\cap V\) with \(V\in\mathcal{T}\). Thus a set can be open as a subset of \(A\) without being open in \(X\). We will state which topology is in use when checking a cover of an interval.

Subcovers and Finite Subcovers

Proposition: A subfamily \(\mathcal{V}\subseteq\mathcal{U}\) is a subcover of \(A\) if and only if \(A\subseteq\bigcup_{V\in\mathcal{V}}V\). It is a finite subcover if and only if this inclusion holds and \(\mathcal{V}\) has finitely many members.

Proof. By definition, a subcover is a subfamily of \(\mathcal{U}\) that covers \(A\). A family covers \(A\) exactly when every point of \(A\) belongs to at least one of its members, which is equivalent to \(A\subseteq\bigcup_{V\in\mathcal{V}}V\). Adding the requirement that \(\mathcal{V}\) have finitely many members gives exactly the definition of a finite subcover. These equivalences include \(A=\varnothing\), for which the empty subfamily satisfies the inclusion. \(\square\)

Worked Example: An Increasing Cover of the Real Line

Give \(\mathbb{R}\) its standard topology and, for \(n\in\mathbb{N}\), let \(U_n=(-n,n)\). The family \(\mathcal{U}=\{U_n:n\in\mathbb{N}\}\) is an open cover of \(\mathbb{R}\). Indeed, each \(U_n\) is open, and for any \(x\in\mathbb{R}\) there is a positive integer \(n>|x|\), so \(-n<x<n\) and \(x\in U_n\).

Now choose any finite nonempty collection of these sets, with indices \(n_1,\ldots,n_k\), and let \(N\) be the largest of those indices. Since \(n_i\le N\), each \((-n_i,n_i)\subseteq(-N,N)\); and \(U_N=(-N,N)\) is among the selected sets. Therefore their union is exactly \((-N,N)\). It does not cover \(\mathbb{R}\): for example, \(N+1\in\mathbb{R}\) and \(N+1\notin(-N,N)\). The empty subfamily does not cover \(\mathbb{R}\) either, since \(\mathbb{R}\ne\varnothing\). Thus this particular cover has no finite subcover.

Check: every real number \(x\) belongs to some \(U_n\), because the positive integers are unbounded above and we can choose \(n>|x|\). In contrast, each finite selection has union \((-N,N)\) for its largest index \(N\), which misses \(N+1\). Both claims follow directly from the endpoints of the intervals.

This example illustrates a useful distinction: the full family covers the space, but a selection of finitely many members need not do so. To test a proposed subcover, check its union rather than just its size or the fact that its members came from a cover.

Open Covers of Intervals

For an interval such as \(I=[0,1]\), we use the subspace topology inherited from \(\mathbb{R}\). A subset \(W\subseteq I\) is open in \(I\) precisely when \(W=I\cap V\) for some open \(V\subseteq\mathbb{R}\). For example, \([0,3/4)=I\cap(-1,3/4)\) and \((1/4,1]=I\cap(1/4,2)\), so both sets are open in \(I\), even though neither is open in \(\mathbb{R}\).

Worked Example: A Finite Open Cover of the Unit Interval

In \(I=[0,1]\) with its subspace topology, consider \(\mathcal{U}=\{U,V\}\), where \(U=[0,3/4)\) and \(V=(1/4,1]\). The relative-openness calculations above show that both sets are open in \(I\).

To check coverage, take \(x\in I\). If \(x<3/4\), then \(x\in U\), since \(x\ge0\). If \(x\ge3/4\), then \(x>1/4\) and \(x\le1\), so \(x\in V\). These two cases include every \(x\in[0,1]\), so \(\mathcal{U}\) covers \(I\). In particular, the endpoint \(0\) belongs to \(U\), the endpoint \(1\) belongs to \(V\), and \(1/2\) belongs to both because \(0\le1/2<3/4\) and \(1/4<1/2\le1\).

The full family \(\{U,V\}\) is a finite subcover of itself. The one-member subfamily \(\{U\}\) is not a subcover, since \(1\in I\) but \(1\notin U\). The one-member subfamily \(\{V\}\) is not a subcover, since \(0\in I\) but \(0\notin V\). Thus neither proper one-member subfamily covers \(I\).

Check: \(U\cup V=[0,1]=I\): the first set covers all \(x<3/4\), including \(0\), and the second covers all \(x>1/4\), including \(1\). The overlap is \((1/4,3/4)\), so in particular the point \(1/2\) is not omitted.

Refinements

Definition: Let \(\mathcal{U}\) and \(\mathcal{V}\) be open covers of the same space \(X\). The cover \(\mathcal{U}\) refines \(\mathcal{V}\) if for every \(U\in\mathcal{U}\), there is some \(V\in\mathcal{V}\) such that \(U\subseteq V\). In this case, \(\mathcal{U}\) is a refinement of \(\mathcal{V}\).

The containment direction matters: each member of the refinement must fit inside a member of the cover it refines. A refinement need not be a subfamily of the original cover; its sets can be new. For example, a cover made of small open intervals can refine one made of larger open intervals, even when none of the small intervals is itself one of the larger ones.

Proposition: If \(\mathcal{U}\) is an open cover of \(X\) and refines an open cover \(\mathcal{V}\) of \(X\), then every point of \(X\) belongs to some member of \(\mathcal{U}\) contained in a member of \(\mathcal{V}\). Moreover, if \(\mathcal{W}\) is an open cover of \(X\) that refines \(\mathcal{U}\), then \(\mathcal{W}\) refines \(\mathcal{V}\).

Proof. For the first assertion, let \(x\in X\). Since \(\mathcal{U}\) covers \(X\), there is \(U\in\mathcal{U}\) with \(x\in U\). Since \(\mathcal{U}\) refines \(\mathcal{V}\), there is \(V\in\mathcal{V}\) with \(U\subseteq V\). Hence \(x\in U\subseteq V\). For the second assertion, take any \(W\in\mathcal{W}\). Because \(\mathcal{W}\) refines \(\mathcal{U}\), there is \(U\in\mathcal{U}\) with \(W\subseteq U\). Because \(\mathcal{U}\) refines \(\mathcal{V}\), there is \(V\in\mathcal{V}\) with \(U\subseteq V\). Thus \(W\subseteq U\subseteq V\), which is exactly the condition that \(\mathcal{W}\) refines \(\mathcal{V}\). \(\square\)

The first assertion follows from having both a cover and the refinement condition. Containment alone would not guarantee coverage: the family \(\{\{a\}\}\) is contained in the member \(\{a,b\}\) of the cover \(\{\{a,b\},\{c\}\}\) of \(\{a,b,c\}\), but the family \(\{\{a\}\}\) does not cover \(\{a,b,c\}\).

Worked Example: Covers and Refinements on a Finite Space

Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},\{b\},\{a,b\},\{b,c\},X\}\). This is a topology: the intersection of any two listed sets is listed, and the union of any subfamily is listed. In particular, the only sets that involve \(c\) are \(\{b,c\}\) and \(X\); their intersections with the other listed sets are \(\varnothing\), \(\{b\}\), \(\{a,b\}\), \(\{b,c\}\), or \(X\), as appropriate.

Consider \(\mathcal{V}=\{\{a,b\},\{b,c\}\}\). Both members are open. The first contains \(a\) and \(b\), and the second contains \(b\) and \(c\); together they cover \(X\). Now let \(\mathcal{U}=\{\{a\},\{b\},\{b,c\}\}\). Its members are also open, and their union is \(X\). The set \(\{a\}\) is contained in \(\{a,b\}\), the set \(\{b\}\) is contained in \(\{a,b\}\), and \(\{b,c\}\) is contained in itself, which belongs to \(\mathcal{V}\). Therefore \(\mathcal{U}\) refines \(\mathcal{V}\).

The family \(\{\{a\},\{b\}\}\) is a subfamily of \(\mathcal{U}\), but it is not a subcover of \(X\): its union is \(\{a,b\}\), which misses \(c\). The full family \(\mathcal{U}\) is a finite subcover of itself, since it has three members and their union is \(\{a,b,c\}\). Here every membership and containment can be checked directly from the displayed sets.

Check: \(\{a\}\subseteq\{a,b\}\), \(\{b\}\subseteq\{a,b\}\), and \(\{b,c\}\subseteq\{b,c\}\); these verify each refinement condition. Also \(\{a\}\cup\{b\}\cup\{b,c\}=\{a,b,c\}\), while \(\{a\}\cup\{b\}=\{a,b\}\ne X\).

A Common Pitfall: Checking Only the Endpoints

A proposed family can fail to be a cover because it misses a point, even when its sets look like adjacent pieces of an interval. For example, in \(I=[0,1]\), the family \(\{[0,1/2),(1/2,1]\}\) misses \(1/2\): the first set contains only numbers strictly less than \(1/2\), and the second contains only numbers strictly greater than \(1/2\). Therefore its union is not \(I\).

Adding the midpoint to the first set repairs coverage, but one must still check openness. The set \([0,1/2]\) is not open in \(I\): if it were \(I\cap V\) for an open \(V\subseteq\mathbb{R}\) containing \(1/2\), then some interval around \(1/2\) would lie in \(V\). That interval would contain points greater than \(1/2\) and close enough to \(1/2\) to belong to \(I\), so \(I\cap V\) could not equal \([0,1/2]\). The valid cover \(\{[0,3/4),(1/4,1]\}\) avoids both errors: its members are open in \(I\), and their union includes every point of \(I\).

Check Your Understanding

For each question, check both the relevant containment and the topology in use.

In standard \(\mathbb{R}\), does the family \(\{(-n,n):n\in\mathbb{N}\}\) have a finite subcover of \(\mathbb{R}\)? Explain why.
In \(I=[0,1]\) with the subspace topology, verify that \(\{[0,2/3),(1/3,1]\}\) is an open cover.
In \(X=\{a,b,c\}\), is \(\{\{a,b\},\{b,c\}\}\) a subcover of \(\{\{a,b\},\{b,c\},\{a\}\}\)?
Does \(\{\{a\},\{b\},\{b,c\}\}\) refine \(\{\{a,b\},\{b,c\}\}\)? State the containing member for each set.
Give a family that covers a nonempty set but has no finite subcover, and verify the claim from its union.