From Open Covers to Closed Families
Compactness is defined through open covers: every open cover must contain a finite subcover. The open-cover definition is useful directly, but it can also be translated into a statement about closed sets. The translation is simple: take complements. An open cover has no finite subcover precisely when the complementary closed sets have nonempty intersections for every finite selection, even though their total intersection is empty.
This connection gives two ways to organize a compactness argument. To prove compactness, one can start with an arbitrary open cover and extract a finite subcover, or show that every family of closed sets with the finite intersection property has nonempty total intersection. To disprove compactness, it is often enough to exhibit a cover with no finite subcover, or a closed family whose finite intersections are nonempty but whose total intersection is empty.
The convention about the empty subfamily matters in proofs. A nonempty finite selection has at least one member and may have a largest index when its members come from an indexed sequence. The empty selection has no largest index; its union is \(\varnothing\), and its intersection is \(X\). We will keep these cases separate when using indexed examples.
The Closed-Set Form of Compactness
The following theorem is the complement-based form of the open-cover definition. It applies to arbitrary topological spaces; no separation axiom is needed. As usual, compactness means that every open cover has a finite subcover.
Proof. First suppose \(X\) is compact, and let \(\{F_\lambda:\lambda\in\Lambda\}\) be a nonempty family of closed subsets with the finite intersection property. Suppose, to obtain a contradiction, that $$ \bigcap_{\lambda\in\Lambda}F_\lambda=\varnothing. $$ Taking complements shows that the open sets \(X\setminus F_\lambda\) cover \(X\). By compactness, finitely many of them cover \(X\), say $$ X=(X\setminus F_{\lambda_1})\cup\cdots\cup(X\setminus F_{\lambda_n}). $$ Taking complements of this finite union gives $$ F_{\lambda_1}\cap\cdots\cap F_{\lambda_n}=\varnothing, $$ contrary to the finite intersection property. Hence the total intersection is nonempty.
Conversely, suppose every nonempty family of closed subsets with the finite intersection property has nonempty total intersection. Let \(\{U_\lambda:\lambda\in\Lambda\}\) be an open cover of \(X\). If \(X\) is empty, the empty subfamily already covers it. If \(X\) is nonempty, the cover cannot be an empty family. Suppose that it has no finite subcover. For each \(\lambda\), put \(F_\lambda=X\setminus U_\lambda\). These sets are closed. Every finite selection \(U_{\lambda_1},\ldots,U_{\lambda_n}\) fails to cover \(X\), so there is a point outside their union; that point belongs to \(F_{\lambda_1}\cap\cdots\cap F_{\lambda_n}\). The empty selection also has nonempty intersection, since that intersection is \(X\). Thus the closed family has the finite intersection property. But the original cover covers \(X\), so $$ \bigcap_{\lambda\in\Lambda}F_\lambda =X\setminus\bigcup_{\lambda\in\Lambda}U_\lambda =\varnothing. $$ This contradicts the assumed closed-set property. Every open cover therefore has a finite subcover, and \(X\) is compact. \(\square\)
The theorem explains why closed families can be a convenient alternative to open covers. The finite intersection property is a finite-level condition: each finite selection must intersect. Compactness says that this finite consistency prevents the total intersection from being empty. Without compactness, the finite intersections can all be nonempty while the intersection of the entire family is empty.
Worked Examples: Finite and Total Intersections
Worked Example: An Open Cover of the Open Interval
For each integer \(n\geq 2\), let \(U_n=(1/n,1)\), an open subset of \(X=(0,1)\). These sets cover \(X\). Indeed, for any \(x\in(0,1)\), choose an integer \(n\geq2\) with \(n>1/x\). Then \(1/n<x<1\), so \(x\in U_n\).
No finite subfamily covers \(X\). For any nonempty finite selection, let \(N\) be its largest index. Since \(U_n\subseteq U_N\) whenever \(n\leq N\), its union is \(U_N=(1/N,1)\). The point \(1/(2N)\) lies in \(X\) and is not in \(U_N\), because \(1/(2N)<1/N\). The empty subfamily also fails to cover \(X\): its union is \(\varnothing\), and \(X\) is nonempty. Thus this cover has no finite subcover, and \((0,1)\) is not compact.
Worked Example: Closed Complements with Empty Total Intersection
Take complements in \(X=(0,1)\) of the sets in the preceding cover. For \(n\geq2\), this gives \(F_n=(0,1/n]\), which is closed in \(X\). The family has the finite intersection property. For any nonempty finite selection, let \(N\) be its largest index. The sets decrease as the index increases, so $$ F_{n_1}\cap\cdots\cap F_{n_k}=F_N=(0,1/N]. $$ This intersection is nonempty; for instance, \(1/(2N)\in(0,1/N]\). For the empty selection, the intersection is \(X=(0,1)\), which is also nonempty.
However, the total intersection is empty. If \(x\in(0,1)\), choose an integer \(n>1/x\). Then \(1/n<x\), so \(x\notin(0,1/n]=F_n\). No point belongs to every \(F_n\). This closed family therefore shows directly how finite intersections can all be nonempty while the total intersection is empty in a noncompact space.
Worked Example: A Closed Family in a Compact Interval
Let \(X=[0,1]\), which is compact, and set \(F_n=[0,1/n]\) for each integer \(n\geq1\). These sets are closed in \(X\). For any nonempty finite selection, let \(N\) be the largest selected index. Since \(1/n_j\geq1/N\) for each selected index \(n_j\), $$ F_{n_1}\cap\cdots\cap F_{n_k}=[0,1/N], $$ which is nonempty. The intersection for the empty selection is \(X=[0,1]\), also nonempty. In this case the total intersection is \(\{0\}\): zero belongs to every \(F_n\), and any \(x>0\) is excluded as soon as \(n>1/x\). The closed-set theorem guarantees a nonempty total intersection, but it does not require that intersection to contain more than one point.
Testing Covers Using a Basis
The family of all open sets can be large, and it may be more efficient to check covers drawn from a basis. Recall that a basis \(\mathcal B\) for a topology has the property that whenever \(x\) belongs to an open set \(U\), there is a basis element \(B\in\mathcal B\) with \(x\in B\subseteq U\). The next result turns that local property into a compactness test.
Proof. If \(X\) is compact, then every cover by basis elements is an open cover, so it has a finite subcover.
For the converse, suppose every cover of \(X\) by basis elements has a finite subcover. Let \(\{U_\lambda:\lambda\in\Lambda\}\) be any open cover of \(X\). Consider the collection $$ \mathcal C=\{B\in\mathcal B:\text{ there is some }\lambda\in\Lambda\text{ such that }B\subseteq U_\lambda\}. $$ This collection covers \(X\). Indeed, if \(x\in X\), choose a member \(U_\lambda\) of the given cover containing \(x\). The basis property gives \(B\in\mathcal B\) with \(x\in B\subseteq U_\lambda\), so \(B\in\mathcal C\). By the hypothesis, finitely many members \(B_1,\ldots,B_m\) of \(\mathcal C\) cover \(X\). For each \(i\), there is a set \(U_{\lambda_i}\) containing \(B_i\). Hence \(U_{\lambda_1},\ldots,U_{\lambda_m}\) cover \(X\), because the \(B_i\) do. We have obtained a finite subcover of the original open cover. If \(X\) is empty, its empty subcover suffices. Thus \(X\) is compact. \(\square\)
This criterion is useful when basis elements have a simpler description than arbitrary open sets. For example, the usual topology on \(\mathbb R\) has a basis consisting of open intervals with rational endpoints. To test compactness in that topology, it is enough to consider covers made from those intervals. The theorem does not say that a particular basis cover is automatically finite; it says that every such cover must have a finite subcover if the space is compact.
Worked Example: Detecting Noncompactness with Basis Elements
Give \(\mathbb Q\) the subspace topology inherited from \(\mathbb R\), and use the basis consisting of sets \(\mathbb Q\cap(a,b)\), where \(a,b\in\mathbb Q\) and \(a<b\). For each integer \(n\geq1\), let \(B_n=\mathbb Q\cap(-n,n)\). Each \(B_n\) is a basis element, and the family covers \(\mathbb Q\), since for any rational \(q\) one can choose an integer \(n>|q|\).
For any nonempty finite selection, let \(N\) be its largest index. The selected sets are nested, so their union is \(B_N\), which does not contain the rational number \(N+1\). The empty selection also fails to cover \(\mathbb Q\), since its union is empty and \(\mathbb Q\) is nonempty. Thus this basis cover has no finite subcover. By the basis-cover theorem, \(\mathbb Q\) is not compact.
Choosing the Right Formulation
Open covers and closed families express the same obstruction from opposite sides. If a cover has no finite subcover, its closed complements have the finite intersection property but empty total intersection. Conversely, such a closed family produces an open cover with no finite subcover by taking complements. A basis can simplify the open-cover side further: to establish compactness it suffices to handle basis covers, while one basis cover without a finite subcover is enough to establish noncompactness.
A common pitfall is to reason only about nonempty finite selections and then claim that every finite selection has a stated largest index. The largest-index argument applies only to a nonempty finite selection. The empty selection must be checked separately: its union is empty, and its intersection is the whole space. This distinction is especially important when proving that a sequence of sets gives a cover with no finite subcover or a closed family with the finite intersection property.
Another point to keep clear is the difference between finite intersections and the total intersection. The finite intersection property asserts that every finite intersection is nonempty; it does not assert that the sets share a common point. Compactness is the additional hypothesis that makes a common point unavoidable for a nonempty closed family with that property.
Check Your Understanding
Use the open-cover and closed-family formulations to answer the following questions.
- What does it mean for a family of sets to have the finite intersection property, and what is the intersection of the empty subfamily?
- Why do the complements of an open cover with no finite subcover have the finite intersection property?
- For \(U_n=(1/n,1)\) with \(n\geq2\), why must every nonempty finite selection have a largest index, and how does that help show there is no finite subcover?
- How does the basis property convert an arbitrary open cover into a cover by basis elements?
- Can a closed family in a compact space have a total intersection consisting of exactly one point? Explain using the interval example.