Finite Subcovers

From Open Covers to Finite Subcovers

An open cover may have infinitely many members, even when the set it covers is simple. Compactness asks whether the cover can be reduced to finitely many of those members without leaving any point uncovered. The key phrase is “of those members”: a finite subcover is a selection from the original family, not an unrelated finite collection that happens to cover the same set.

Recall from “What Is Compactness?” that a set is compact when every open cover of it has a finite subcover. This tutorial focuses on what that selection means in practice and how to use it. A cover can be infinite and still have a finite subcover; it can also cover a set while no finite selection does. The distinction is exactly what compactness measures.

Definition: Let $\{U_\alpha:\alpha\in A\}$ be an open cover of $E\subseteq\mathbb{R}$. A finite subcover is a finite subfamily $\{U_\alpha:\alpha\in A_0\}$, where $A_0\subseteq A$ is finite, such that $E\subseteq\bigcup_{\alpha\in A_0}U_\alpha$. The selected members must come from the original cover.

The finite index set $A_0$ may be empty. The empty family covers the empty set, since the empty set is contained in every set, including the empty union. But the empty family cannot cover a nonempty set. Thus, when the set being covered is nonempty, any finite subcover must select at least one member.

A finite subcover does not have to use every member of the original family, and it need not be the smallest possible selection. If a cover is indexed in a way that repeats the same open set, a finite selection of indices may include repetitions of that set; the repetitions do not affect coverage. What matters is that finitely many selected members still cover every point of the target set.

Worked Examples: Finding and Ruling Out Finite Subcovers

Worked Example: Selecting Members from an Infinite Cover

Let $E=\{-1,2\}$. Consider the open sets $$ U_1=(-2,0),\qquad U_2=(1,3),\qquad V_n=(-10,10)\quad(n\geq1). $$ This is an infinite family of open sets. It covers $E$: $-1\in U_1$, because $-2<-1<0$, and $2\in U_2$, because $1<2<3$. The family also contains many other members that cover both points, but they are not needed.

The subfamily $\{U_1,U_2\}$ is finite and covers $E$, so it is a finite subcover. In particular, finding a finite subcover does not require examining every member of an infinite family: it is enough to identify a finite selection and verify that every point of the set lies in at least one selected member.

Worked Example: An Infinite Cover with No Finite Subcover

Let $E=(0,1)$, and for each integer $n\geq1$ define $$ U_n=\left(\frac{1}{n},2\right). $$ Each $U_n$ is open. These sets cover $E$. Indeed, if $x\in(0,1)$, the Archimedean property gives an integer $n\geq1$ with $n>1/x$. Since $x>0$, this implies $1/n<x$, and since $x<1<2$, we also have $x<2$. Thus $x\in U_n$.

Now take any finite selection of the sets $U_n$. If the selection is empty, it does not cover $E$, because $E$ is nonempty. If the selection is nonempty, it has a largest index $N$. For every selected index $n\leq N$, we have $1/n\geq1/N$, and consequently $U_n\subseteq U_N$. Therefore the union of the selected sets is contained in $U_N$.

The point $1/(N+1)$ belongs to $E$, since $0<1/(N+1)<1$. But $1/(N+1)<1/N$, so it does not belong to $U_N=(1/N,2)$, and hence it does not belong to the selected union. This holds for every nonempty finite selection, while the empty selection also fails. Therefore this open cover of $(0,1)$ has no finite subcover.

Worked Example: A Finite Selection That Covers a Closed Interval

Let $E=[0,2]$, and consider the open cover consisting of $$ U_n=\left(-1,2-\frac{1}{n}\right)\quad(n\geq1), \qquad W=\left(\frac{3}{2},3\right). $$ The subfamily $\{U_3,W\}$ covers $E$. In fact, $U_3=(-1,5/3)$, so every $x\in[0,5/3)$ lies in $U_3$, because $-1<0\leq x<5/3$. For $x\in[5/3,2]$, we have $3/2<5/3\leq x\leq2<3$, so $x\in W$. These two ranges together contain all of $[0,2]$.

The selected sets are members of the original family: $U_3$ is one of the $U_n$, and $W$ is also in that family. Thus $\{U_3,W\}$ is a finite subcover. Notice that $U_3$ does not contain the right endpoint $2$, and $W$ supplies coverage there. Checking endpoints explicitly is often essential when the target set is a closed interval.

Compactness and Closed Subsets

A finite subcover can be obtained for a subset by first covering a larger compact set. The main idea is to add one open set that covers the part of the larger set outside the subset. Closedness ensures that this additional set is open, so compactness applies to the resulting cover.

Theorem: If $K\subseteq\mathbb{R}$ is compact and $F\subseteq K$ is closed in $\mathbb{R}$, then $F$ is compact.

Proof. If $F=\varnothing$, then every open cover of $F$ has the empty family as a finite subcover, so $F$ is compact. Suppose instead that $F\neq\varnothing$, and let $\{U_\alpha:\alpha\in A\}$ be any open cover of $F$. Because $F$ is closed in $\mathbb{R}$, its complement $\mathbb{R}\setminus F$ is open.

The family consisting of $\mathbb{R}\setminus F$ together with all the $U_\alpha$ covers $K$. To see this, take $x\in K$. If $x\notin F$, then $x\in\mathbb{R}\setminus F$. If $x\in F$, the given cover of $F$ contains some $U_\alpha$ with $x\in U_\alpha$. Since $K$ is compact, this open cover of $K$ has a finite subcover.

The finite selection must include at least one of the $U_\alpha$. Otherwise it would consist only of $\mathbb{R}\setminus F$, which does not contain any point of the nonempty set $F\subseteq K$, and therefore could not cover $K$. Discard $\mathbb{R}\setminus F$, if it was selected. The remaining finitely many $U_\alpha$ still cover $F$: every point of $F$ is not in $\mathbb{R}\setminus F$, so it must be covered by one of the selected $U_\alpha$. Thus the original cover of $F$ has a finite subcover. Since the cover was arbitrary, $F$ is compact. $\square$

The requirement that $F$ be closed in $\mathbb{R}$ is part of the theorem. A subset can be closed relative to $K$ without being closed in the whole real line, but the argument above specifically uses the open set $\mathbb{R}\setminus F$ in $\mathbb{R}$. In applications, check which meaning of “closed” is being used before applying this result.

Worked Example: Covering a Closed Part of a Compact Set

Take $K=[-2,3]$ and $F=[0,2]$. The interval $K$ is compact by the result for closed bounded intervals (proved in the later tutorial “Closed Intervals” and assumed here), and $F$ is closed in $\mathbb{R}$. The theorem therefore implies that $F$ is compact.

The mechanism is visible for any open cover $\{U_\alpha\}$ of $[0,2]$. Add the open set $\mathbb{R}\setminus[0,2]=(-\infty,0)\cup(2,\infty)$. The resulting family covers $[-2,3]$: the added set covers the points of $K$ outside $[0,2]$, and the $U_\alpha$ cover the points in $[0,2]$. Compactness of $K$ gives a finite selection covering all of $K$. After removing the added complement, the selected cover members still cover $[0,2]$.

Finite Unions of Compact Sets

A second useful construction starts with separate finite subcovers. If a set is a finite union of compact pieces, then an open cover of the whole union restricts to a cover of each piece. Compactness supplies finitely many members for each piece, and the selections can then be combined.

Theorem: If $K_1,\ldots,K_m\subseteq\mathbb{R}$ are compact and $m$ is a positive integer, then $\bigcup_{j=1}^{m}K_j$ is compact.

Proof. Let $\{U_\alpha:\alpha\in A\}$ be an open cover of $\bigcup_{j=1}^{m}K_j$. For each $j$, this same family covers $K_j$, because $K_j\subseteq\bigcup_{i=1}^{m}K_i$. Since $K_j$ is compact, there is a finite subfamily $\mathcal{V}_j$ of the given cover that covers $K_j$. If $K_j$ is empty, take $\mathcal{V}_j$ to be the empty family.

The family $\mathcal{V}_1\cup\cdots\cup\mathcal{V}_m$ is finite: it is a union of finitely many finite families. It covers $\bigcup_{j=1}^{m}K_j$, because any point $x$ in that union belongs to some $K_j$, and $\mathcal{V}_j$ covers that $K_j$. Every member in the combined family comes from the original open cover. Hence it is a finite subcover of the union. Since the open cover was arbitrary, the union is compact. $\square$

The finiteness of the number of sets is essential to this proof: finitely many finite selections have a finite union. For infinitely many compact sets, combining one finite selection for each set can produce infinitely many cover members. Compactness of each piece alone does not make that combined selection finite.

Worked Example: A Cover of Two Compact Pieces

Let $K_1=[-3,-1]$ and $K_2=[1,4]$. Each is a closed bounded interval and hence compact. Suppose $\{U_\alpha:\alpha\in A\}$ is any open cover of $K_1\cup K_2$. Its members cover $K_1$, so compactness gives finitely many members covering $K_1$. Its members also cover $K_2$, so compactness gives finitely many members covering $K_2$.

Taking both finite selections together gives a finite subfamily of the original cover. Every point in $K_1\cup K_2$ lies in $K_1$ or $K_2$, and so lies in a selected member from the corresponding finite selection. Thus $K_1\cup K_2$ is compact. The intervals happen to be disjoint, but disjointness is not needed: the proof also works if the compact pieces overlap.

How to Check a Finite-Subcover Argument

A proof that a particular family has a finite subcover should identify the selected members and verify their coverage. A proof that no finite subcover exists must address every finite selection, not just one plausible selection. For an infinite indexed cover, useful structure may make this manageable: in the interval example above, any nonempty finite selection has a largest index, and all its members are contained in the one with that index. One point outside that largest member then defeats the entire selection.

The empty selection deserves separate attention whenever the covered set is nonempty. It cannot cover that set, but it has no largest index, so an argument based on a largest selected index applies only after the empty case has been excluded. Similarly, endpoint claims require checking the strict inequalities in the open intervals. These small checks are often where an otherwise sound finite-subcover argument succeeds or fails.

Name the selection.
Specify the finite set of indices, or explicitly name the finite subfamily being chosen.

Verify coverage point by point or by cases.
Check that each point of the target lies in at least one selected member, including endpoints when they occur.

When ruling out finite selections, include the empty case.
For a nonempty selection, use its finite structure to find a point of the target that remains uncovered.

For compactness proofs, keep selections within the given cover.
Combine the finite subfamilies supplied by compactness, and confirm that their union is still finite.

Check Your Understanding

Use the definition and results in this tutorial to answer the following questions.

What two conditions make a finite family a finite subcover of a given open cover?
Why must the empty selection be considered separately in the argument for the cover of $(0,1)$?
In the closed-subset theorem, why is $\mathbb{R}\setminus F$ added to the cover of $F$?
Why does the proof for a finite union of compact sets combine to a finite family of cover members?
What part of the proof for a finite union would not automatically work for an infinite union?

What You'll Learn