Open Covers

What an Open Cover Records

Compactness is defined using open covers, so it is useful to understand exactly what a cover records before asking whether it can be reduced to finitely many members. The previous tutorial defined an open cover of a set $E\subseteq\mathbb{R}$ as a family of open subsets of $\mathbb{R}$ whose union contains $E$. The central requirement is containment: every point of $E$ must lie in at least one member of the family.

Definition: An open cover of $E\subseteq\mathbb{R}$ is a family $\{U_\alpha:\alpha\in A\}$ of open subsets of $\mathbb{R}$ such that $E\subseteq\bigcup_{\alpha\in A}U_\alpha$. The family need not be finite, its members need not be disjoint, and their union need not equal $E$.

The index set $A$ is part of the way the family is described. Two different indices may refer to the same open set; this causes no difficulty. A point may belong to several cover members, or to just one. Nor is a cover a partition: overlap is allowed, and the open sets can extend beyond the set being covered.

These distinctions matter when selecting or comparing cover members. A cover of $E$ is not required to consist of subsets of $E$. For example, open intervals in $\mathbb{R}$ can cover a closed interval while extending past its endpoints. Also, the union of a cover may include many points that are not in $E$. The purpose of a cover is to ensure that no point of $E$ is missed, not to describe $E$ exactly by disjoint pieces.

Worked Examples: Reading Covers Carefully

Worked Example: A Cover of a Closed Interval

Let $E=[0,1]$, and consider the two open intervals $$ U_1=\left(-\frac{1}{3},\frac{2}{3}\right), \qquad U_2=\left(\frac{1}{2},\frac{4}{3}\right). $$ Both are open subsets of $\mathbb{R}$. If $x\in[0,1]$ and $x\leq 2/3$, then $-1/3<x\leq2/3$, so $x\in U_1$ when $x<2/3$; the endpoint $x=2/3$ belongs to $U_2$, since $1/2<2/3<4/3$. If $2/3<x\leq1$, then $1/2<x<4/3$, so $x\in U_2$. Thus every point of $E$ lies in at least one of $U_1,U_2$.

Therefore $\{U_1,U_2\}$ covers $[0,1]$. Its union is not $[0,1]$: for instance, $-1/4$ belongs to $U_1$ but not to $[0,1]$. This is still an open cover, because the required inclusion is $[0,1]\subseteq U_1\cup U_2$.

Worked Example: Overlap Is Allowed

Consider $E=(0,1)$ and the family $$ U_1=(-1,7/10),\qquad U_2=(3/10,2). $$ If $x\in(0,1)$ and $x<7/10$, then $x\in U_1$. If $x\geq7/10$, then $3/10<x<1<2$, so $x\in U_2$. Hence the two intervals cover $E$.

Their overlap is $(3/10,7/10)$, which is nonempty. For example, $1/2$ belongs to both intervals. The overlap does not interfere with the cover: a point being covered more than once is just as acceptable as a point being covered once.

Worked Example: An Infinite Cover of a Sequence of Points

Let $E=\{1/n:n\geq1\}$. For each integer $n\geq2$, define $$ U_n=\left(\frac{1}{n+1},\frac{1}{n-1}\right), $$ and define $U_1=(1/2,3/2)$. Each set is an open interval. The point $1$ lies in $U_1$, since $1/2<1<3/2$. For $n\geq2$, the inequalities $n-1<n<n+1$, with all three integers positive, imply $$ \frac{1}{n+1}<\frac{1}{n}<\frac{1}{n-1}. $$ Thus $1/n\in U_n$. Every point of $E$ is therefore contained in a member of the family, so $\{U_n:n\geq1\}$ is an open cover of $E$.

This example illustrates one way to certify a cover: identify, for each point of the set, a particular member that contains it. There is no need to show that the cover members are disjoint, or that a point belongs only to the member chosen for it.

Ambient and Relative Open Covers

Sometimes it is convenient to describe open sets from the point of view of the set being covered. If $E\subseteq\mathbb{R}$, a subset $W\subseteq E$ is called relatively open in $E$ when $W=E\cap O$ for some open set $O\subseteq\mathbb{R}$. The previous tutorial noted that using relatively open sets gives the same cover criterion as using open sets in $\mathbb{R}$.

Here is how to translate between the two descriptions. If $\{O_\alpha\}$ is a family of ambient open sets covering $E$, then the sets $E\cap O_\alpha$ are relatively open in $E$, and they cover $E$: for each $x\in E$, some $O_\alpha$ contains $x$, so $x\in E\cap O_\alpha$. In the other direction, suppose relatively open sets $W_\alpha$ cover $E$. Write each one as $W_\alpha=E\cap O_\alpha$, where $O_\alpha$ is open in $\mathbb{R}$. Since $W_\alpha\subseteq O_\alpha$, the ambient open sets $O_\alpha$ cover $E$ as well. This translation lets us use the ambient open sets in the definition without losing the relative viewpoint.

Worked Example: Translating a Relative Cover

Take $E=[0,1]$, and let $O_1=(-1,3/5)$ and $O_2=(2/5,2)$. The relative sets are $$ W_1=E\cap O_1=[0,3/5), \qquad W_2=E\cap O_2=(2/5,1]. $$ The first equality follows because the points of $[0,1]$ that are also in $(-1,3/5)$ are exactly those in $[0,3/5)$. The second follows because the points of $[0,1]$ in $(2/5,2)$ are exactly those in $(2/5,1]$.

These relative sets cover $E$. If $x\in[0,1]$ and $x<3/5$, then $x\in W_1$; if $x\geq3/5$, then $2/5<x\leq1$, so $x\in W_2$. The ambient intervals extend outside $E$, while their intersections with $E$ are the relatively open cover members.

Replacing a Cover by Intervals

Open intervals are especially convenient in $\mathbb{R}$. A basic fact established earlier in the course is that every open subset of $\mathbb{R}$ is a union of open intervals. Applied point by point, this gives an interval version of any open cover: each point can be covered by a small open interval lying inside one of the original cover members.

Definition: A family $\mathcal{V}$ is a refinement of a family $\mathcal{U}$ if every member of $\mathcal{V}$ is contained in some member of $\mathcal{U}$. A refinement of an open cover of $E$ that still covers $E$ is called an open-cover refinement of that cover.

Theorem: Every open cover of $E\subseteq\mathbb{R}$ has an open-cover refinement consisting of open intervals.

Proof. Let $\{U_\alpha:\alpha\in A\}$ be an open cover of $E$. Consider the family $\mathcal{I}$ of all open intervals $I$ for which $I\subseteq U_\alpha$ for at least one $\alpha\in A$. Each member of $\mathcal{I}$ is contained in a member of the original cover, so $\mathcal{I}$ refines that cover.

It remains to show that $\mathcal{I}$ covers $E$. Take any $x\in E$. Since the original family covers $E$, there is an index $\alpha\in A$ such that $x\in U_\alpha$. The set $U_\alpha$ is open, so there exists $r>0$ such that $(x-r,x+r)\subseteq U_\alpha$. This interval belongs to $\mathcal{I}$ and contains $x$. Since this works for every $x\in E$, the interval family covers $E$. Thus $\mathcal{I}$ is the required refinement. If $E$ is empty, the empty family is an interval refinement that covers it. $\square$

The theorem produces a family of intervals, not necessarily a finite family. It also does not say that one interval works for all points of $E$. The interval is chosen locally: the cover member containing a point supplies room for an interval around that point. The benefit is that arguments about a general open cover can sometimes be reduced to arguments about intervals while retaining the original covering structure.

Combining Covers

Open covers can also be assembled from pieces. If a set is a union of several subsets, a cover of each subset can be pooled to cover the whole union. This elementary operation is useful when a set is naturally divided into regions and each region has its own convenient cover.

Theorem: Let $\{E_j:j\in J\}$ be a family of subsets of $\mathbb{R}$. For each $j\in J$, let $\mathcal{U}_j$ be a family of open subsets of $\mathbb{R}$ that covers $E_j$. Then $\bigcup_{j\in J}\mathcal{U}_j$, understood as the family containing all members of all the $\mathcal{U}_j$, covers $\bigcup_{j\in J}E_j$.

Proof. Take any $x\in\bigcup_{j\in J}E_j$. By the definition of union, there is some $j\in J$ such that $x\in E_j$. Since $\mathcal{U}_j$ covers $E_j$, there is a member $U\in\mathcal{U}_j$ with $x\in U$. That same $U$ is a member of $\bigcup_{j\in J}\mathcal{U}_j$, so $x$ belongs to the union of the pooled family. Every point of $\bigcup_{j\in J}E_j$ is covered, as required. All members of the pooled family are open because they are members of one of the open families $\mathcal{U}_j$. $\square$

The converse statement also holds in a simple form: if a family covers $E$, then it covers every subset of $E$. Indeed, if $A\subseteq E$, each point of $A$ is already a point of $E$, so it lies in some member of the cover. These operations do not assert compactness or produce a finite subcover. They describe how the basic covering relation behaves under restriction and union.

Why These Distinctions Matter

An open cover is a flexible description, not a special kind of partition. When checking that a proposed family is a cover, focus on whether every point of the target set lies in at least one open member. Do not mistakenly require that the members be disjoint, lie inside the target set, or have union exactly equal to it. Conversely, showing that the union contains many points is not enough unless it contains every point of the set being covered.

Refinements provide a second useful viewpoint. An interval refinement retains coverage while replacing broad open sets by smaller ones that are easier to locate around individual points. Covering a union provides a third: one may prove coverage region by region and then gather the families. In later applications of compactness, the challenge will be to extract finitely many members from a cover. These preparatory ideas help keep clear which family is covering which set and how its members are related.

Identify the set being covered.
Write down the target $E$ and check that each of its points lies in at least one family member.

Translate the viewpoint if useful.
Intersect ambient open sets with $E$ to obtain relatively open sets, or choose intervals inside the ambient cover members.

Use the structure of the target.
If $E$ is a union of subsets, covers of those subsets can be combined into a cover of $E$.

Check Your Understanding

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

Does an open cover of $E$ have to have union equal to $E$, or is containment enough?
Why does overlap between two members of an open cover cause no problem?
How can a cover by relatively open subsets of $E$ be translated into a cover by ambient open subsets of $\mathbb{R}$?
In the interval-refinement theorem, why can an interval be chosen around each $x\in E$ inside a member of the original cover?
If separate families cover each of two sets, what family covers their union?

What You'll Learn