Compactness as the Finite-Subcover Property

Compactness As a Statement About Covers

Compactness turns a potentially large collection of local sets into a finite selection that still covers the whole space. Its definition applies to open covers, not to arbitrary collections of subsets: whether a set is open depends on the topology. To use the definition reliably, it helps to make explicit both the coverage condition and the finite-selection condition.

Let $(X,\mathcal{T})$ be a topological space. An indexed family $(U_i)_{i\in I}$ means that for each index $i$ in a set $I$, a set $U_i$ has been specified. Different indices are allowed to specify the same set. The family is an open cover of $X$ if every $U_i$ belongs to $\mathcal{T}$ and every point of $X$ belongs to at least one of them.

Definition: An indexed family $(U_i)_{i\in I}$ is an open cover of $X$ when $U_i\in\mathcal{T}$ for every $i\in I$ and $$ \bigcup_{i\in I} U_i=X. $$ A finite subcover is a choice of a finite set of indices $J\subseteq I$ such that $\bigcup_{j\in J}U_j=X$. The space $(X,\mathcal{T})$ is compact if every open cover of $X$ has a finite subcover.

In quantifiers, the coverage requirement says that for each point there is some index whose set contains it. Compactness then says there is one finite set of indices that works for all points:

\forall I\ \forall (U_i)_{i\in I},\ \left[ \left(\forall i\in I,\ U_i\in\mathcal{T}\right) \ \text{and}\ \left(\forall x\in X\ \exists i\in I:\ x\in U_i\right) \right] \Longrightarrow \left[ \exists J\subseteq I\ \left(J\text{ finite}\right)\ \forall x\in X\ \exists j\in J:\ x\in U_j \right].

The order of the quantifiers matters. First an open cover is given; then compactness asserts that some finite choice of its indices works for every point. It does not assert that the same finite choice works for every possible cover.

The empty space is compact under this definition. The empty family, indexed by $I=\varnothing$, covers $X=\varnothing$, and the empty set of indices is a finite subcover. If $X\ne\varnothing$, the empty family is not a cover, since there is no index whose set contains any point of $X$. A finite subcover means finitely many members of the cover; those members themselves need not be finite sets.

Indexed Families and Quantified Tests

Some covers are naturally listed by indices, such as the positive integers; others are given simply as a collection of open sets. These viewpoints give the same definition. The indexed form is useful when the sets repeat or when their labels matter, while the collection form can be more concise.

Proposition: A space is compact if and only if every collection of open sets whose union is $X$ has a finite subcollection whose union is $X$.

Proof. Suppose $X$ is compact and $\mathcal{C}$ is a collection of open sets with union $X$. Index the family by the members of $\mathcal{C}$, so that the set indexed by $C\in\mathcal{C}$ is $C$ itself. Compactness gives a finite set of indices, and the corresponding members form a finite subcollection covering $X$. Conversely, suppose every open cover given as a collection has a finite subcover. For an indexed open cover $(U_i)_{i\in I}$, its range $\{U_i:i\in I\}$ is a collection of open sets with union $X$. A finite subcollection of this range covers $X$. For each member of that finite subcollection, choose one index that specifies it; the resulting set of indices is finite and gives a finite subcover of the indexed family. This also works for the empty collection: it covers only when $X=\varnothing$, in which case the empty subcollection suffices. $\square$

Negating the compactness condition gives a useful way to describe failure. A space is not compact exactly when there is an open cover for which every finite choice misses at least one point. The quantifiers are important: the missed point may depend on the finite choice.

Theorem: A space $(X,\mathcal{T})$ is not compact if and only if there exist an index set $I$ and an open cover $(U_i)_{i\in I}$ of $X$ such that $$ \forall J\subseteq I\ \left(J\text{ finite}\right)\ \exists x\in X:\ x\notin\bigcup_{j\in J}U_j. $$

Proof. If $X$ is not compact, the definition supplies an open cover with no finite subcover. For every finite $J\subseteq I$, the sets indexed by $J$ therefore fail to cover $X$. Failure to cover means that there is some $x\in X$ not in their union, as required. Conversely, suppose an open cover satisfies the displayed condition. No finite $J\subseteq I$ covers $X$, because for each such $J$ the condition gives a point outside the union. Thus this cover has no finite subcover, so $X$ is not compact. If $X=\varnothing$, the condition cannot hold: for $J=\varnothing$, it would require a point $x\in\varnothing$. This is consistent with the fact that the empty space is compact. $\square$

Working with Particular Covers

Worked Example: An Infinite Cover with a Finite Subcover

Give $[0,1]$ the subspace topology inherited from $\mathbb{R}$ with its standard topology. Consider the family $$ U_n=(-1,1-1/n)\cap[0,1]\quad(n\ge2),\qquad V=(1/2,2)\cap[0,1]. $$ Each set is open in the subspace topology: it is the intersection of $[0,1]$ with an open interval in $\mathbb{R}$.

This family covers $[0,1]$. The point $1$ belongs to $V$. If $x\in[0,1)$, choose an integer $n\ge2$ with $n>1/(1-x)$. Then $1/n<1-x$, so $x<1-1/n$, and therefore $x\in U_n$.

In fact, just $U_3$ and $V$ cover. Since $1-1/3=2/3$, we have $U_3=[0,2/3)$. If $x\in[0,1]$ and $x<2/3$, then $x\in U_3$. If $x\ge2/3$, then $x>1/2$, so $x\in V$. Thus $\{U_3,V\}$ is a finite subcover. This verifies a finite-subcover claim for this particular cover; it does not by itself establish that every open cover of $[0,1]$ has a finite subcover.

Worked Example: A Finite Cover of the Real Line

Give $\mathbb{R}$ its standard topology and consider $U=(-\infty,1)$ and $V=(0,\infty)$. Both sets are open intervals or open rays, so both are open. If $x\le0$, then $x<1$, giving $x\in U$. If $x>0$, then $x\in V$. These cases include every real number, so $U\cup V=\mathbb{R}$.

The family $\{U,V\}$ is already finite; it is a finite subcover of itself. In the quantified formulation, take the index set $I=\{1,2\}$, with $U_1=U$ and $U_2=V$, and choose $J=I$. The example illustrates the definition directly, but a single finite cover says nothing about whether all open covers of $\mathbb{R}$ have finite subcovers.

Worked Example: An Open Cover with No Finite Subcover

Give $(0,1)$ the subspace topology inherited from the standard topology on $\mathbb{R}$. For each $n\in\mathbb{N}$, set $U_n=(1/n,1)$, interpreted as a subset of $(0,1)$. Each $U_n$ is open in this subspace because $U_n=(1/n,1)\cap(0,1)$, an intersection of open subsets of $\mathbb{R}$.

The family covers $(0,1)$. Given $x\in(0,1)$, choose $n\in\mathbb{N}$ with $n>1/x$. Then $1/n<x<1$, so $x\in U_n$. Now let $J\subseteq\mathbb{N}$ be finite and nonempty, and let $N$ be its largest index. If $n\in J$, then $n\le N$, hence $1/n\ge1/N$ and $U_n\subseteq U_N$. Therefore $$ \bigcup_{n\in J}U_n=U_N=(1/N,1). $$ The point $x=1/(2N)$ lies in $(0,1)$, since $N\ge1$, but $1/(2N)<1/N$, so it does not belong to $U_N$. If $J=\varnothing$, its union is empty and also fails to cover $(0,1)$. Thus no finite subfamily covers $(0,1)$, and this cover witnesses that $(0,1)$ is not compact.

How the Topology Affects Compactness

Since compactness quantifies over open covers, changing the topology changes which families must be tested. One useful comparison follows directly from the definition.

Proposition: Suppose $\mathcal{T}_1\subseteq\mathcal{T}_2$ are topologies on the same set $X$. If $(X,\mathcal{T}_2)$ is compact, then $(X,\mathcal{T}_1)$ is compact.

Proof. Let $(U_i)_{i\in I}$ be any open cover of $(X,\mathcal{T}_1)$. Since $\mathcal{T}_1\subseteq\mathcal{T}_2$, every $U_i$ is also open in $\mathcal{T}_2$, and the family still covers the same set $X$. Compactness of $(X,\mathcal{T}_2)$ gives a finite set $J\subseteq I$ such that $\bigcup_{j\in J}U_j=X$. The same finite family is a subcover in $\mathcal{T}_1$. Since the $\mathcal{T}_1$-open cover was arbitrary, $(X,\mathcal{T}_1)$ is compact. $\square$

The argument uses only that every open set for the coarser topology $\mathcal{T}_1$ is open for the finer topology $\mathcal{T}_2$. It does not establish a converse: a topology with fewer open sets has fewer open covers to test, so compactness in that topology alone does not yield a finite subcover for covers that are open only in a finer topology.

Common pitfall. A finite subcover is a finite number of sets, not a finite number of points. For instance, in $\mathbb{R}$ with its standard topology, $U_n=(-n,n)$, $n\in\mathbb{N}$, is an open cover: for each $x\in\mathbb{R}$, choose $n>|x|$, and then $-n<x<n$. A mistaken argument might say that $U_1$ and $U_2$ suffice because each contains many real numbers. But $U_1\cup U_2=(-2,2)$, which misses $3$. More generally, for any nonempty finite index set $J$, let $M$ be its largest index. Since $U_n\subseteq U_M$ whenever $n\in J$, the finite union is $(-M,M)$, and it misses $M+1$. The full cover therefore has no finite subcover.

Check Your Understanding

For each question, identify the open cover first, then state what finite selection would have to accomplish.

Write the definition of compactness using quantifiers over points $x$, indices $i$, and a finite set $J$.
Why does the empty space count as compact under the finite-subcover definition?
For the cover of $(0,1)$ by $U_n=(1/n,1)$, what point is missed by the finite subfamily with largest index $N$?
Explain why the finite subcover in the example on $[0,1]$ does not prove that $[0,1]$ is compact.
If $\mathcal{T}_1\subseteq\mathcal{T}_2$, which direction of implication between compactness for the two topologies follows from the definition?

Compactness as the Finite-Subcover Property

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