Compactness Through Open Covers
In a metric space, compactness is a condition on every open cover, not just on the shape of the space or on one convenient cover. The key question is whether, from each cover, one can select finitely many of its members and still cover the whole space. This distinction matters: a cover can have infinitely many members and nevertheless contain a finite subcover.
Let \((X,d)\) be a metric space, with the topology consisting of unions of open metric balls. An open cover is a family of open sets whose union contains every point of \(X\). A subcover is a selection from that family that still covers \(X\). Families may be indexed, so different indices can even refer to the same open set.
The definition does not say that the given cover has finitely many members. It says that a finite selection suffices, even when the original cover is infinite. Nor is it enough that a space has some finite open cover: every space is covered by the one open set \(X\), so that fact alone gives no information about other covers.
Every Finite Metric Space Is Compact
Proof. Let \(X\) be a finite metric space and let \((U_\alpha)_{\alpha\in J}\) be an open cover of \(X\). If \(X=\varnothing\), the empty subfamily covers it. Otherwise, list its points as \(x_1,\ldots,x_m\), where \(m\) is a positive integer. For each \(i\in\{1,\ldots,m\}\), the cover property gives an index \(\alpha_i\in J\) with \(x_i\in U_{\alpha_i}\). The subfamily indexed by \(\{\alpha_1,\ldots,\alpha_m\}\) is finite and covers every listed point, hence all of \(X\). Thus every open cover has a finite subcover, as required. \(\square\)
Worked Example: A Three-Point Metric Space
Let \(X=\{a,b,c\}\) have the discrete metric \(d(x,y)=0\) when \(x=y\) and \(d(x,y)=1\) when \(x\ne y\). This is a metric: if \(x\ne z\), then \(d(x,z)=1\), and at least one of \(d(x,y)\) or \(d(y,z)\) is \(1\), so \(d(x,z)\le d(x,y)+d(y,z)\); if \(x=z\), the triangle inequality holds because its left side is \(0\).
The ball \(B_d(x,1/2)\) is \(\{x\}\), since \(d(x,x)=0<1/2\), whereas \(d(x,y)=1\not<1/2\) for \(y\ne x\). Thus each singleton is open, and every subset of \(X\), being a union of singletons, is open.
Consider the open cover \(U_1=\{a,b\}\), \(U_2=\{b,c\}\), \(U_3=\{a,c\}\). It covers \(X\), and in fact the two members \(U_1,U_2\) already cover it: \(a\in U_1\), \(b\in U_1\cap U_2\), and \(c\in U_2\). So \(\{U_1,U_2\}\) is a finite subcover. The theorem proves more: for any open cover, even one with many redundant indexed members, choosing a cover member for each of \(a,b,c\) produces a subcover with at most three members.
Check: \(U_1\cup U_2=\{a,b\}\cup\{b,c\}=\{a,b,c\}=X\), so the selected pair really does cover every point.
The Open Unit Interval Is Not Compact
To prove that a space is not compact, it is enough to give one open cover for which every finite selection fails to cover the space. In a subspace such as \((0,1)\subseteq\mathbb R\), “open” means open in the subspace topology: a set has the form \(V\cap(0,1)\) for some open \(V\subseteq\mathbb R\). In particular, an interval open in \(\mathbb R\) and contained in \((0,1)\) is open in the subspace.
Worked Example: An Open Cover With No Finite Subcover
Give \((0,1)\), with the metric inherited from the standard metric on \(\mathbb R\), the cover \(U_n=(1/n,1)\) for integers \(n\ge2\). Each \(U_n\) is open in \((0,1)\).
To check that these sets cover, take any \(x\in(0,1)\). Since \(1/x\) is a real number, the Archimedean property gives an integer \(n\ge2\) with \(n>1/x\). Multiplying by the positive number \(x/n\), or equivalently taking reciprocals of positive numbers, gives \(1/n<x\). Also \(x<1\), so \(x\in U_n\). Hence \(\bigcup_{n\ge2}U_n=(0,1)\).
Now select finitely many members, with indices \(n_1,\ldots,n_k\). There is at least one selected member if the selection is to cover the nonempty interval. Let \(N=\max\{n_1,\ldots,n_k\}\). For every selected \(n_i\le N\), we have \(1/n_i\ge1/N\), so \(U_{n_i}\subseteq U_N\). Therefore the union of the selected sets is contained in \(U_N=(1/N,1)\). The point \(x=1/(2N)\) satisfies \(0<1/(2N)<1/N<1\), so it belongs to \((0,1)\) but not to \(U_N\) or to the selected union. Thus no finite subfamily covers \((0,1)\), and \((0,1)\) is not compact.
Check: the strict inequality in \(U_N=(1/N,1)\) matters: \(1/(2N)<1/N\), so the point used to defeat the finite selection is indeed excluded.
This argument uses an expanding family of intervals: as \(n\) increases, \(1/n\) decreases, and \(U_n\) grows toward the missing left endpoint \(0\). Any finite selection has a largest index and therefore still misses points sufficiently close to \(0\). The example verifies noncompactness directly from the open-cover definition; no conclusion is being drawn merely from the fact that \((0,1)\) is not closed in \(\mathbb R\).
A Compact Interval Can Have an Infinite Cover
The contrasting example \([0,1]\) illustrates both the strength of the definition and the distinction between a cover and a finite subcover. We first prove compactness directly, using the least-upper-bound property of \(\mathbb R\): every nonempty subset of \(\mathbb R\) that is bounded above has a supremum.
Proof. Let \((U_\alpha)_{\alpha\in J}\) be an open cover of \([0,1]\), with each \(U_\alpha\) open in the subspace topology. Define \(S\) to be the set of \(t\in[0,1]\) such that \([0,t]\) is covered by finitely many members of this cover. The cover has a member \(U_\alpha\) containing \(0\). By relative openness, there is \(\delta>0\) such that \((-\delta,\delta)\cap[0,1]\subseteq U_\alpha\). Taking \(t=\min(\delta/2,1/2)>0\), we have \([0,t]\subseteq U_\alpha\). Thus \(S\ne\varnothing\). Since \(S\subseteq[0,1]\), it is bounded above, so let \(s=\sup S\).
We show that \(s=1\). Suppose instead that \(s<1\), and choose a cover member \(U_\beta\) containing \(s\). Relative openness gives \(\delta>0\) such that \((s-\delta,s+\delta)\cap[0,1]\subseteq U_\beta\). Because \(s=\sup S\), there is \(t\in S\) with \(t>s-\delta/2\); otherwise \(s-\delta/2\) would be an upper bound for \(S\) smaller than \(s\). Choose \(r=s+\min(\delta/2,(1-s)/2)\). Then \(s<r\le1\), and \(t\le s\). Every point of \([t,r]\) lies in \((s-\delta,s+\delta)\cap[0,1]\): the left endpoint satisfies \(t>s-\delta/2>s-\delta\), and the right endpoint satisfies \(r<s+\delta\). A finite cover of \([0,t]\), which exists because \(t\in S\), together with \(U_\beta\) therefore covers \([0,r]\). This means \(r\in S\), contradicting \(r>s=\sup S\). Hence \(s=1\).
Finally, choose a cover member \(U_\gamma\) containing \(1\). There is \(\delta>0\) such that \((1-\delta,1+\delta)\cap[0,1]\subseteq U_\gamma\). Since \(\sup S=1\), there is \(t\in S\) with \(t>1-\delta\). A finite collection covers \([0,t]\), and \(U_\gamma\) covers \([t,1]\). Together they give a finite subcover of \([0,1]\). Since the original cover was arbitrary, \([0,1]\) is compact. \(\square\)
Worked Example: An Infinite Cover With a Finite Subcover
For each \(n\in\mathbb N\), let \(V_n=(-1/n,1+1/n)\), an open subset of \(\mathbb R\), and consider these sets as a cover of \([0,1]\). They are distinct: if \(n<m\), then \(-1/n<-1/m\), so their left endpoints differ. Each contains \([0,1]\), because every \(x\in[0,1]\) satisfies \(-1/n<x<1+1/n\). Thus the family has infinitely many distinct members and covers the interval.
Nevertheless, \(V_1=(-1,2)\) alone contains all of \([0,1]\). Therefore \(\{V_1\}\) is a finite subcover. The original cover is infinite; the selected subcover is finite. These statements are compatible, and this is exactly the distinction the definition requires.
Check: the endpoint inequalities are strict: \(-1<0\le x\le1<2\) for every \(x\in[0,1]\), so \(V_1\) does cover both endpoints as well as all interior points.
Common Pitfall: Counting the Given Cover
A common mistake is to decide compactness by counting the members of one given cover. For the family \((V_n)_{n\in\mathbb N}\) above, the incorrect calculation is: “There are infinitely many \(V_n\), so \([0,1]\) is not compact.” That conclusion does not follow. The correction is to test whether a finite selection covers: here \(\{V_1\}\) works, since \([0,1]\subseteq(-1,2)\). By contrast, the cover \(U_n=(1/n,1)\) of \((0,1)\) has no finite subcover, as the point \(1/(2N)\) shows for every finite selection with largest index \(N\).
These examples capture the practical method. To prove compactness, begin with an arbitrary open cover and show how to extract finitely many members. To disprove compactness, exhibit one open cover and show that every finite selection misses a point. A finite cover is certainly already a finite subcover of itself; but compactness requires the finite-subcover conclusion for covers that may have infinitely many members.
Check Your Understanding
For each question, use the open-cover definition and keep track of whether the selected family is finite.
- Why does the one-member cover \(\{X\}\) fail to establish that an arbitrary metric space \(X\) is compact?
- For the cover \(U_n=(1/n,1)\), \(n\ge2\), what point is missed by a finite selection whose largest index is \(N\)? Verify both inequalities needed for membership in \((0,1)\) and exclusion from \(U_N\).
- Explain why the finite metric space theorem also covers a space with no points.
- For \(V_n=(-1/n,1+1/n)\), verify directly that \(V_1\) covers \([0,1]\), although the family has infinitely many distinct members.
- State the difference between an open cover being finite and an open cover having a finite subcover.