The Open-Cover Proof for Closed Intervals

From an Open Cover to a Supremum

Compactness is defined using arbitrary open covers, so a proof for a real interval must start with an arbitrary cover rather than with a convenient choice of intervals. The key is to measure how far a finite part of the cover reaches from the left endpoint. The supremum of those reachable points cannot stop short: openness at the supremum lets a finite cover reach farther.

Throughout, \(\mathbb{R}\) has its standard topology, and \([a,b]\) has the subspace topology inherited from \(\mathbb{R}\). Thus a set \(U\subseteq[a,b]\) is open in \([a,b]\) precisely when \(U=O\cap[a,b]\) for some open set \(O\subseteq\mathbb{R}\). An open cover of \([a,b]\) is a family of sets open in this subspace whose union contains every point of \([a,b]\). A finite subcover is a finite selection from that family which still covers \([a,b]\). A space is compact if every open cover has a finite subcover.

The proof uses the least-upper-bound property of \(\mathbb{R}\): every nonempty subset of \(\mathbb{R}\) that is bounded above has a supremum. The degenerate case \(a=b\) is handled separately, and the main argument applies when \(a<b\).

The Supremum Proof

Proof. Let \(\mathcal{U}\) be an arbitrary open cover of \([a,b]\). If \(a=b\), some \(U\in\mathcal{U}\) contains \(a\); then the one-set family \(\{U\}\) covers \([a,b]=\{a\}\). Now assume \(a<b\).

Define \(E\) to be the set of right endpoints up to which a finite selection from \(\mathcal{U}\) covers the interval starting at \(a\):

The set \(E\) is nonempty. Indeed, choose a cover member \(U_a\) containing \(a\). Since \(U_a\) is open in the subspace, there is an open \(O\subseteq\mathbb{R}\) with \(U_a=O\cap[a,b]\). Since \(a\in O\) and \(O\) is open, some \(\varepsilon>0\) satisfies \((a-\varepsilon,a+\varepsilon)\subseteq O\). Choose \(t=\min(\varepsilon/2,(b-a)/2)\). Then \(t>0\), \(a+t\le b\), and \([a,a+t]\subseteq U_a\). Hence \(a+t\in E\). Also \(E\subseteq[a,b]\), so \(E\) is bounded above by \(b\). Its supremum \(c=\sup E\) therefore exists and satisfies \(a<c\le b\).

Choose a member \(U_c\in\mathcal{U}\) containing \(c\). As above, relative openness gives an \(\varepsilon>0\) such that \((c-\varepsilon,c+\varepsilon)\cap[a,b]\subseteq U_c\). Since \(c=\sup E\), there is \(x\in E\) with \(c-\varepsilon/2<x\le c\). Otherwise \(c-\varepsilon/2\) would be an upper bound for \(E\) smaller than its supremum. By the definition of \(E\), finitely many cover members cover \([a,x]\).

If \(c=b\), then \([x,c]\subseteq U_c\), because \(x>c-\varepsilon\) and \(c\in U_c\). The finite cover of \([a,x]\), together with \(U_c\), covers \([a,b]\), as required. If \(c<b\), let \(y=c+\min(\varepsilon/2,(b-c)/2)\). Then \(c<y\le b\), and \([x,y]\subseteq(c-\varepsilon,c+\varepsilon)\cap[a,b]\subseteq U_c\). Combining \(U_c\) with the finite cover of \([a,x]\) gives a finite cover of \([a,y]\). Thus \(y\in E\), contradicting \(y>c=\sup E\). The case \(c<b\) is impossible, so \(c=b\), and the preceding endpoint argument gives a finite subcover of \([a,b]\). Since \(\mathcal{U}\) was arbitrary, \([a,b]\) is compact. \(\square\)

The argument does not assume the cover members are intervals. It uses only that a cover member containing \(c\) contains a small real open interval around \(c\), after intersecting with \([a,b]\). That local interval is what carries a finite cover past a proposed stopping point.

Worked Cover Selections

Why the Argument Needs Care

Common pitfall. In the first example, it is tempting to select \(U_1\) and \(V\) and declare that they cover \([0,1]\). But \(U_1=(-1,\frac12)\), while \(V=(\frac34,2)\). The point \(\frac58\) is in neither: \(\frac58>\frac12\), so it is not in \(U_1\), and \(\frac58<\frac34\), so it is not in \(V\). The selection leaves a gap.

The correction is to compute the endpoint of the chosen \(U_n\) and check the overlap with \(V\). With \(n=4\), the endpoint is \(\frac45\), and \(\frac45>\frac34\). Therefore \(U_4\) reaches into \(V\), with no uncovered point between them. In a general proof there is no fixed pair of sets to guess; the supremum argument supplies a cover member at the limiting point and uses its openness to make the same overlap reasoning work.

The endpoint cases also matter. A cover member containing \(c\) provides a neighborhood on both sides in \(\mathbb{R}\), but only the part inside \([a,b]\) is relevant. If \(c=b\), there is no need to extend beyond \(b\): the finite cover already reaches the right endpoint. If \(c<b\), the neighborhood does extend a positive distance to the right inside the interval, which contradicts the definition of \(c\). Separating these cases avoids assuming the interval continues past its endpoint.