The Shape of a Noncompactness Proof
Compactness is defined by what must happen for every open cover: some finite collection of members covers the space. To disprove compactness, it is enough to find one open cover for which no finite collection does the job. The useful work is therefore in designing the cover. A successful construction has two separate checks: every point must be covered, and every finite selection must leave at least one point uncovered.
Throughout, a topological space is written \((X,\mathcal{T})\). An open cover of \(X\) is a family of open subsets of \(X\) whose union is \(X\). A finite subcover is a finite subfamily whose union is still \(X\). Compactness means that every open cover has a finite subcover. When \(A\subseteq\mathbb{R}\), we give \(A\) the subspace topology: its open sets are the intersections \(A\cap U\), where \(U\) is open in \(\mathbb{R}\).
This is exactly the negation of the definition of compactness: “for every open cover, there is a finite subcover” becomes “there is an open cover for which no finite subcover exists.” The quantifiers matter. Showing that one particular cover has a finite subcover says nothing about whether other covers do; showing that a family is infinite does not by itself show that it lacks a finite subcover.
Expanding Intervals Detect Unboundedness
Proof. For each \(n\in\mathbb{N}\), define \(U_n=A\cap(-n,n)\). The interval \((-n,n)\) is open in \(\mathbb{R}\), so \(U_n\) is open in the subspace \(A\). Every \(a\in A\) is a real number, so there is a positive integer \(n>|a|\); hence \(a\in U_n\). Thus the family \(\{U_n:n\in\mathbb{N}\}\) covers \(A\). Now take any finite subfamily. If it is nonempty, let \(N\) be the largest index used. Since the intervals are nested, its union is \(A\cap(-N,N)\). Because \(A\) is unbounded, it is not contained in \((-N,N)\), so some \(a\in A\) satisfies \(|a|\geq N\) and is not in this union. If the finite subfamily is empty, its union is empty and cannot cover the nonempty unbounded set \(A\). Therefore no finite subfamily covers \(A\), proving that \(A\) is not compact. \(\square\)
The construction works because its sets grow as the index grows. Each point eventually enters one of them, while any finite collection has a largest index and therefore remains inside one bounded interval. Unboundedness guarantees a point outside that interval. Notice that the proof does not require \(A\) to be closed or to contain any particular endpoint.
Worked Example: The Nonnegative Real Line
Let \(A=[0,\infty)\) with the subspace topology inherited from \(\mathbb{R}\). For each \(n\in\mathbb{N}\), set \(U_n=A\cap(-n,n)\). Since \(n>0\), this is \(U_n=[0,n)\), open in \(A\): it is the intersection of \(A\) with the open real interval \((-n,n)\).
To check that the family covers \(A\), take any \(x\in[0,\infty)\). Choose \(n\in\mathbb{N}\) with \(n>x\). Then \(0\leq x<n\), so \(x\in[0,n)=U_n\). Now take a nonempty finite subfamily and let \(N\) be its largest index. Its union is \([0,N)\), because \(U_n\subseteq U_N\) whenever \(n\leq N\). The point \(N+1\) belongs to \(A\) but not to \([0,N)\). Thus no finite subfamily covers \(A\).
Check: the proposed sets really cover every nonnegative real number, and the point \(N+1\) witnesses failure for every finite selection with largest index \(N\). Therefore \(A\) is not compact.
Worked Example: The Integers
Give \(A=\mathbb{Z}\) the subspace topology from \(\mathbb{R}\), and define \(U_n=\mathbb{Z}\cap(-n,n)\) for \(n\in\mathbb{N}\). Each \(U_n\) is open in \(\mathbb{Z}\) by the definition of the subspace topology. For any integer \(k\), choose \(n>|k|\); then \(k\in U_n\). So \(\{U_n:n\in\mathbb{N}\}\) is an open cover.
For a nonempty finite selection, let \(N\) be its largest index. The selected sets are nested, so their union is \(U_N=\{k\in\mathbb{Z}:-N<k<N\}\). The integer \(N\) is not in \(U_N\), since membership would require \(N<N\). Thus the finite selection misses a point of \(\mathbb{Z}\). The empty selection also fails to cover \(\mathbb{Z}\), which is nonempty. Hence \(\mathbb{Z}\) is not compact.
Check: the endpoints are excluded in \((-N,N)\), so \(N\) is indeed missed. In particular, replacing a strict inequality by a weak one in the membership test would give the wrong conclusion about this witness.
Isolated Points Can Supply a Different Cover
A point \(x\in X\) is isolated if its singleton \(\{x\}\) is open in \(X\). When every point of an infinite space is isolated, the family of all singletons is an open cover. A finite selection from that family covers only finitely many points, so it cannot cover the whole infinite space. This gives a direct construction that does not depend on unboundedness or on intervals.
Proof. Let \(X\) be an infinite discrete space. Each singleton \(\{x\}\), for \(x\in X\), is open, so \(\{\{x\}:x\in X\}\) is an open cover of \(X\). A finite subfamily consists of singletons for only finitely many points. Since \(X\) is infinite, there is a point of \(X\) not among those finitely many points, and that point is not covered. The cover therefore has no finite subcover, so \(X\) is not compact. \(\square\)
Worked Example: A Bounded Subspace with Isolated Points
Consider \(A=\{1/n:n\in\mathbb{N}\}\subseteq\mathbb{R}\), with the subspace topology. This set is bounded, so the expanding-interval argument by itself does not establish noncompactness. Instead, we construct an open cover using its isolated points.
For \(n=1\), let \(V_1=A\cap(1/2,2)\). The only member of \(A\) in \((1/2,2)\) is \(1\), so \(V_1=\{1\}\). For each integer \(n\geq2\), let \(V_n=A\cap(1/(n+1),1/(n-1))\). The sequence \(1/k\) decreases as \(k\) increases. If \(k\leq n-1\), then \(1/k\geq1/(n-1)\), so \(1/k\) is not in this open interval. If \(k\geq n+1\), then \(1/k\leq1/(n+1)\), so it is not in the interval either. The remaining index is \(k=n\), and \(1/(n+1)<1/n<1/(n-1)\). Therefore \(V_n=\{1/n\}\).
Each \(V_n\) is open in \(A\), since it is the intersection of \(A\) with an open interval in \(\mathbb{R}\). The family \(\{V_n:n\in\mathbb{N}\}\) covers \(A\), because it contains \(\{1/n\}\) for every \(n\). Any finite subfamily covers only the finitely many points whose indices occur in that subfamily. Since \(A\) contains \(1/n\) for every positive integer \(n\), some point remains uncovered. Hence \(A\) is not compact, despite being bounded.
Check: for \(n=2\), the interval is \((1/3,1)\), which contains \(1/2\) and excludes both \(1\) and every \(1/k\) with \(k\geq3\). Thus \(V_2=\{1/2\}\), as required by the general calculation.
Common Pitfalls in Cover Construction
A proposed family can fail in either of the two essential checks. First, it may not cover the space. For example, the family of singleton sets \(\{\{x\}:x\in A\}\) is not automatically an open cover in the subspace topology. In \(A=(0,1)\) with its usual subspace topology, \(\{1/2\}\) is not open: every open real interval containing \(1/2\) contains other points of \((0,1)\). Thus the family of singletons is not a family of open sets in this space. The correction is to verify openness in the stated topology, or to choose neighborhoods that are open in the subspace, as in the preceding example.
Second, an open cover may have a finite subcover even if it has infinitely many members. For instance, let \(A=[0,1]\) and \(U_n=A\cap(-n,n)\). These sets form an open cover, but \(U_2=[0,1]\), so the single set \(U_2\) is already a finite subcover. The unboundedness hypothesis in the theorem prevents this: a finite selection of the expanding intervals cannot reach beyond its largest index. For bounded sets, a different construction may be needed.
Check Your Understanding
For each question, check separately that the proposed sets are open and that the finite selections fail to cover.
- For an unbounded \(A\subseteq\mathbb{R}\), explain why \(A\cap(-n,n)\) is open in \(A\) for every \(n\in\mathbb{N}\).
- In the cover \(U_n=[0,n)\) of \([0,\infty)\), what point is missed by a finite subfamily whose largest index is \(N\)?
- Why does the family of singleton subsets prove that an infinite discrete space is not compact, but not prove the same statement for every infinite topological space?
- For \(A=\{1/n:n\in\mathbb{N}\}\), verify directly that \(A\cap(1/4,1/2)=\{1/3\}\).
- Why does the expanding-interval cover fail to prove that \([0,1]\) is noncompact?