A Different Obstruction to Compactness
The previous tutorial showed why boundedness is necessary for compactness: points of an unbounded set can escape every fixed bounded region. Boundedness does not, however, guarantee compactness. A bounded set can still omit a point that its members approach arbitrarily closely. We will see how this missing point produces an open cover with no finite subcover.
Recall that a set \(E\subseteq\mathbb{R}\) is closed if it contains all its limit points. Equivalently, \(E\) is closed if and only if \(\overline{E}=E\), by the earlier characterization of closure. Thus, when \(E\) is not closed, there is a point \(p\in\overline{E}\) with \(p\notin E\). The Neighborhood Characterization of Closure says that every open ball centered at \(p\) meets \(E\). Since \(p\) itself is not in \(E\), those intersections provide points of \(E\) close to \(p\) but never equal to it.
A Cover Built Around a Missing Point
For each positive integer \(n\), take all real numbers whose distance from \(p\) is greater than \(1/n\). These sets omit a smaller and smaller neighborhood of \(p\) as \(n\) increases.
Each \(U_n\) is open: it is the complement of a closed interval. The sets are nested, with \(U_n\subseteq U_{n+1}\), because \(1/(n+1)<1/n\). Together, they cover \(E\). Indeed, for any \(x\in E\), we have \(x\neq p\), so \(|x-p|>0\). By the Archimedean property, there is a positive integer \(n\) such that \(n>1/|x-p|\). It follows that \(1/n<|x-p|\), and hence \(x\in U_n\).
Proof. We have already shown that each \(U_n\) is open and that the family covers \(E\). Consider any finite selection \(U_{n_1},\ldots,U_{n_m}\), and let \(N=\max\{n_1,\ldots,n_m\}\). Since the sets are nested, their union is \(U_N\). Because \(p\in\overline{E}\), the Neighborhood Characterization of Closure gives a point \(x\in E\cap(p-1/N,p+1/N)\). Since \(p\notin E\), this point satisfies \(x\neq p\). In particular, \(|x-p|<1/N\), so \(x\notin U_N\). The finite selection therefore does not cover \(E\). This applies to every finite selection, proving that there is no finite subcover. \(\square\)
The key is that all the sets in a finite selection leave out at least one common neighborhood around \(p\). The full cover succeeds because the omitted neighborhoods shrink as \(n\) increases; no single finite selection shrinks them enough to include every point of \(E\).
Worked Example: The Open Interval \((0,1)\)
Take \(E=(0,1)\) and \(p=0\). Then \(p\in\overline{E}\setminus E\). The corresponding cover is \(\displaystyle U_n=(-\infty,-1/n)\cup(1/n,\infty)\). It covers \(E\): if \(x\in(0,1)\), choose a positive integer \(n\) with \(n>1/x\). Then \(1/n<x\), so \(x\in U_n\).
Now choose any finite collection and let \(N\) be its largest index. Its union is \(U_N\). The point \(x=1/(2N)\) belongs to \((0,1)\), since \(0<1/(2N)\leq1/2<1\). But \(0<x<1/N\), so \(x\notin U_N\). Thus no finite selection covers \((0,1)\), and this open cover proves that the interval is not compact.
Why the Obstruction Applies Even to Bounded Sets
The cover just constructed consists of unbounded open sets. That is enough to show that a nonclosed set is not compact, but a bounded set allows a more pointed version: we can make every member of the cover bounded as well. Then the failure of a finite subcover cannot be attributed to unboundedness of the set or to unbounded members of the cover.
Proof. Since \(E\) is nonclosed, choose \(p\in\overline{E}\setminus E\). Since \(E\) is bounded, choose \(R>0\) such that \(E\subseteq(-R,R)\). Define \(V_n=U_n\cap(-R-1,R+1)\), where \(U_n\) is the open set constructed above. Each \(V_n\) is open as the intersection of two open sets, and it is bounded because \(V_n\subseteq(-R-1,R+1)\).
The sets \(V_n\) cover \(E\): the family \(\{U_n\}\) covers \(E\), and every point of \(E\) also lies in \((-R-1,R+1)\). For any finite selection, let \(N\) be its largest index. The selected sets have union \(V_N\), since both component families are nested. Because \(p\in\overline{E}\), choose \(x\in E\cap(p-1/N,p+1/N)\). This point belongs to \((-R,R)\), but it does not belong to \(U_N\), and therefore does not belong to \(V_N\). No finite selection covers \(E\). \(\square\)
In particular, every bounded nonclosed subset of \(\mathbb{R}\) is noncompact. The Heine–Borel Theorem, established earlier in this module, states that a subset of \(\mathbb{R}\) is compact if and only if it is closed and bounded. The theorem above gives a direct open-cover explanation for why closedness cannot be omitted from that criterion.
Worked Example: A Reciprocal Sequence Without Its Limit
Let \(E=\{1/k:k\geq1\}\), so \(E\) is bounded and \(0\in\overline{E}\setminus E\). Consider the bounded open sets \(\displaystyle V_n=(-2,-1/n)\cup(1/n,2)\). They are open and bounded. They cover \(E\): for each \(1/k\in E\), choose \(n=k+1\). Then \(1/(k+1)<1/k<2\), so \(1/k\in(1/n,2)\subseteq V_n\).
For a finite selection, let \(N\) be its largest index. The union of the selected sets is \(V_N\). The point \(1/(N+1)\) belongs to \(E\), and \(\displaystyle 0<\frac{1}{N+1}<\frac{1}{N}<2\). Thus \(1/(N+1)\notin V_N\). No finite selection covers \(E\), which shows directly that this bounded set is not compact.
Worked Example: Removing an Interior Point from a Closed Interval
Let \(E=[0,1]\setminus\{1/2\}\), and put \(p=1/2\). The point \(p\) lies in \(\overline{E}\) but not in \(E\). For each positive integer \(n\), use the bounded open set \(\displaystyle V_n=(-2,\tfrac12-\tfrac1n)\cup(\tfrac12+\tfrac1n,2)\). Every point \(x\in E\) lies in \((-2,2)\) and has \(|x-1/2|>0\). Choose \(n\) so large that \(1/n<|x-1/2|\). Then \(x\in V_n\), so these sets cover \(E\).
Given a finite selection with largest index \(N\), its union is \(V_N\). The point \(x=\tfrac12+\tfrac{1}{2(N+1)}\) belongs to \(E\): it is greater than \(1/2\) and at most \(3/4\), hence is in \([0,1]\) and is not the removed point. Moreover, \(\displaystyle 0<x-\tfrac12=\frac{1}{2(N+1)}<\frac{1}{N}\). Therefore \(x\notin V_N\). This finite selection misses a point of \(E\), as must every finite selection.
What Closedness Controls
Closedness is not a restriction on how far points can travel; boundedness addresses that issue. Instead, closedness ensures that if points of a set approach a real number, the limit is not missing from the set. When a point \(p\) is missing but belongs to the closure, the cover above detects points of \(E\) that come arbitrarily close to \(p\). A finite selection can only remove one fixed neighborhood of \(p\), leaving some of those points uncovered.
A common pitfall is to suppose that a set must be closed because it contains many points close to a particular number. The relevant question is whether it contains the limit point itself. For instance, \(\{1/n:n\geq1\}\) has points arbitrarily close to \(0\), but it does not contain \(0\); its closure therefore contains a point missing from the set. Conversely, merely omitting an arbitrary real number does not show that a set is nonclosed. The omitted point must belong to the closure.
This argument also clarifies why closedness is necessary even when all points stay in a bounded region. Boundedness prevents escape to infinity, but it does not prevent approach to a finite point outside the set. Compactness requires both protections: boundedness rules out escape along the real line, and closedness rules out a missing limit point.
For a nonclosed set \(E\), choose \(p\in\overline{E}\setminus E\).
Use open sets that omit a neighborhood of \(p\), with the omitted neighborhoods shrinking as the index increases.
The largest index determines the smallest omitted neighborhood; closure guarantees that this neighborhood still contains a point of \(E\).
Check Your Understanding
Use the missing-point cover construction to answer the following questions.
- Why is \(p\notin E\) important when showing that the sets \(U_n\) cover \(E\)?
- For a finite selection \(U_{n_1},\ldots,U_{n_m}\), why is the union equal to \(U_N\) when \(N=\max\{n_1,\ldots,n_m\}\)?
- Which property of \(p\) guarantees that \(E\) has a point inside \((p-1/N,p+1/N)\)?
- Why can the cover for a bounded nonclosed set be chosen to consist of bounded open sets?
- Give an example of a real number omitted by a set that does not, by itself, prove the set is nonclosed. What additional condition would be needed?