From Compactness to a Convergent Subsequence
The Heine-Borel Theorem connects compactness in \(\mathbb R\) with closedness and boundedness. The Bolzano-Weierstrass Theorem gives a related sequence-level conclusion: bounded real sequences have convergent subsequences. Here we examine the mechanism that turns compactness itself into a convergent subsequence.
The key is to find a point whose every neighborhood contains terms with infinitely many indices. Once such a point is found, choosing increasingly accurate terms at increasingly large indices produces a convergent subsequence. The infinite-indices condition matters: a neighborhood may contain the same value many times, and those repetitions can still provide a subsequence.
“Infinitely many indices” does not mean that the sequence takes infinitely many different values in the neighborhood. For instance, if a sequence is constantly equal to \(p\), then every term is the same value, but every neighborhood of \(p\) contains terms at infinitely many indices.
The Infinite-Visits Criterion
Proof. First suppose \(p\) is a subsequential limit. Then there is a strictly increasing sequence of indices \((n_j)\) such that \(x_{n_j}\to p\). Fix \(\varepsilon>0\). By convergence, there is a positive integer \(J\) such that \(d(x_{n_j},p)<\varepsilon\) whenever \(j\geq J\). The indices \(n_J,n_{J+1},\ldots\) are all distinct, so infinitely many indices \(n\) satisfy \(d(x_n,p)<\varepsilon\).
Conversely, suppose that for every \(\varepsilon>0\), infinitely many indices \(n\) satisfy \(d(x_n,p)<\varepsilon\). We choose indices recursively. Since infinitely many indices satisfy \(d(x_n,p)<1\), choose one and call it \(n_1\). After \(n_{j-1}\) has been chosen, there are infinitely many indices \(n\) for which \(d(x_n,p)<1/j\). Only finitely many indices are at most \(n_{j-1}\), so at least one of these infinitely many indices is larger than \(n_{j-1}\). Choose such an index as \(n_j\). Thus \(n_1<n_2<\cdots\), and \(d(x_{n_j},p)<1/j\) for every \(j\). Given \(\varepsilon>0\), choose \(J\) such that \(1/J<\varepsilon\). For every \(j\geq J\), \(d(x_{n_j},p)<1/j\leq1/J<\varepsilon\). Therefore \(x_{n_j}\to p\), proving that \(p\) is a subsequential limit. \(\square\)
The construction addresses a small but important issue: finding a term close to \(p\) is not enough. The indices must increase, or the selected terms may not form a subsequence. Infinitely many qualifying indices guarantee that one can always choose the next index beyond the ones already selected.
Worked Example: Two Subsequential Limits in a Compact Interval
Define a sequence in \([0,1]\) by \(x_{2n-1}=1/n\) and \(x_{2n}=1-1/n\) for every positive integer \(n\). These terms do lie in \([0,1]\): \(0<1/n\leq1\), and \(0\leq1-1/n<1\).
The odd-indexed terms converge to \(0\), since \(d(x_{2n-1},0)=1/n\to0\). More explicitly, for any \(\varepsilon>0\), every integer \(n>1/\varepsilon\) satisfies \(d(x_{2n-1},0)=1/n<\varepsilon\). There are infinitely many such \(n\), so every neighborhood of \(0\) contains terms at infinitely many indices. The even-indexed terms converge to \(1\), because \(d(x_{2n},1)=|(1-1/n)-1|=1/n\to0\). Thus both \(0\) and \(1\) are subsequential limits, as the Infinite-Visits Criterion also confirms.
This example shows why a sequence need not have just one subsequential limit, even though a convergent sequence has at most one limit. The full sequence oscillates between terms close to opposite ends of the interval.
Compactness Produces an Infinite-Visits Point
The next argument uses compactness through its open-cover definition. It does not first select a bounded subsequence. Instead, it rules out the possibility that every point of the compact space has some neighborhood containing terms at only finitely many indices.
Proof. Suppose, to the contrary, that no such point exists. Then for every \(x\in K\), there is a radius \(r_x>0\) such that the ball \(B(x,r_x)\), considered in \(K\), contains terms \(x_n\) for only finitely many indices. The collection of these balls, one for each \(x\in K\), is an open cover of \(K\), since each \(x\) belongs to its own ball.
By compactness, finitely many of the balls cover \(K\), say \(B(x_1,r_{x_1}),\ldots,B(x_m,r_{x_m})\). For each \(i\), define \(I_i=\{n\in\mathbb N:x_n\in B(x_i,r_{x_i})\}\). Each \(I_i\) is finite by the choice of the ball. But every term \(x_n\) lies in \(K\), and the selected balls cover \(K\). Consequently every positive integer \(n\) belongs to at least one of the sets \(I_1,\ldots,I_m\). This would make \(\mathbb N=I_1\cup\cdots\cup I_m\) a finite union of finite sets, hence finite, which is impossible. The supposition was false, so some \(p\in K\) has infinitely many sequence indices in every neighborhood. \(\square\)
The finite-subcover step is the decisive one. Each point may have its own neighborhood that catches only finitely many terms, but compactness reduces the entire space to finitely many such neighborhoods. Their union could then catch only finitely many indices altogether, even though every term of the sequence must be caught.
Proof. If the space is empty, there is no sequence in it, so the statement is vacuous. Otherwise, apply the proposition to the given sequence and obtain \(p\in K\) such that every ball centered at \(p\) contains terms at infinitely many indices. The Infinite-Visits Criterion gives a subsequence converging to \(p\). Since \(p\in K\), the subsequence converges to a point of the space. \(\square\)
In particular, the corollary applies to compact subsets of \(\mathbb R\) with the usual distance. By the Heine-Borel Theorem, every closed and bounded subset of \(\mathbb R\) is compact; hence every sequence in such a set has a subsequence converging to a point of that set. The proposition explains the open-cover mechanism behind this sequential conclusion.
Worked Examples: Applying the Criterion
Worked Example: A Subsequence Can Converge to a Value Never Taken
Let \(x_n=1/(n+1)\), regarded as a sequence in \([0,1]\). Every term is positive, so \(x_n\neq0\). Nevertheless, \(0\) is a subsequential limit. Indeed, for any \(\varepsilon>0\), if \(n+1>1/\varepsilon\), then \(d(x_n,0)=1/(n+1)<\varepsilon\). Infinitely many indices satisfy this inequality, so the Infinite-Visits Criterion applies.
Here the entire sequence, not merely a selected subsequence, converges to \(0\). The example emphasizes that a subsequential limit need not occur as a term of the sequence; what matters is that every neighborhood contains terms at infinitely many indices.
Worked Example: Approaching a Missing Endpoint
Consider \(K=(0,1)\) and the sequence \(y_n=1-1/(n+1)\). For every positive integer \(n\), we have \(0<1-1/(n+1)<1\), so \(y_n\in K\). Also, \(d(y_n,1)=|1-1/(n+1)-1|=1/(n+1)\to0\).
Every subsequence of \((y_n)\) also converges to \(1\): for a subsequence with indices \(n_j\), the indices satisfy \(n_j\to\infty\), so \(1/(n_j+1)\to0\). If any subsequence converged to a point \(q\in K\), it would therefore converge both to \(q\) and to \(1\). Uniqueness of limits in a metric space would give \(q=1\), contradicting \(q\in(0,1)\). Thus this sequence has no subsequence converging to a point of \(K\). This is consistent with the fact that \((0,1)\) is not compact.
Worked Example: Repeated Terms Still Give Infinitely Many Visits
Let \(K=\{0,1\}\) and define \(z_n=0\) for odd \(n\), \(z_n=1\) for even \(n\). For every \(\varepsilon>0\), the ball of radius \(\varepsilon\) around \(0\) contains \(z_n=0\) at every odd index, and the ball around \(1\) contains \(z_n=1\) at every even index. Both sets of indices are infinite.
The odd-indexed subsequence is constantly \(0\), and the even-indexed subsequence is constantly \(1\). Thus both points are subsequential limits. The sequence takes only two values, but each limit is supported by infinitely many indices. This illustrates why the criterion counts indices rather than distinct values.
What the Proof Uses—and What It Does Not
The proof begins with a sequence already contained in the compact space. Compactness then produces a point whose neighborhoods are visited infinitely often, and the Infinite-Visits Criterion converts those visits into a convergent subsequence. No claim is made that every bounded sequence in every metric space has such a subsequence: boundedness alone is not enough outside settings such as \(\mathbb R\) where additional theorems apply.
For subsets of \(\mathbb R\), Heine-Borel supplies a convenient route. If \(K\) is closed and bounded, it is compact, so every sequence in \(K\) has a subsequence converging to a point of \(K\). The endpoint example explains the role of closedness: a sequence can approach a boundary point that is missing from the set. The compactness proof ensures that the subsequential limit stays inside the space under consideration.
Use compactness: if every point had a neighborhood containing only finitely many terms, a finite subcover would account for only finitely many indices.
At stage \(j\), choose an index larger than the preceding one with distance less than \(1/j\) from the selected point.
The selected distances are less than \(1/j\), which tends to zero, so the selected subsequence converges to a point in the compact space.
Check Your Understanding
Use the infinite-visits criterion and the compactness argument to answer the following questions.
- Why does convergence of a subsequence imply infinitely many indices lie in every ball around its limit?
- In the reverse direction of the Infinite-Visits Criterion, why can the next selected index be required to exceed the previous one?
- What contradiction follows if finitely many neighborhoods, each containing terms at only finitely many indices, cover the space containing the whole sequence?
- Can a subsequential limit fail to be a value of the sequence? Give an example from the tutorial.
- Why does the sequence \(y_n=1-1/(n+1)\) fail to have a subsequence converging to a point of \((0,1)\)?