3.10: Cluster Points. Convergent Sequences
This page is a draft and is under active development.
Consider the set
\[A=\left\{1, \frac{1}{2}, \ldots, \frac{1}{m}, \ldots\right\};\]
we may as well let \(A\) denote the sequence \(x_{m}=1 / m\) in \(E^{1.1}\) Plotting it on the axis, we observe a remarkable fact: The points \(x_{m}\) "cluster" close to 0, approaching 0 as \(m\) increases-see Figure 12.
To make this more precise, take any globe about 0 in \(E^{1}\), \(G_{0}(\varepsilon)=(-\varepsilon, \varepsilon)\). No matter how small, it contains infinitely many (even all but finitely many) points \(x_{m}\), namely, all from some \(x_{k}\) onward, so that
\[(\forall m>k) \quad x_{m} \in G_{0}(\varepsilon).\]
Indeed, take \(k>1 / \varepsilon\), so \(1 / k<\varepsilon\). Then
\[(\forall m>k) \quad \frac{1}{m}<\frac{1}{k}<\varepsilon;\]
i.e., \(x_{m} \in(-\varepsilon, \varepsilon)=G_{0}(\varepsilon)\).
This suggests the following generalizations.
A set, or sequence, \(A \subseteq(S, \rho)\) is said to cluster at a point \(p \in S\) (not necessarily \(p \in A )\), and \(p\) is called its cluster point or accumulation point, iff every globe \(G_{p}\) about \(p\) contains infinitely many points (respectively, terms of \(A\). (Thus only infinite sets can cluster.
Note 1. In sequences (unlike sets) an infinitely repeating term counts as infinitely many terms. For example, the sequence \(0,1,0,1,\) clusters at 0 and 1 (why?); but its range, \(\{0,1\}\), has no cluster points (being finite). This distinction is, however, irrelevant if all terms \(x_{m}\) are distinct, i.e., different from each other. Then we may treat sequences and sets alike.
A sequence \(\left\{x_{m}\right\} \subseteq(S, \rho)\) is said to converge or tend to a point \(p\) in \(S\), and \(p\) is called its limit, iff every globe \(G_{p}(\varepsilon)\) about \(p\) (no matter how small) contains all but finitely many terms \(x_{m} .^{2}\) In symbols,
\[(\forall \varepsilon>0)(\exists k)(\forall m>k) \quad x_{m} \in G_{p}(\varepsilon), \text { i.e., } \rho\left(x_{m}, p\right)<\varepsilon\]
If such a \(p\) exists, we call \(\left\{x_{m}\right\}\) a convergent sequence in \((S, \rho))\); otherwise, a divergent one. The notation is
\[x_{m} \rightarrow p, \text { or } \lim x_{m}=p, \text { or } \lim _{m \rightarrow \infty} x_{m}=p.\]
In \(E^{n}\), \(\rho\left(\overline{x}_{m}, \overline{p}\right)=\left|\overline{x}_{m}-\overline{p}\right|\); thus formula (1) turns into
\[\overline{x}_{m} \rightarrow \overline{p} \text { in } E^{n} \text { iff }(\forall \varepsilon>0)(\exists k)(\forall m>k) \quad\left|\overline{x}_{m}-\overline{p}\right|<\varepsilon\]
Since "all but finitely many" (as in Definition 2) implies "infinitely many" (as in Definition 1 ), any limit is also a cluster point. Moreover, we obtain the following result.
If \(x_{m} \rightarrow p\), then \(p\) is the unique cluster point of \(\left\{x_{m}\right\}\). (Thus a sequence with two or more cluster points, or none at all, diverges.) For if \(p \neq q\), the Hausdorff property (Theorem 1 of §12) yields an \(\varepsilon\) such that
\[G_{p}(\varepsilon) \cap G_{q}(\varepsilon)=\emptyset.\]
As \(x_{m} \rightarrow p\), \(G_{p}(\varepsilon)\) leaves out at most finitely many \(x_{m}\), and only these can possibly be in \(G_{q}(\varepsilon)\). (Why?) Thus \(q\) fails to satisfy Definition 1 and hence is no cluster point. Hence \(\lim x_{m}\) (if it exists) is unique.
(i) We have \(x_{m} \rightarrow p \text { in }(S, \rho)\) iff \(\rho\left(x_{m}, p\right) \rightarrow 0\) in \(E^{1}\).
Hence
(ii) \(\overline{x}_{m} \rightarrow \overline{p}\) in \(E^{n}\) iff \(\left|\overline{x}_{m}-\overline{p}\right| \rightarrow 0\) and
(iii) \(\overline{x}_{m} \rightarrow \overline{0}\) in \(E^{n}\) iff \(\left|\overline{x}_{m}\right| \rightarrow 0\).
- Proof
-
By (2), we have \(\rho\left(x_{m}, p\right) \rightarrow 0\) in \(E^{1}\) if
\[(\forall \varepsilon>0)(\exists k)(\forall m>k) \quad\left|\rho\left(x_{m}, p\right)-0\right|=\rho\left(x_{m}, p\right)<\varepsilon.\]By \((1),\) however, this means that \(x_{m} \rightarrow p,\) proving our first assertion. The rest easily follows from it, since \(\rho\left(\overline{x}_{m}, \overline{p}\right)=\left|\overline{x}_{m}-\overline{p}\right|\) in \(E^{n} . \square\)
If \(x_{m}\) tends to \(p,\) then so does each subsequence \(x_{m_{k}}\)
For \(x_{m} \rightarrow p\) means that each \(G_{p}\) leaves out at most finitely many \(x_{m} .\) This certainly still holds if we drop some terms, passing to \(\left\{x_{m_{k}}\right\} .\)
Note 2.
A similar argument shows that the convergence or divergence of \(\left\{x_{m}\right\},\) and its limit or cluster points, are not affected by dropping or adding
a finite number of terms; similarly for cluster points of sets. For example, if \(\left\{x_{m}\right\}\) tends to \(p,\) so does \(\left\{x_{m+1}\right\}\) (the same sequence without \(x_{1} )\).
We leave the following two corollaries as exercises.
If \(\left\{x_{m}\right\}\) splits into two subsequences, each tending to the same limit \(p,\) then also \(x_{m} \rightarrow p\).
If \(\left\{x_{m}\right\}\) converges in \((S, \rho),\) it is bounded there.
Of course, the convergence or divergence of \(\left\{x_{m}\right\}\) and its clustering depend on the metric \(\rho\) and the space \(S .\) Our theory applies to any \((S, \rho) .\) In particular, it applies to \(E^{*},\) with the metric \(\rho^{\prime}\) of Problem 5 in §11. Recall that under that metric, globes about \(\pm \infty\) have the form \((a,+\infty]\) and \([-\infty, a),\) respectively. Thus limits and cluster points in \(\left(E^{*}, \rho^{\prime}\right)\) coincide with those defined in Chapter 2, §13, (formulas \((1)-(3)\) and Definition 2 there). Our theory then applies to infinite limits as well, and generalizes Chapter 2, §13.
(a) Let
\[x_{m}=p \quad \text{ for all } m\]
(such sequences are called constant). As \(p \in G_{p},\) any \(G_{p}\) contains all \(x_{m} .\) Thus \(x_{m} \rightarrow p,\) by Definition \(2 .\) We see that each constant sequence converges to the common value of its terms.
(b) In our introductory example, we showed that
\[\lim _{m \rightarrow \infty} \frac{1}{m}=0 \quad \text{ in } E^{1}\]
and that 0 is the (unique) cluster point of the set \(A=\left\{1, \frac{1}{2}, \ldots\right\} .\) Here 0\(\notin A .\)
(c) The sequence
\[0,1,0,1, \ldots\]
has two cluster points, 0 and \(1,\) so it diverges by Corollary \(1 .\) (It "oscillates" from 0 to \(1 . )\) This shows that \(a\) bounded sequence may diverge. The converse to Corollary 5 fails.
(d) The sequence
\[x_{m}=m\]
(or the set \(N\) of all naturals) has \(n o\) cluster points in \(E^{1},\) for a globe of radius \(<\frac{1}{2}\) (with any center \(p \in E^{1} )\) contains at most one \(x_{m},\) and hence no \(p\) satisfies Definition 1 or 2.
However, \(\left\{x_{m}\right\}\) does cluster in \(\left(E^{*}, \rho^{\prime}\right),\) and even has a limit there,
namely \(+\infty .(\) Prove it! \()\)
(e) The set \(R\) of all rationals in \(E^{1}\) clusters at each \(p \in E^{1} .\) Indeed, any globe
\[G_{p}(\varepsilon)=(p-\varepsilon, p+\varepsilon)\]
contains infinitely many rationals (see Chapter 2, §10, Theorem 3), and this means that each \(p \in E^{1}\) is a cluster point of \(R .\)
(f) The sequence
\[1,1,2, \frac{1}{2}, 3, \frac{1}{3}, \ldots \quad\left( \text{with } x_{2 k}=\frac{1}{k} \text{ and } x_{2 k-1}=k\right)\]
has only one cluster point, \(0,\) in \(E^{1} ;\) yet it diverges, being unbounded (see Corollary 5\() .\) In \(\left(E^{*}, \rho^{\prime}\right),\) it has two cluster points, 0 and \(+\infty .\) (Verify!)
(g) The lim and lim of any sequence in \(E^{*}\) are cluster points (cf. Chapter 2, §13, Theorem 2 and Problem 4). Thus in \(E^{*},\) all sequences cluster.
(h) Let
\[A=[a, b], \quad a<b.\]
Then \(A\) clusters exactly at all its points, for if \(p \in A,\) then any globe
\[G_{p}(\varepsilon)=(p-\varepsilon, p+\varepsilon)\]
overlaps with \(A\) (even with \((a, b) )\) and so contains infinitely many points of \(A,\) as required. Even the endpoints a and \(b\) are cluster points of \(A\) (and of \((a, b),(a, b],\) and \([a, b) ) .\) On the other hand, no point outside \(A\) is a cluster point. (Why?)
(i) In a discrete space (§11, Example (3)), no set can cluster, since small globes, such as \(G_{p}\left(\frac{1}{2}\right),\) are singletons. (Explain!)
Example \((\mathrm{h})\) shows that a set \(A\) may equal the set of its cluster points \((\mathrm{call}\) it \(A^{\prime} ) ; \mathrm{i.e.}\)
\[A=A^{\prime}.\]
Such sets are said to be perfect. Sometimes we have \(A \subseteq A^{\prime}, A^{\prime} \subseteq A, A^{\prime}=S\) \((\) as in Example \((\mathrm{e})),\) or \(A^{\prime}=\emptyset .\) We conclude with the following result.
\(A\) set \(A \subseteq(S, \rho)\) clusters at p iff each globe \(G_{p}\) (about p) contains at least one point of \(A\) other than \(p\).
Indeed, assume the latter. Then, in particular, each globe
\[G_{p}\left(\frac{1}{n}\right), \quad n=1,2, \ldots\]
contains some point of \(A\) other than \(p ;\) call it \(x_{n} .\) We can make the \(x_{n}\) distinct by choosing each time \(x_{n+1}\) closer to \(p\) than \(x_{n}\) is. It easily follows that each \(G_{p}(\varepsilon)\) contains infinitely many points of \(A\) (the details are left to the reader), as required. The converse is obvious.