3.10: Cluster Points. Convergent Sequences

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.

Definition: cluster at a point

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.

Definition

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.

corollary $\PageIndex{1}$

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.

corollary $\PageIndex{2}$

(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 $E^{n} . \square$

corollary $\PageIndex{3}$

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.

corollary $\PageIndex{4}$

If $\left\{x_{m}\right\}$ splits into two subsequences, each tending to the same limit $p,$ then also $x_{m} \rightarrow p$.

corollary $\PageIndex{5}$

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.

Example $\PageIndex{1}$

(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.

corollary $\PageIndex{6}$

$A$ 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.