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# 3.10: Cluster Points. Convergent Sequences

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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.