# 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.

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.