
# 3.12: More on Cluster Points and Closed Sets. Density


I. The notions of cluster point and closed set (§§12, 14) can be characterized in terms of convergent sequences. We start with cluster points.

Theorem $$\PageIndex{1}$$

(i) A sequence $$\left\{x_{m}\right\} \subseteq(S, \rho)$$ clusters at a point $$p \in S$$ iff it has a subsequence $$\left\{x_{m_{n}}\right\}$$ converging to $$p.$$

(ii) A set $$A \subseteq(S, \rho)$$ clusters at $$p \in S$$ iff p is the limit of some sequence $$\left\{x_{n}\right\}$$ of points of $$A$$ other than $$p ;$$ if so, the terms $$x_{n}$$ can be made distinct.

Proof

(i) If $$p=\lim _{n \rightarrow \infty} x_{m_{n}},$$ then by definition each globe about $$p$$ contains all but finitely many $$x_{m_{n}},$$ hence infinitely many $$x_{m} .$$ Thus $$p$$ is a cluster point.

Conversely, if so, consider in particular the globes

$G_{p}\left(\frac{1}{n}\right), \quad n=1,2, \ldots$

By assumption, $$G_{p}(1)$$ contains some $$x_{m} .$$ Thus fix

$x_{m_{1}} \in G_{p}(1).$

Next, choose a term

$x_{m_{2}} \in G_{p}\left(\frac{1}{2}\right)\text{ with } m_{2}>m_{1}.$

(Such terms exist since $$G_{p}\left(\frac{1}{2}\right)$$ contains infinitely many $$x_{m} . )$$ Next, fix

$x_{m_{3}} \in G_{p}\left(\frac{1}{3}\right),\text{ with } m_{3}>m_{2}>m_{1},$

and so on.

Thus, step by step (inductively), select a sequence of subscripts

$m_{1}<m_{2}<\cdots<m_{n}<\cdots$

that determines a subsequence (see Chapter 1, §8) such that

$(\forall n) \quad x_{m_{n}} \in G_{p}\left(\frac{1}{n}\right),\text{ i.e., } \rho\left(x_{m_{n}}, p\right)<\frac{1}{n} \rightarrow 0,$

whence $$\rho\left(x_{m_{n}}, p\right) \rightarrow 0,$$ or $$x_{m_{n}} \rightarrow p .$$ (Why?) Thus we have found a subsequence $$x_{m_{n}} \rightarrow p,$$ and assertion (i) is proved.

Assertion (ii) is proved quite similarly - proceed as in the proof of Corollary 6 in §§14; the inequalities $$m_{1}<m_{2}<\cdots$$ are not needed here. $$\square$$

Example $$\PageIndex{1}$$

(a) Recall that the set $$R$$ of all rationals clusters at each $$p \in E^{1}$$ (§§14, Example (e)). Thus by Theorem 1(ii), each real $$p$$ is the limit of a sequence of rationals. See also Problem 6 of §§12 for $$\overline{p}$$ in $$E^{n}$$.

(b) The sequence

$0,1,0,1, \ldots$

has two convergent subsequences,

$x_{2 n}=1 \rightarrow 1\text{ and } x_{2 n-1}=0 \rightarrow 0.$

Thus by Theorem 1(i), it clusters at 0 and 1.

Interpret Example (f) and Problem 10(a) in §14 similarly.

As we know, even infinite sets may have no cluster points (take $$N$$ in $$E^{1})$$. However, a bounded infinite set or sequence in $$E^{n}$$ (*or $$C^{n}$$) must cluster. This important theorem (due to Bolzano and Weierstrass) is proved next.

Theorem $$\PageIndex{1}$$ (Bolzano-Weierstrass).

(i) Each bounded infinite set or sequence $$A$$ in $$E^{n}$$ (* or $$C^{n}$$) has at least one cluster point $$\overline{p}$$ there (possibly outside $$A.$$

(ii) Thus each bounded sequence in $$E^{n}$$ (* or $$C^{n}$$) has a convergent subsequence.

Proof

Take first a bounded sequence $$\left\{z_{m}\right\} \subseteq[a, b]$$ in $$E^{1} .$$ Let

$p=\overline{\lim } z_{m}.$

By Theorem 2(i) of Chapter 2, §13, {$$z_{m}$$} clusters at $$p.$$ Moreover, as

$a \leq z_{m} \leq b,$

we have

$a \leq \inf z_{m} \leq p \leq \sup z_{m} \leq b$

by Corollary 1 of Chapter 2, §13. Thus

$p \in[a, b] \subseteq E^{1},$

and so {$$z_{m}$$} clusters in $$E^{1}$$.

Assertion (ii) now follows - for $$E^{1}-$$ by Theorem 1(i) above.

Next, take

$\left\{\overline{z}_{m}\right\} \subseteq E^{2}, \overline{z}_{m}=\left(x_{m}, y_{m}\right) ; x_{m}, y_{m} \in E^{1}.$

If $$\left\{\overline{z}_{m}\right\}$$ is bounded, all $$\overline{z}_{m}$$ are in some square $$[\overline{a}, \overline{b}] .$$ (Why?) Let

$\overline{a}=\left(a_{1}, a_{2}\right)\text{ and } \overline{b}=\left(b_{1}, b_{2}\right).$

Then

$a_{1} \leq x_{m} \leq b_{1}\text{ and } a_{2} \leq y_{m} \leq b_{2}\text{ in } E^{1}.$

Thus by the first part of the proof, $$\left\{x_{m}\right\}$$ has a convergent subsequence

$x_{m_{k}} \rightarrow p_{1}\text{ for some } p_{1} \in\left[a_{1}, b_{1}\right].$

For simplicity, we henceforth write $$x_{m}$$ for $$x_{m_{k}}, y_{m}$$ for $$y_{m_{k}},$$ and $$\overline{z}_{m}$$ for $$\overline{z}_{m_{k}}$$. Thus $$\overline{z}_{m}=\left(x_{m}, y_{m}\right)$$ is now a subsequence, with $$x_{m} \rightarrow p_{1},$$ and $$a_{2} \leq y_{m} \leq b_{2}$$, as before.

We now reapply this process to $$\left\{y_{m}\right\}$$ and obtain a subsubsequence

$y_{m_{i}} \rightarrow p_{2}\text{ for some } p_{2} \in\left[a_{2}, b_{2}\right].$

The corresponding terms $$x_{m_{i}}$$ still tend to $$p_{1}$$ by Corollary 3 of §14. Thus we have a subsequence

$\overline{z}_{m_{i}}=\left(x_{m_{i}}, y_{m_{i}}\right) \rightarrow\left(p_{1}, p_{2}\right) \quad\text{ in } E^{2}$

by Theorem 2 in §15. Hence $$\overline{p}=\left(p_{1}, p_{2}\right)$$ is a cluster point of $$\left\{\overline{z}_{m}\right\}.$$ Note that $$\overline{p} \in[\overline{a}, \overline{b}]$$ (see above). This proves the theorem for sequences in $$E^{2}$$ (hence in $$C ).$$

The proof for $$E^{n}$$ is similar; one only has to take subsequences n times. (*The same applies to $$C^{n}$$ with real components replaced by complex ones.)

Now take a bounded infinite set $$A \subset E^{n}\left(^{*} C^{n}\right).$$ Select from it an infinite sequence $$\left\{\overline{z}_{m}\right\}$$ of distinct points (see Chapter 1, §9, Problem 5). By what was shown above, $$\left\{\overline{z}_{m}\right\}$$ clusters at some point $$\overline{p},$$ so each $$G_{\overline{p}}$$ contains infinitely many distinct points $$\overline{z}_{m} \in A.$$ Thus by definition, $$A$$ clusters at $$\overline{p} . \square$$

Note 1. We have also proved that if $$\left\{\overline{z}_{m}\right\} \subseteq[\overline{a}, \overline{b}] \subset E^{n},$$ then $$\left\{\overline{z}_{m}\right\}$$ has a cluster point in $$[\overline{a}, \overline{b}] .$$ (This applies to closed intervals only.)

Note 2. The theorem may fail in spaces other than $$E^{n}\left(^{*} C^{n}\right) .$$ For example, in a discrete space, all sets are bounded, but no set can cluster.

II. Cluster points are closely related to the following notion.

Definition

The closure of a set $$A \subseteq(S, \rho),$$ denoted $$\overline{A},$$ is the union of $$A$$ and the set of all cluster points of $$A$$ call it $$A^{\prime}$$. Thus $$\overline{A}=A \cup A^{\prime} .$$

Theorem $$\PageIndex{1}$$

We have $$p \in \overline{A}$$ in $$(S, \rho)$$ iff each globe $$G_{p}(\delta)$$ about p meets $$A$$, i. e.,

$(\forall \delta>0) \quad A \cap G_{p}(\delta) \neq \emptyset.$

Equivalently, $$p \in \overline{A}$$ iff

$p=\lim _{n \rightarrow \infty} x_{n}\text{ for some } \left\{x_{n}\right\} \subseteq A.$

Proof

The proof is as in Corollary 6 of §14 and Theorem 1. (Here, however, the $$x_{n}$$ need not be distinct or different from $$p.$$ ) The details are left to the reader.

This also yields the following new characterization of closed sets (cf. §12).

Theorem $$\PageIndex{1}$$

A set $$A \subseteq(S, \rho)$$ is closed iff one of the following conditions holds.

(i) $$A$$ contains all its cluster points (or has none); i.e., $$A \supseteq A^{\prime}$$.

(ii) $$A=\overline{A}$$.

(iii) $$A$$ contains the limit of each convergent sequence $$\left\{x_{n}\right\} \subseteq A$$ (if any).

Proof

Parts (i) and (ii) are equivalent since

$A \supseteq A^{\prime} \Longleftrightarrow A=A \cup A^{\prime}=\overline{A} . \quad\text{(Explain!)}$

Now let $$A$$ be closed. If $$p \notin A,$$ then $$p \in-A;$$ therefore, by Definition 3 in §12, some $$G_{p}$$ fails to meet $$A\left(G_{p} \cap A=\emptyset\right).$$ Hence no $$p \in-A$$ is a cluster point, or the limit of a sequence $$\left\{x_{n}\right\} \subseteq A.$$ (This would contradict Definitions 1 and 2 of §14.) Consequently, all such cluster points and limits must be in $$A,$$ as claimed.

Conversely, suppose $$A$$ is not closed, so $$-A$$ is not open. Then $$-A$$ has a noninterior point $$p;$$ i.e., $$p \in-A$$ but $$n o G_{p}$$ is entirely in $$-A.$$ This means that each $$G_{p}$$ meets $$A.$$ Thus

$p \in \overline{A}\text{ (by Theorem 3),}$

and

$p=\lim _{n \rightarrow \infty} x_{n}\text{ for some } \left\{x_{n}\right\} \subseteq A\text{ (by the same theorem),}$

even though $$p \notin A($$ for $$p \in-A)$$.

We see that (iii) and (ii), hence also (i), fail if $$A$$ is not closed and hold if $$A$$ is closed. (See the first part of the proof.) Thus the theorem is proved. $$\square$$

Corollary 1. $$\overline{\emptyset}=\emptyset$$.

Corollary 2. $$A \subseteq B \Longrightarrow \overline{A} \subseteq \overline{B}$$.

Corollary 3. $$\overline{A}$$ is always a closed set $$\supseteq A$$.

Corollary 4. $$\overline{A \cup B}=\overline{A} \cup \overline{B}$$ (the closure of $$A \cup B$$ equals the union of $$\overline{A}$$ and $$\overline{B} )$$.

III. As we know, the rationals are dense in $$E^{1}$$ (Theorem 3 of Chapter 2, §10). This means that every globe $$G_{p}(\delta)=(p-\delta, p+\delta)$$ in $$E^{1}$$ contains rationals. Similarly (see Problem 6 in §12), the set $$R^{n}$$ of all rational points is dense in $$E^{n}.$$ We now generalize this idea for arbitrary sets in a metric space $$(S, \rho).$$

Definition

Given $$A \subseteq B \subseteq(S, \rho),$$ we say that $$A$$ is dense in $$B$$ iff each globe $$G_{p}$$ $$p \in B,$$ meets $$A.$$ By Theorem $$3,$$ this means that each $$p \in B$$ is in $$\overline{A};$$ i.e.,

$p=\lim _{n \rightarrow \infty} x_{n} \quad\text{ for some } \left\{x_{n}\right\} \subseteq A.$

Equivalently, $$A \subseteq B \subseteq \overline{A} .^{3}$$.

Summing up, we have the following:

$A\text{ is open iff } A=A^{0}.$

$A\text{ is closed iff } A=\overline{A}\text{; equivalently, iff } A \supseteq A^{\prime}.$

$A\text{ is dense in } B\text{ iff } A \subseteq B \subseteq \overline{A}.$

$A\text{ is perfect iff } A=A^{\prime}.$