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

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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:

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

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

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

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