📂MetricSpace

Closure and Derived Set in Metric Space

Definitions

Let’s say $(X,d)$ is a metric space. Suppose $p \in X$ and $E \subset X$.

• A set that contains all $q$ satisfying $d(q,p)<r$ is defined as the neighborhood of point $p$, denoted by $N_{r}(p)$. In this case, $r$ is called the radius of $N_{r}(p)$. If it’s permissible to omit the distance, it can also be denoted as $N_{p}$.

• If every neighborhood of $p$ includes a $q$ that is $q\ne p$ and $q\in E$, then $p$ is called a limit point of $E$.

• If all limit points of $E$ are included in $E$, then $E$ is said to be closed.

• If there exists a neighborhood $N$ satisfying $N\subset E$ for $p$, then $p$ is called an interior point of $E$.

• If every point of $E$ is an interior point of $E$, then $E$ is said to be open.

• The set of all limit points of $E$ is called the derived set of $E$, denoted by $E^{\prime}$.

• The union of $E$ and $E^{\prime}$ is called the closure, denoted by $\overline{E}=E\cup E^{\prime}$.

Theorem 1

For $A,B\subset X$, the following equations hold:

(1a) $A\subset B \implies A^{\prime} \subset B^{\prime}$

(1b) $(A\cup B)^{\prime}=A^{\prime}\cup B^{\prime}$

(1c) $(A \cap B)^{\prime} \subset A^{\prime}\cap B^{\prime}$

Proof

(1a)

Assume $A\subset B$. And let $p\in A^{\prime}$. Then $p$ is a limit point of $A$, so by the definition of a limit point, every neighborhood $N$ of $p$ includes a $q$ that is $q\ne p$ and $q\in A$. Assuming $A\subset B$, the statement means that every neighborhood $N$ of $p$ includes a $q$ that is $q\ne p$ and $q\in B$. Therefore, by the definition of a limit point, $p \in B^{\prime}$ is established.

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(1b)

• part 1. $A^{\prime} \cup B^{\prime} \subset (A\cup B)^{\prime}$

  Since $A\subset A\cup B$ and $B \subset A\cup B$, by $(a1)$, it follows that:

  $$A^{\prime} \subset (A\cup B)^{\prime} \quad \text{and} \quad B^{\prime} \subset (A \cup B)^{\prime}$$

  Therefore,

  $$A^{\prime} \cup B^{\prime} \subset (A\cup B)^{\prime}$$

• part 2. $(A\cup B)^{\prime} \subset A^{\prime}\cup B^{\prime}$

  Let’s say $p \in (A\cup B)^{\prime}$. Then, by the definition of a limit point, every neighborhood $N$ of $p$ includes a $q$ that is $q\ne p$ and $q\in A\cup B$. Rewriting $q\in A\cup B$ as $q\in A \text{ or } q\in B$ means $p \in A^{\prime} \text{ or } p\in B^{\prime}$. Therefore, $p\in A^{\prime}\cup B^{\prime}$, so it follows that:

  $$(A\cup B)^{\prime} \subset A^{\prime}\cup B^{\prime}$$

• part 3.

  Combining the above results yields:

  $$A^{\prime}\cup B^{\prime} = (A\cup B)^{\prime}$$

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(1c)

Since $A\cap B \subset A$ and $A\cap B \subset B$, by (1a), it follows that:

$$(A\cap B)^{\prime} \subset A^{\prime} \quad \text{and} \quad (A\cap B)^{\prime} \subset B^{\prime}$$

Therefore,

$$(A\cap B)^{\prime} \subset A^{\prime}\cap B^{\prime}$$

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Theorem 2

For $A,B \subset X$, the following equations hold:

(2a) $A\subset B \implies \overline{A} \subset \overline{B}$

(2b) $\overline{A\cup B} = \overline{A}\cup \overline{B}$

(2c) $\overline{A\cap B} \subset \overline{A}\cap \overline{B}$

Proof

(2a)

Assuming $A \subset B$, by (1a), $A^{\prime} \subset B^{\prime}$ is established. Therefore,

$$\overline{A} = A\cup A^{\prime} \subset B \cup B^{\prime} = \overline{B}$$

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(2b)

• part 1. $\overline{A\cup B}\subset \overline{A}\cup \overline{B}$

  Let’s say $p \in \overline{A\cup B}$. This means $p\in A\cup B$ or $p \in (A\cup B)^{\prime}$.

  • case 1-1. $p \in A\cup B$

    In this case, $p \in A$ or $p \in B$. But since $A \subset \overline{A}$ and $B \subset \overline{B}$,

    $$p\in \overline{A}\ \text{or} \ p \in \overline{B}\implies p \in \overline{A}\cup \overline{B}$$

  • case 1-2. $p\in (A\cup B)^{\prime}$

    By (1b), $p\in (A\cup B)^{\prime}=A^{\prime}\cup B^{\prime}$ is established. This means $p\in A^{\prime}$ or $p\in B^{\prime}$. But since $A^{\prime} \subset \overline{A}$ and $B^{\prime} \subset \overline{B}$, similarly to the previous case,

    $$p\in \overline{A}\ \text{or} \ p \in \overline{B}\implies p \in \overline{A}\cup\overline{B}$$

  By case 1-1, 1-2, the following is established:

  $$\overline{A\cup B}\subset \overline{A}\cup \overline{B}$$

• part 2. $\overline{A}\cup \overline{B} \subset \overline{A\cup B}$

  Since $A \subset A\cup B$ and $B\subset A\cup B$, by $(b1)$, the following is established:

  $$\overline{A} \subset \overline{A\cup B}\quad \text{and} \quad \overline{B}\subset \overline{A\cup B}$$

  Therefore,

  $$\overline{A}\cup \overline{B} \subset \overline{A\cup B}$$

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(2c)

Let’s say $p \in \overline{A\cap B}$. Then, $p\in A\cap B$ or $p\in (A \cap B)^{\prime}$.

• case 1. $p\in A\cap B$

  In this case, $p \in A$ while $p \in B$. But since $A\subset \overline{A}$ and $B\subset \overline{B}$,

  $$p\in A \ \text{and} \ \ p \in B \implies p\in \overline{A} \ \text{and} \ p \in \overline{B} \implies p\in \overline{A}\cap \overline{B}$$

• case 2. $p \in (A\cap B)^{\prime}$

  By (1a), $(A\cap B)^{\prime}\subset A^{\prime}$ and $(A\cap B)^{\prime} \subset B^{\prime}$ are established. But since $A^{\prime}\subset \overline{A}$ and $B^{\prime} \subset \overline{B}$,

  $$p\in A^{\prime} \ \text{and} \ p\in B^{\prime} \implies p\in \overline{A}\quad \text{and} \quad p\in \overline{B}\implies p\in \overline{A}\cap \overline{B}$$

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Theorem 3

For metric spaces $(X,d)$ and $E \subset X$, the following facts hold:

(3a) $\overline{E}$ is closed.

(3b) $E=\overline{E}$ being equivalent to $E$ being closed.

(3c) For a closed set $F\subset X$ satisfying $E\subset F$, $\overline{E} \subset F$ holds.

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(3a) and (3c) imply that $\overline{E}$ is the smallest closed subset of $X$ that includes $E$.

Proof

(3a)

Let’s say $p \in X$ and $p \notin \overline{E}$. In other words, $p \in (\overline{E})^{c}$. Then $p$ is neither a point of $E$ nor of $E^{\prime}$. Therefore, by the definition of an accumulation point, $p$ has at least one neighborhood $N$ where $N\cap E=\varnothing$ is true. Therefore, since $N\subset (\overline{E})^{c}$ and $p$ was any point of $(\overline{E})^{c}$, by the definition of an interior point, every point of $(\overline{E})^{c}$ is an interior point, which means $(\overline{E})^{c}$ is an open set. Since $(\overline{E})^{c}$ is an open set, $\overline{E}$ is a closed set.¹

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(3b)

• $(\implies)$

  Since $E=\overline{E}=E \cup E^{\prime}$, all accumulation points of $E$ are elements of $E$. This is the definition of a closed set, so $E$ is closed. Or it can be immediately understood from the definitions of closure and being closed.

• $(\impliedby)$

  By the definition of a closed set, all accumulation points of $E$ are included in $E$. Therefore, $\overline{E}=E\cup E^{\prime}=E$

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(3c)

Let $F$ be a closed set with $E\subset F \subset X$. Then, by (3b), $F^{\prime} \subset \overline{F}=F$ is true. Also, by (2a), $E^{\prime} \subset F^{\prime} \subset F$ is true. Therefore, the following holds:

$$E \subset F \quad \text{and} \quad E^{\prime}\subset F$$

Therefore,

$$E\cup E^{\prime} =\overline{E} \subset F$$

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Theorem 4

Let $E$ be a non-empty set of real numbers and suppose it is bounded above. Let $y=\sup E$. Then, $y \in \overline{E}$ is true. Moreover, if $E$ is closed, then $y \in E$ is true.

Proof

If $y \in \overline{E}$ holds, then the subsequent propositions are trivial by (3a), so we will only prove $y \in \overline{E}$. The proof is divided into two cases.

• case 1. $y \in E$

  $$y \in E \subset \overline{E}$$

  Therefore, it holds.

• case 2. $y \notin E$

  Then, for every positive number $h>0$, there exists $x\in E$ satisfying $y-h<x<y$. This means that every neighborhood of $y$, which is $N_{h}(y)$, must contain an element of $E$. Therefore, by definition, $y$ is an accumulation point of $E$. Thus, $y\in E\cup E^{\prime}=\overline{E}$

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