Point Set

1. Metric Space

Definition 1.1    Suppose \(X\) is a set, and \(d: X \times X \to \mathbb{R}\) satisfies

    (1) \(\forall x, y \in X, d(x, y) \geq 0\);

    (2) \(\forall x, y \in X, d(x, y)=d(y, x)\);

    (3) \(\forall x, y, z \in X, d(x, y) \leq d(x, z)+ d(z, y)\).

We define \((X, d)\) as a metric space.

Definition 1.2    Suppose \(A\) and \(B\) are non-empty point sets. The distance between \(A\) and \(B\) is \[d(A, B)=\inf_{p \in A, q \in B} d(p, q).\]

Definition 1.3    Suppose \(E\) is a non-empty point set. The diameter of \(E\) is \[\delta(E)=\sup_{p \in E, q \in E} d(p, q).\]

Definition 1.4    Suppose \(E \subset \mathbb{R}^n\). If \(\delta(E)<\infty\), then \(E\) is a bounded point set.

Definition 1.5    In a metric space \((X, d)\), a set is a neighborhood of a point \(p\) if there exists an open ball with centre \(p\) and radius \(r>0\), such that \[B_r(p)=\{x \in X: d(x, p)<r\}.\]

Property 1.1    The neighborhood has the following basic properties:

    (1) \(p \in B_r(p)\);

    (2) \(\forall r_1, r_2>0, \exists r_3>0\) s.t. \(B_{r_3}(p) \subset B_{r_1}(p) \cap B_{r_2}(p)\);

    (3) \(\forall q \in B_{r_1}(p), \exists B_{r'}(q) \subset B_r(p)\);

    (4) If \(p \neq q, \exists B_{r_1}(p), B_{r_2}(q)\) s.t. \(B_{r_1}(p) \cap B_{r_2}(q)=\varnothing\).

2. Cluster Point, Interior Point and Boundary Point

Definition 2.1    A point \(p\) is an interior point of \(E\) if and only if \[\exists B_\delta(p)\ \text{s.t.}\ B_\delta(p) \subset E.\]

Definition 2.2    A point \(p\) is an exterior point of \(E\) if and only if \[\exists B_\delta(p)\ \text{s.t.}\ B_\delta(p) \subset E^c.\]

Definition 2.3    A point \(p\) is a boundary point of \(E\) if and only if \[\forall B_\delta(p): B_\delta(p) \cap E \neq \varnothing, B_\delta(p) \cap E^c \neq \varnothing.\]

Definition 2.4    The set of all interior points of \(E\), denoted \(E^\circ\), is interior or open kernel. The set of all boundary points of \(E\), denoted \(\partial E\), is boundary.

Definition 2.5    A point \(p\) is a cluster point of \(E\) if and only if \[\forall B_r(p), \exists q \neq p, q \in B_r(p)\ \text{s.t.}\ q \in E.\]

Definition 2.6    A point \(p\) is a isolated point of \(E\) if and only if \[p \in E, \exists B_r(p)\ \text{s.t.}\ \forall q \neq p \in B_r(p), q \notin E.\]

The boundary point is either a cluster point or an isolated point.

Theorem 2.1    The following statements are equivalent:

    (1) \(p\) is a cluster point of \(E\);

    (2) Any neighborhood of \(p\) includes infinite points that belong to \(E\);

    (3) There exists a sequence \(\{p_n\}\) including mutually different points in \(E\) such that \(\displaystyle \lim_{n \to \infty} p_n=p\).

Proof. We need to show (1) \(\Rightarrow\) (2) \(\Rightarrow\) (3) \(\Rightarrow\) (1).

(i) Suppose \(\exists B_\delta(p)\) s.t. \(B_\delta(p) \cap E=\{q_1, \ldots, q_k\}\). We can assume that \(q_1, \ldots, q_k \neq p\). Let \[\delta_0=\frac{1}{2}\min\{\delta, d(p, q_1), \ldots, d(p, q_k)\},\] then \[\forall p^* \in B_{\delta_0}(p)-\{p\}: p^* \notin \{p, q_1, \ldots, q_k\},\] i.e., \(\exists B_{\delta_0}(p)\) s.t. \(\forall p^* \neq p \in B_{\delta_0}(p): p^* \notin E\), which is a contradiction. Hence, (1) \(\Rightarrow\) (2).

(ii) Suppose \(p_1 \in B_1(p) \cap E\), where \(p_1 \neq p\). Let \[\delta_1=\frac{1}{2}\min\left\{d(p_1, p), \frac{1}{2}\right\},\] then we can find a \(p_2 \in B_{\delta_1} \cap E\), where \(p_2 \neq p\). We can keep constructing such neighborhood: \[\delta_k=\frac{1}{2}\min\left\{d(p_k, p), \frac{1}{k}\right\}\] and \(p_k \in B_{\delta_k} \cap E\), where \(p_k \neq p\). Therefore we can find a sequence \(\{p_n\}\) including mutually points in \(E\). Besides, \[\forall \varepsilon>0, \exists N \in \mathbb{N}\ \text{s.t.}\ N>\frac{1}{\varepsilon},\] and thus \[\forall n>N, d(p_n, p)<\frac{1}{n}<\frac{1}{N}<\varepsilon,\] i.e., \(\displaystyle \lim_{n \to \infty} p_n=p\). Hence, (2) \(\Rightarrow\) (3).

(iii) Take an arbitrary \(B_\delta(p)\). Since \(\displaystyle \lim_{n \to \infty} p_n=p\), then \[\exists N \in \mathbb{N}\ \text{s.t.}\ n>N, d(p_n, p)<\delta.\] Since \(\{p_n\}\) includes mutually different points, then \[\exists m>N\ \text{s.t}\ p_m \neq p,\] where \(p_m \in B_\delta(p_n) \cap E\). Hence, \(p\) is a cluster point of \(E\), i.e., (3) \(\Rightarrow\) (1).

\(\square\)

Definition 2.7    The set of all cluster points of \(E\), denoted \(E'\), is derived set. The closure of \(E\), denoted \(\overline{E}\), is \(E \cup E'\).

\(\overline{E}=E \cup \partial E=E^\circ \cup \partial E=E' \cup \{\text{All isolated points of}\ E\}\).

Theorem 2.2    The closure and open kernel have the following relationships:

    (1) \((E^\circ)^c=\overline{E^c}\);

    (2) \((\overline{E})^c=(E^c)^\circ\);

    (3) \((A \cup B)'=A' \cup B'\).

Theorem 2.3    If \(E \subset \mathbb{R}^n \neq \varnothing\) and \(E \neq \mathbb{R}^n\), then \(\partial E \neq \varnothing\).

Hint of proof. We can use the connectivity in \(\mathbb{R}^n\).

\(\square\)

Theorem 2.4 (Bolzano-Weierstrass Theorem)    Suppose \(E \subset \mathbb{R}^n\) is a bounded and infinite set, then \(E\) includes at least one cluster point.

3. Open Set and Closed Set

Definition 3.1    If \(\forall x \in E\), \(x\) are interior point, then \(E\) is an open set. If \(\forall x \in E': x \in E\), then \(E\) is a closed set.

Theorem 3.1    \(E\) is an open set if and only if \(E=E^\circ\).

Theorem 3.2    \(E\) is a closed set if and only if \(E=\overline{E}\).

Claim. \(E^\circ\) is an open set, and \(E^\circ\) is the "largest" (\(\cup\)) open set that lies in \(E\). \(\overline{E}\) is a closed set, and \(\overline{E}\) is the "smallest" (\(\cap\)) closed set that includes \(E\).

Proof. Let \(E_o\) be the set of all open sets of \(E\).

Suppose \(x \in E^\circ\). Since \(E^\circ \subset E\) and \(E^\circ\) is an open set, then \(\displaystyle x \in \bigcup_{G \subset E_o} G\). Suppose \(\displaystyle x \in \bigcup_{G \subset E_o} G\), then \(\exists G \subset E\) s.t. \(x \in G\), where \(G\) is an open set of \(E\).

Hence, \(\exists \delta>0\) s.t. \(B_\delta(x) \subset G \subset E\), i.e., \(x \in E^\circ\), and thus \(\displaystyle E^\circ=x \in \bigcup_{G \subset E_o} G\), i.e., \(E^\circ\) is the "largest" open set that lies in \(E\).

Similarly, we can show that \(\overline{E}\) is the "smallest" closed set that includes \(E\).

\(\square\)

Theorem 3.3    If \(E\) is an open set, then \(E^c\) is a closed set. If \(E\) is a closed set, then \(E^c\) is an open set.

Proof. Suppose \(E\) is an open set. Take an arbitrary cluster point \(x\) of \(E^c\), then \(\forall B_r(x), \exists x' \neq x, x' \in B_r(x)\) s.t. \(x' \in E^c\), then \(x\) is not an interior point of \(E\), i.e., \(x \in E^c\). Therefore, \(E^c\) is a closed set.

Suppose \(E\) is a closed set. Take an arbitrary \(x \in E^c\). Suppose \(x\) is not an interior point of \(E^c\), then \(\forall B_r(x): B_r(x) \cap E \neq \varnothing\). Since \(x \notin E\), then \(\forall B_r(x), \exists x' \neq x\) s.t. \(x' \in E\). Therefore, \(x\) is a cluster point of \(E\) and. thus \(x \in E\), which is a contradiction. Hence, \(x\) is an interior point of \(E^c\), and \(E^c\) is an open set.

Theorem 3.4    The union of any number of open sets is an open set. The intersection of a finite number of open sets is an open set. The intersection of any number of closed sets is a closed set. The union of a finite number of closed sets is a closed set.

Proof. Suppose \(G_\lambda\) are open sets, where \(\lambda \in \Lambda\), and \(\Lambda\) is an arbitrary indicator set. We have \[\forall x \in \bigcup_{\lambda \in \Lambda} G_\lambda, \exists \lambda \in \Lambda\ \text{s.t.}\ x \in G_\lambda,\] and thus \[\exists \delta>0\ \text{s.t.}\ B_\delta(x) \subset G_\lambda \subset \bigcup_{\lambda \in \Lambda} G_\lambda.\] Therefore, \(x\) is an interior point of \(\displaystyle \bigcup_{\lambda \in \Lambda} G_\lambda\), and thus \(\displaystyle \bigcup_{\lambda \in \Lambda} G_\lambda\) is an open set.

Suppose \(G_i\) are open sets, where \(i=1, \ldots, n.\) If \(\displaystyle \bigcap_{i=1}^n G_i=\varnothing\), then \(\displaystyle \bigcap_{i=1}^n G_i\) is an open set. If \(\displaystyle \bigcap_{i=1}^n G_i \neq \varnothing\), then take an arbitrary \(\displaystyle x \in \bigcap_{i=1}^n G_i\), we have \(x \in G_i\). Hence, \(\exists \delta_i>0\) s.t. \(B_{\delta_i}(x) \subset G_i\). Take \(\delta=\min\{\delta_1, \ldots, \delta_n\}\), which is greater than \(0\), then \(B_\delta(x) \subset G_i\), i.e., \(\displaystyle B_\delta(x) \subset \bigcap_{i=1}^n G_i\). Therefore, \(x\) is an interior point of \(\displaystyle \bigcap_{i=1}^n G_i\), and thus \(\displaystyle \bigcap_{i=1}^n G_i\) is an open set.

As a consequence, \[\left(\bigcup_{\lambda \in \Lambda}G_\lambda\right)^c=\bigcap_{\lambda \in \Lambda}G_\lambda^c\] is a closed set, where \(G_\lambda^c\) is a closed set; \[\left(\bigcap_{i=1}^n G_i\right)^c=\bigcup_{i=1}^n G_i^c\] is a closed set, where \(G_i^c\) is a closed set.

\(\square\)

The intersection of any number of open sets is not necessarily an open set. For example, \[G_n=\left(-1-\frac{1}{n}, 1+\frac{1}{n}\right),\] where \(G_n\) are open sets. However, \[\bigcap_{n=1}^\infty G_n=[-1, 1],\] which is a closed set.

The union of any number of closed sets is not necessarily a closed set. For example, \[F_n=\left[-1+\frac{1}{n}, 1-\frac{1}{n}\right],\] where \(F_n\) are closed sets. However, \[\bigcup_{n=1}^\infty F_n=(-1, 1),\] which is an open set.

Example 3.1    Suppose \(F_1, F_2 \subset \mathbb{R}^n\) are closed sets, and \(F_1 \cap F_2=\varnothing\).

4. Compact Set and Complete Set

Definition 4.1    \(M\) is a compact set if every open cover of \(M\) has a finite sub-cover.

Theorem 4.1    In \(\mathbb{R}^n\), if set \(F\) is bounded and closed, then \(F\) is compact.

Lemma 4.1.1    The closed subset of a compact set is compact.

Proof. Suppose \(M\) is compact, and \(F \subset M\), where \(F\) is closed. Take any open cover of \(F\), denoted \(\{U_\lambda\}_{\lambda \in \Lambda}\). We know \(\{U_\lambda\}_{\lambda \in \Lambda} \cup F^c\) is an open cover of \(M\), then there exists a finite sub-cover: \(U_1, \ldots, U_n, F^c\). Hence, \(\{U_i\}_{i=1}^n\) is the finite sub-cover of \(F\), and thus \(F\) is compact.

\(\square\)

Proof. Suppose \(F\) is bounded and closed. Since \(F \subset \mathbb{R}^n\) is bounded, then \(\exists T_0=[-a, a]^n\) s.t. \(F \subset T_0\). Suppose \(T_0\) is not compact, then there exists an open cover of \(T_0\) such that it does not have finite sub-cover. We divide \(T_0\) equally into a closed sets with edge length \(a\), and at least one set does not have finite sub-cover, denoted as \(T_1\). We divide \(T_1\) equally into a closed sets with edge length \(\displaystyle \frac{a}{2}\), and at least one set does not have finite sub-cover, denoted as \(T_2\). We repeat this process, and get \[T_1 \supset T_2 \supset \cdots \supset T_n \supset \cdots,\] where \(T_n\) has edge length \(\displaystyle \frac{a}{2^{n-1}}\).

Take an arbitrary \(x_n \in T_{n-1}-T_n\), then \(\{x_n\}\) is bounded and infinite. Because of Bolzano-Weierstrass theorem, \(\{x_n\}\) includes at least one cluster point, denoted as \(x\), i.e., \(\exists \{x_{k_s}\}\) s.t. \(x_{k_s} \to x \in T_j\) for some \(j\). Hence, there exists an open set \(G\) in the sub-cover such that \(x \in G\), i.e., \(\exists \delta>0\) s.t. \(B_\delta(x) \subset G\).

Take \(s \to \infty\) such that the edge length of \(T_{k_s}\) is less than \(\displaystyle \frac{\delta}{2}\) and \(\displaystyle d(x_{k_s}, x)<\frac{\delta}{2}\), and thus \(T_{k_s} \subset B_\delta(x) \subset G\), which is a contradiction. As a consequence, \(T_0\) is compact and \(F \subset T_0\) is compact.

\(\square\)

Theorem 4.2    In \(\mathbb{R}^n\), if set \(M\) is compact, then \(M\) is bounded and closed.

Proof. Since \(M\) is compact, then we can find an open cover such that \(\displaystyle M \subset \bigcup_{p \in M} B_1(p)\), and thus there exists a finite sub-cover: \(B_1(p_1), \ldots, B_1(p_n)\). Since \(B_1(p_i)\) is bounded, then \(\displaystyle \bigcup_{i=1}^n B_1(p_i)\) is bounded, and thus \(\displaystyle M \subset \bigcup_{i=1}^n B_1(p_i)\) is bounded.

Take arbitrary \(q \in M^c\), then \(\forall p \in M, \exists \delta_p>0\) s.t. \(B_{\delta_p}(p) \cap B_{\delta_p}(q)=\varnothing\). We know \(\displaystyle M \subset \bigcup_{p \in M} B_{\delta_p}(p)\), and thus there exists a finite sub-cover: \(B_{\delta_{p_1}}, \ldots, B_{\delta_{p_n}}\).

Take \(\delta=\min\{\delta_{p_1}, \ldots, \delta_{p_n}\}\), then \(B_\delta(q) \cap B_{\delta_{p_i}}(p_i)=\varnothing\) and thus \(B_\delta(q) \cap M=\varnothing\), i.e., \(B_\delta(q) \subset M^c\). Therefore, \(M^c\) is open, and \(M\) is closed.

\(\square\)

在\(\mathbb{R}^n\)中，紧集和有界闭集是等价的. 但在一般空间中，不一定有\(n\)维区间，而点\(p \neq q\)不一定能被开集分开，因此上述两个定理无法推广到一般度量空间或一般拓扑空间中. 在一般度量空间中，我们可以找到反例说明有界闭集不是紧集. 在一般拓扑空间中，我们可以找到反例说明紧集不是闭集.

Definition 4.2    If \(E \subset E'\) (i.e., every point of \(E\) is a cluster point, or \(E\) does not have any isolated points), then \(E\) is dense-in-itself or crowded. If \(E=E'\) (i.e., \(E\) is a closed set without any isolated points), then \(E\) is complete.

Example 4.1    \(\varnothing\) is complete. \(\mathbb{Q}\) is dense-in-itself. \([a, b]\) is complete.

5. Construction of Open Set, Closed Set, and Complete Set on \(\mathbb{R}\)

Theorem 5.1    The set of all disjoint intervals on \(\mathbb{R}\) is at most countable.

Proof. Take an arbitrary interval \(I\). Take a rational number \(q_I \in I\), then we can find an injection: \(I \mapsto q_I\), i.e., the set of all disjoint intervals on \(\mathbb{R}\) is at most countable.

\(\square\)

Definition 5.1    Suppose \(G \subset \mathbb{R}\) is open. If \((\alpha, \beta) \subset G\), and \(\alpha, \beta \notin G\), then \((\alpha, \beta)\) is the component interval of \(G\).

Theorem 5.2    Suppose \(G \subset \mathbb{R}\) is non-empty and open, then \(G\) can be constructed by the at most countable union of disjoint component intervals.

Proof. Suppose \((\alpha_1, \beta_1)\) and \((\alpha_2, \beta_2)\) are different component intervals of \(G\) and \((\alpha_1, \beta_1) \cap (\alpha_2, \beta_2) \neq \varnothing\). Assume, without loss of generality, that \(\alpha_1 \in (\alpha_2, \beta_2) \subset G\), then \(\alpha_1 \in G\), which is a contradiction. Therefore, different component intervals of \(G\) are disjoint. By Theorem 5.1, we can let \[\widetilde{G}=\bigcup_{\substack{(\alpha, \beta) \subset G \\ \alpha \notin G, \beta \notin G}} (\alpha, \beta)\] be the at most countable union of disjoint intervals.

It is obvious that \(\widetilde{G} \subset G\). Take an arbitrary \(x_0 \in G\), and let \[A_{x_0}=\{(x_0-\delta, x_0+\delta): (x_0-\delta, x_0+\delta) \subset G\} \neq \varnothing.\] Let \[\alpha=\inf_{(x_0-\delta, x_0+\delta) \in A_{x_0}} x_0-\delta, \beta=\sup_{(x_0-\delta, x_0+\delta) \in A_{x_0}} x_0+\delta.\] It is obvious that \(x_0 \in (\alpha, \beta)\). Take an arbitrary \(y \in (\alpha, \beta)\). When \(\alpha<y \leq x_0\), since \(\alpha\) is an infimum, then \(\exists \delta\) s.t. \(\alpha \leq x_0-\delta<y\) and \((x_0-\delta, x_0+\delta) \in A_{x_0}\), then \(y \in (x_0-\delta, x_0+\delta) \subset G\). Similarly, when \(x_0 \leq y<\beta\), we can show that \(y \in G\). Hence, \((\alpha, \beta) \subset G\).

Suppose \(\alpha \in G\), then \(\exists \delta_{\alpha}>0\) s.t. \((\alpha-\delta_\alpha, \alpha+\delta_\alpha) \subset G\), then \((\alpha-\delta_\alpha, \beta) \subset G\) and \(\alpha-\delta_\alpha<\alpha\), which is a contradiction, i.e., \(\alpha \notin G\). Similarly, we can prove that \(\beta \notin G\). Therefore \((\alpha, \beta)\) is the component interval of \(G\), and thus \(G=\widetilde{G}\).

\(\square\)

Definition 5.2    Suppose \(A \subset \mathbb{R}\) is closed. The component interval of \(A^c\) is the complementary interval of \(A\).

Theorem 5.3    A closed set \(F\) on \(\mathbb{R}\) is either a full straight line, or deleting at most countable disjoint open intervals (i.e., the complementary interval of \(F\)) from the straight line.

In \(\mathbb{R}\), an isolated point is the common endpoint of two complementary intervals. As a consequence, a complete set is a closed set without adjacent complementary intervals.

In \(\mathbb{R}^n\), an open set is a union of at most countable semi-open-closed intervals.

5.1. Cantor Ternary Set

Suppose \(E_0=[0, 1]\). One starts by deleting the open middle third \(\displaystyle \left(\frac{1}{3}, \frac{2}{3}\right)\) from \(E_0\), leaving \(\displaystyle E_1=\left[0, \frac{1}{3}\right] \cup \left[\frac{2}{3}, 1\right]\). Next, the open middle third of each of remaining closed intervals is deleted, leaving \[\displaystyle E_2=\left[0, \frac{1}{9}\right] \cup \left[\frac{2}{9}, \frac{1}{3}\right] \cup \left[\frac{2}{3}, \frac{7}{9}\right] \cup \left[\frac{8}{9}, 1\right].\] This process is continued ad infinitum, and thus the Cantor ternary set \(P\) is \[P=\lim_{n \to \infty} E_n=\bigcap_{n=0}^\infty E_n.\]

Property 5.4    \(P\) is complete. Since we delete countable disjoint open intervals, then \(P\) is a closed set without adjacent complementary intervals.

Property 5.5    \(P\) does not have interior point. Take an arbitrary \(x \in P\), we know \(B_{3^{-n}}(x)\) always includes a point that does not belong to \(P\). Since \(3^{-n} \to 0\), then we cannot find a \(B_\delta(x) \subset P\).

Definition 5.3    A set is nowhere dense or rare if its closure has empty interior.

Property 5.6    \(\overset{=}{P}=c\).

Property 5.7    The length of \(P\) is 0.

不难发现，\(P\)是一个非常“古怪”的集合，因此引申出一个分形（Fractal）的概念.