해석학 I 한·영 대조 개념 정리Basic Topology — Bilingual Study Guide

Let $A$ be countable, $E \subset A$ infinite. Arrange elements of $A$ in a sequence $(x_n)$. Let $n_1$ be the smallest positive integer with $x_{n_1} \in E$. Inductively, let $n_k$ be the smallest integer greater than $n_{k-1}$ with $x_{n_k} \in E$. This is possible for all $k$ since $E$ is infinite. By putting $f(k) = x_{n_k}$, we obtain a 1-1 correspondence between $\mathbb{N}$ and $E$. $\square$

Since each $E_n$ is countable, arrange as $E_n = \{x_{nk}\}$ ($k = 1,2,3,\ldots$). We can count all elements of $\bigcup_{n=1}^{\infty} E_n$ using the diagonal arrangement: $x_{11};\; x_{21}, x_{12};\; x_{31}, x_{22}, x_{13};\; x_{41}, x_{32}, x_{23}, x_{14};\; \ldots$ If any two of the sets $E_n$ have elements in common, these will appear more than once. Hence there is a subset $T$ of $\mathbb{N}$ such that $T \sim \bigcup_{n=1}^{\infty} E_n$. Since $E_1 \subset \bigcup E_n$ and $E_1$ is an infinite set, $\bigcup E_n$ is an infinite set. Therefore by Theorem 2.13, $\bigcup E_n$ is countable. $\square$

Obviously if $n = 1$, then $B_1 = A$ is countable. We use mathematical induction. Suppose $B_{n-1}$ is countable for some $n \in \mathbb{N}$ with $n \geq 2$. The elements of $B_n$ are of the form $(b, a)$ ($b \in B_{n-1}$, $a \in A$). For every fixed $b$, $\{(b,a) : a \in A\} \sim A$. Thus $\{(b,a) : a \in A\}$ is countable. Observe that $B_n = \bigcup_{b \in B_{n-1}} \{(b,a) : a \in A\}$. In other words, $B_n$ is the countable union of countable sets. Therefore, by Theorem 2.17, $B_n$ is countable. $\square$

$\mathbb{Z}$ is countable (Example 2.9). By Thm 2.21, $\mathbb{Z} \times \mathbb{Z}$ is countable. Every rational is $a/b$ where $a,b \in \mathbb{Z}$, $b \neq 0$. So $A \subset \mathbb{Z} \times \mathbb{Z}$ with $A \sim \mathbb{Q}$. Since $\mathbb{Q}$ is infinite, it's countable by Thm 2.13. $\square$

Suppose $A$ is countable. Then the members of $A$ can be arranged in a sequence $s_1, s_2, s_3, \ldots$, where each $s_n = (s_{n,1},\, s_{n,2},\, s_{n,3},\, \ldots)$.

Construct a sequence $\bar{s} = (\bar{s}_1, \bar{s}_2, \bar{s}_3, \ldots)$ by $$\bar{s}_n = \begin{cases} 1 & \text{if } s_{n,n} = 0 \\ 0 & \text{if } s_{n,n} = 1 \end{cases}$$

Then $\bar{s} \in A$, and $\bar{s} \neq s_n$ for every $n$ (since $\bar{s}_n \neq s_{n,n}$). This contradicts the assumption that every member of $A$ appears in the sequence. This is Cantor's diagonal process. $\square$

Let $E = N_r(p)$. Let $q \in N_r(p)$. Set $h = r - d(p,q) > 0$. For all $s \in N_h(q)$, by triangle inequality $d(p,s) \leq d(p,q) + d(q,s) < d(p,q) + h = r$. So $N_h(q) \subset N_r(p)$, meaning $q$ is an interior point. $\square$

Suppose some $N_r(p)$ contains only finitely many points $q_1,\ldots,q_n$ of $E$ distinct from $p$. Let $r' = \min\{d(p,q_i)\}$. Then $N_{r'}(p)$ contains no point of $E$ other than possibly $p$, contradicting $p$ being a limit point. $\square$

If $E = \emptyset$, all statements are obvious. So assume $E$ is not empty. Set $n = |E|$. By $a_1, a_2, \ldots, a_n$, we denote the elements of $E$.

(i) Set $M = \max\{d(a_1, a_i) : i = 1,\ldots,n\} + 1$. Then $d(a_1, a_i) < M$ for all $i$. Therefore $E$ is bounded.

(ii) Assume that $E$ has a limit point $p$. By Theorem 2.34, $N_r(p)$ contains infinitely many points of $E$. It is contradiction and thus (ii) is proved.

(iii) By (ii), the set of all limit points of $E$ is empty set. Since $\emptyset \subset E$, $E$ is closed. $\square$

If $p$ is a limit point of $E$, then every neighborhood of $p$ contains a point $q \neq p$ such that $q \in E$. Since $x$ is not a limit point of $E$, not every neighborhood of $x$ contains a point $q \neq x$ such that $q \in E$. Thus there exists a neighborhood of $x$ which does not contain a point $q \neq x$ such that $q \in E$. Since every neighborhood of $x$ can be represented by $N_r(x)$ with $r > 0$, there exists a $r > 0$ such that $N_r(x) \cap E = \{x\}$ or $N_r(x) \cap E = \emptyset$. Since $x \notin E$ by the assumption, there exists a $r > 0$ such that $N_r(x) \cap E = \emptyset$. $\square$

($\Rightarrow$) Suppose $E$ is open. Let $x$ be a limit point of $E^c$. If $x \in E$, then $x$ is interior to $E$, so $N_r(x) \subset E$ for some $r$, meaning $N_r(x) \cap E^c = \emptyset$, contradicting $x$ being a limit point. Thus $x \in E^c$.

($\Leftarrow$) Suppose $E^c$ is closed. Let $x \in E$. Then $x \notin E^c$ and $x$ is not a limit point of $E^c$. By Lem 2.39, $N_r(x) \cap E^c = \emptyset$ for some $r$. Hence $N_r(x) \subset E$. $\square$

Due to Theorem 2.40, $F^c$ is open if and only if $(F^c)^c$ is closed. Since $(F^c)^c = F$, the corollary is proved. $\square$

We only prove (a) because of the similarity of the proof. First we prove $\left(\bigcup_{\alpha \in A} E_\alpha\right)^c \subset \bigcap_{\alpha \in A} E_\alpha^c$. Let $x \in \left(\bigcup_{\alpha \in A} E_\alpha\right)^c$. Then $x \notin \bigcup_{\alpha \in A} E_\alpha$ and thus $x \notin E_\alpha$ for all $\alpha \in A$. By the definition of $E_\alpha^c$, $x \in E_\alpha^c$ for all $\alpha \in A$. Therefore $x \in \bigcap_{\alpha \in A} E_\alpha^c$.

Next we prove $\left(\bigcup_{\alpha \in A} E_\alpha\right)^c \supset \bigcap_{\alpha \in A} E_\alpha^c$. Let $x \in \bigcap_{\alpha \in A} E_\alpha^c$. Then $x \in E_\alpha^c$ for all $\alpha \in A$ and thus $x \notin E_\alpha$ for all $\alpha \in A$. Therefore $x \notin \bigcup_{\alpha \in A} E_\alpha$ and finally we have $x \in \left(\bigcup_{\alpha \in A} E_\alpha\right)^c$. $\square$

(a) It suffices to show that for any $x \in \bigcup_{\alpha \in A} G_\alpha$, $x$ is an interior point. Let $x \in \bigcup G_\alpha$. Then by the definition of the union, there exists $\alpha_0 \in A$ such that $x \in G_{\alpha_0}$. Since $G_{\alpha_0}$ is open, $x$ is an interior point of $G_{\alpha_0}$. Thus there exists $r > 0$ such that $N_r(x) \subset G_{\alpha_0} \subset \bigcup G_\alpha$.

(b) Due to Corollary 2.41, it suffices to show that $\left(\bigcap_{\alpha \in A} F_\alpha\right)^c$ is open. By Theorem 2.40, $F_\alpha^c$ is open for all $\alpha \in A$. Thus by (a), $\bigcup_{\alpha \in A} F_\alpha^c$ is also open. Moreover, due to Lemma 2.42, $\left(\bigcap F_\alpha\right)^c = \bigcup F_\alpha^c$. Therefore $\left(\bigcap F_\alpha\right)^c$ is open.

(c) It suffices to show that any point $x \in \bigcap_{i=1}^{n} G_i$ is an interior point. Let $x \in \bigcap_{i=1}^{n} G_i$. Then $x \in G_i$ for all $i = 1,\ldots,n$. Since each $G_i$ is open, $x$ is an interior point and thus for each $i$, there exists $r_i > 0$ such that $N_{r_i}(x) \subset G_i$. Put $r = \min\{r_1,\ldots,r_n\}$. Then obviously $r > 0$ and $N_r(x) \subset N_{r_i}(x)$ for all $i = 1,\ldots,n$. Therefore $N_r(x) \subset \bigcap_{i=1}^{n} G_i$.

(d) Since the proof of (d) is similar to that of (b), we skip the details. $\square$

It suffices to show that $E' \cap N_r(x) = \emptyset$, i.e. there is no limit point of $E$ in $N_r(x)$. We use the reduction to absurdity. Assume that there is a limit point $p$ of $E$ in $N_r(x)$. Then $d(p,x) < r$ and define $h := r - d(p,x) > 0$. Following the proof of Theorem 2.33, we can check that $N_h(p) \subset N_r(x)$.

However, since $p$ is a limit point of $E$, there exists a $q \in E$ such that $q \neq p$ and $q \in N_h(p)$. Thus there exists a $q \in E \cap N_h(p) \subset E \cap N_r(x)$. It is contradiction to the assumption that $E \cap N_r(x) = \emptyset$. $\square$

Since $x \notin E$ and $x$ is not a limit point of $E$, by Lemma 2.39, there exists $r > 0$ such that $N_r(x) \cap E = \emptyset$. By Lemma 2.48, $\bar{E} \cap N_r(x) = \emptyset$. $\square$

(a) Due to Corollary 2.41, it suffices to show that $\bar{E}^c$ is open, i.e. for any $x \in \bar{E}^c$, $x$ is an interior point of $\bar{E}^c$. Let $x \in \bar{E}^c$. Then $x \notin \bar{E}$. Thus $x \notin E$ and $x \notin E'$. Therefore by Corollary 2.49, there exists a $r > 0$ such that $N_r(x) \cap \bar{E} = \emptyset$, which implies $N_r(x) \subset \bar{E}^c$ and $x$ is an interior point of $\bar{E}^c$.

(b) First we prove the only if part ($\Rightarrow$). If $E = \bar{E}$, then by (a) $E$ is closed. Next we prove the if part ($\Leftarrow$). Since $\bar{E} = E \cup E'$, obviously $E \subset \bar{E}$. Moreover, if $E$ is closed, then $E' \subset E$. Therefore, if $E$ is closed, then we have $\bar{E} = E$.

(c) It suffices to show that $E' \subset F$ since $E \subset F$. Observe that if $E \subset F$, then any limit point of $E$ is also a limit point of $F$, i.e. $E' \subset F'$. Moreover since $F$ is closed, $F' \subset F$. Therefore, we have $E' \subset F' \subset F$. $\square$

($\Rightarrow$) Assume that $E \subset Y$ and $E$ is open relative to $Y$. Then for any $y \in E$, there exists a $r_y > 0$ such that $N_{r_y}^Y(y) \subset E$. Define $G := \bigcup_{y \in E} N_{r_y}^X(y)$ where $N_{r_y}^X(y) := \{q \in X : d(y,q) < r_y\}$. Then by Theorem 2.43(a), $G$ is open in $X$. Obviously, $E \subset G \cap Y$. Moreover, since $N_{r_y}^X(y) \cap Y = N_{r_y}^Y(y)$, we have $G \cap Y = \bigcup_{y \in E}\left(N_{r_y}^X(y) \cap Y\right) = \bigcup_{y \in E} N_{r_y}^Y(y) \subset E$.

($\Leftarrow$) Assume that $G$ is open in $X$ and $E = G \cap Y$. Let $x \in E$. Then since $x \in G$, there exists a $r_x > 0$ such that $N_{r_x}^X(x) \subset G$. Then since $N_{r_x}^X(x) \cap Y = N_{r_x}^Y(x)$, we have $N_{r_x}^Y(x) \subset G \cap Y = E$. Therefore $x$ is an interior point of $E$ in $Y$ and it implies $E$ is open relative to $Y$. $\square$

($\Rightarrow$) Assume $K$ is compact relative to $X$. Let $\{G_\alpha^Y : \alpha \in A\}$ be an open cover of $K$ relative to $Y$. Due to Theorem 2.55, for each $\alpha$, there exist an open set $G_\alpha$ relative to $X$ such that $G_\alpha^Y = G_\alpha \cap Y$. Since $\{G_\alpha^Y\}$ is an open cover of $K$ relative to $Y$, we have $K \subset \bigcup G_\alpha^Y \subset \bigcup G_\alpha$. Thus $\{G_\alpha\}$ is an open cover of $K$ relative to $X$. By the assumption, there exist $\alpha_1,\ldots,\alpha_n$ such that $K \subset G_{\alpha_1} \cup \cdots \cup G_{\alpha_n}$. Therefore, since $K \subset Y$, $K \subset (G_{\alpha_1} \cup \cdots \cup G_{\alpha_n}) \cap Y = G_{\alpha_1}^Y \cup \cdots \cup G_{\alpha_n}^Y$.

($\Leftarrow$) Assume $K$ is compact relative to $Y$. Let $\{G_\alpha\}$ be an open cover of $K$ relative to $X$. Then $\{G_\alpha \cap Y : \alpha \in A\}$ is an open cover of $K$ relative to $Y$. Since $K$ is compact relative to $Y$, there exist $\alpha_1,\ldots,\alpha_n$ such that $K \subset (G_{\alpha_1} \cap Y) \cup \cdots \cup (G_{\alpha_n} \cap Y)$. Since obviously $G_{\alpha_i} \cap Y \subset G_{\alpha_i}$, we have $K \subset G_{\alpha_1} \cup \cdots \cup G_{\alpha_n}$. $\square$

It suffices to show that $K^c$ is open. Let $p \in K^c$. It is sufficient to show that $p$ is an interior point of $K^c$. Then for any $q \in K$, $d(p,q) > 0$ since $p \neq q$. For $q \in K$, consider $N_{\frac{1}{2}d(p,q)}(p)$ and $N_{\frac{1}{2}d(p,q)}(q)$. Then $N_{\frac{1}{2}d(p,q)}(p) \cap N_{\frac{1}{2}d(p,q)}(q) = \emptyset$ (2.3). Thus obviously $K \subset \bigcup_{q \in K} N_{\frac{1}{2}d(p,q)}(q)$, and thus $\{N_{\frac{1}{2}d(p,q)}(q) : q \in K\}$ is an open cover of $K$. Therefore, since $K$ is compact, there exist $q_1,\ldots,q_n$ such that $K \subset N_{\frac{1}{2}d(p,q_1)}(q_1) \cup \cdots \cup N_{\frac{1}{2}d(p,q_n)}(q_n)$.

Next set $V_p := N_{\frac{1}{2}d(p,q_1)}(p) \cap \ldots \cap N_{\frac{1}{2}d(p,q_n)}(p)$. Then $V_p$ is open and due to (2.3), $V_p \cap (N_{\frac{1}{2}d(p,q_1)}(q_1) \cup \ldots \cup N_{\frac{1}{2}d(p,q_n)}(q_n)) = \emptyset$. Thus since $K \subset N_{\frac{1}{2}d(p,q_1)}(q_1) \cup \ldots \cup N_{\frac{1}{2}d(p,q_n)}(q_n)$, we have $V_p \cap K = \emptyset$ and it implies $V_p \subset K^c$. Moreover, since $V_p$ is open and $p \in V_p$, there exists $N_r(p)$ such that $N_r(p) \subset V_p$. Therefore $p$ is an interior point of $K^c$. $\square$

Let $\{V_\alpha : \alpha \in A\}$ be an open cover of $F$. Then since $F$ is closed, $F^c$ is open. Since $K \subset X = F \cup F^c \subset \left(\bigcup_{\alpha \in A} V_\alpha\right) \cup F^c$, thus the collection $\{V_\alpha : \alpha \in A\} \cup \{F^c\}$ is an open cover of $K$. Since $K$ is compact, there exist $\alpha_1,\ldots,\alpha_n$ such that $K \subset V_{\alpha_1} \cup \cdots \cup V_{\alpha_n} \cup F^c$. Since $F \subset K$ and $F \cap F^c = \emptyset$, we have $F \subset V_{\alpha_1} \cup \cdots \cup V_{\alpha_n}$. In other words, $\{V_{\alpha_1},\ldots, V_{\alpha_n}\}$ is a finite open cover of $F$. Therefore $F$ is compact. $\square$

By Theorem 2.63, $K$ is closed. Thus by Theorem 2.43(b), $F \cap K$ is closed. Since $F \cap K \subset K$, $F \cap K$ is a closed subset of $K$. Therefore by Theorem 2.65, $F \cap K$ is compact. $\square$

We use the reduction to absurdity. Assume that $\bigcap_{\alpha \in A} K_\alpha = \emptyset$. For each $K_\alpha$, put $G_\alpha := K_\alpha^c$. Fix $K_{\alpha_0} \in \{K_\alpha : \alpha \in A\}$. Then due to the assumption and the generalization of De Morgan's law, $K_{\alpha_0} \subset \left(\bigcap_{\alpha \in A - \{\alpha_0\}} K_\alpha\right)^c = \bigcup_{\alpha \in A - \{\alpha_0\}} G_\alpha$.

Thus $\{G_\alpha : \alpha \in A - \{\alpha_0\}\}$ is an open cover of $K_{\alpha_0}$. Since $K_{\alpha_0}$ is compact, there exist a finite set $B = \{\alpha_1,\ldots,\alpha_n\}$ such that $K_{\alpha_0} \subset G_{\alpha_1} \cup \cdots \cup G_{\alpha_n}$. By De Morgan's law, $G_{\alpha_1} \cup \cdots \cup G_{\alpha_n} = (K_{\alpha_1} \cap \cdots \cap K_{\alpha_n})^c$. Therefore, we have $K_{\alpha_0} \cap K_{\alpha_1} \cap \cdots \cap K_{\alpha_n} = \emptyset$ and it is contradiction to our assumption that every finite subcollection of $\{K_\alpha\}$ is nonempty. $\square$

We use the reduction to absurdity. Assume that there is no limit point of $E$ in $K$. Then for each $q \in K$, there exists a neighborhood $N_{r_q}(q)$ such that $N_{r_q}(q) \cap E = \emptyset$ or $N_{r_q}(q) \cap E = \{q\}$. Obviously, $K \subset \bigcup_{q \in K} N_{r_q}(q)$. Thus the collection $\{N_{r_q}(q) : q \in K\}$ is an open cover of $K$. Since $K$ is compact, there exists $q_1,\ldots,q_n \in K$ such that $K \subset N_{r_{q_1}}(q_1) \cup \cdots \cup N_{r_{q_n}}(q_n)$.

Since $E \subset K$, $E \subset K \cap E \subset \left(N_{r_{q_1}}(q_1) \cup \cdots \cup N_{r_{q_n}}(q_n)\right) \cap E$. Finally, since $N_{r_q}(q) \cap E = \emptyset$ or $N_{r_q}(q) \cap E = \{q\}$, we have $E \subset \{q_1,\ldots,q_n\}$ and it is contradiction to the assumption that $E$ is infinite. $\square$

Since each $I_n$ is a closed interval, there exist $a_n \leq b_n$ such that $I_n = [a_n, b_n]$. Define $E = \{a_n : n \in \mathbb{N}\}$. Due to the assumption that $I_n \supset I_{n+1}$, for all positive integers $n, m$ we have $a_n \leq a_{n+m} \leq b_{n+m} \leq b_n$. In particular, $a_n \leq b_m$ for all $n, m \in \mathbb{N}$.

Hence for all $m \in \mathbb{N}$, $b_m$ is an upper bound of $E$ and by the least upper bound property of $\mathbb{R}$, there exists $\sup E$. Since each $a_n \in E$, obviously $a_n \leq \sup E$ for all $n \in \mathbb{N}$. Moreover, since each $b_n$ is an upper bound of $E$, we have $\sup E \leq b_n$ for all $n \in \mathbb{N}$. Therefore, $a_n \leq \sup E \leq b_n$ for all $n$ and it implies $\sup E \in [a_n, b_n] = I_n$ for all $n \in \mathbb{N}$. $\square$

Since each $I_n$ is a $k$-cell, there exist real numbers $a_{n,j} \leq b_{n,j}$ such that $I_n = [a_{n,1}, b_{n,1}] \times [a_{n,2}, b_{n,2}] \times \cdots \times [a_{n,k}, b_{n,k}]$. For each $n \in \mathbb{N}$ and $j = 1,2,\ldots,k$, we define $I_{n,j} := [a_{n,j}, b_{n,j}]$. Then $I_n = I_{n,1} \times I_{n,2} \times \cdots \times I_{n,k}$.

Moreover, due to the assumption $I_n \supset I_{n+1}$, for each $j$, we have $I_{n,j} \supset I_{n+1,j}$ for all $n \in \mathbb{N}$. Thus by Theorem 2.73, for each $j = 1,\ldots,k$, $\bigcap_{n=1}^{\infty} I_{n,j} \neq \emptyset$ and we can choose $x_j^* \in \bigcap_{n=1}^{\infty} I_{n,j}$. Set $x^* = (x_1^*,\ldots,x_k^*)$. Then $x^* \in \bigcap_{n=1}^{\infty} I_n$ and it implies that $\bigcap I_n \neq \emptyset$. $\square$

Let $I$ be a $k$-cell. Then by the definition, there exist $a_j \leq b_j$ ($j = 1,\ldots,k$) such that $I = [a_1, b_1] \times \cdots \times [a_k, b_k]$. Put $\delta = \left(\sum_{j=1}^{k}(b_j - a_j)^2\right)^{1/2}$. Then obviously $|x - y| \leq \delta$ for all $x, y \in I$.

We use the reduction to absurdity. Assume that $I$ is not compact. Then there exists an open cover $\{G_\alpha : \alpha \in A\}$ of $I$ which does not have a finite subcover of $I$. For $j = 1,\ldots,k$, put $c_j = (a_j + b_j)/2$. By subdividing each interval into $[a_j, c_j]$ and $[c_j, b_j]$, $I$ is divided into $2^k$ $k$-cells. Since $\{G_\alpha\}$ does not have a finite subcover of $I$, there exist a $k$-cell $I_1$ such that $I \supset I_1$, $I_1$ is not covered by any finite subcollection of $\{G_\alpha\}$, and $|x - y| \leq \delta/2$ for all $x, y \in I_1$.

Inductively, we can find $I_1, I_2, I_3, \ldots$ such that $I \supset I_1 \supset I_2 \supset I_3 \supset \cdots$, $I_n$ is not covered by any finite subcollection of $\{G_\alpha\}$, and $|x - y| \leq \delta/2^n$ for all $x, y \in I_n$, for all $n \in \mathbb{N}$.

By Theorem 2.75, there exists $x^* \in \bigcap_{n=1}^{\infty} I_n$. Thus $x^* \in I_n$ for all $n \in \mathbb{N}$. Since $\{G_\alpha\}$ is an open cover of $I$, there exists $G_\alpha$ such that $x^* \in G_\alpha$. Since $G_\alpha$ is an open set, there exists a neighborhood $N_r(x^*)$ such that $N_r(x^*) \subset G_\alpha$. Choose a $n \in \mathbb{N}$ large enough so that $2^{-n}\delta < r$. Then since $x^* \in I_n$, $|x^* - y| \leq \delta/2^n < r$ for all $y \in I_n$. Thus $I_n \subset N_r(x^*) \subset G_\alpha$. It is contradiction to the fact that $I_n$ is not covered by any finite subcollection of $\{G_\alpha\}$. $\square$