# Open and closed set-(2)

## Theorem 3.1.13

$U \subset \mathbb{R}$ is open $\Rightarrow$ \(\exists \{I_n\}\): a finite or countable family of pairwise disjoint union of open intervals such that $U=\cup_n I_n.$

<*proof*>
Let $x \in U$ be given. Since $U$ is open, there is $\epsilon >0$ such that $(x-\epsilon, x+\epsilon) \subset U.$ Define sets and supremum/infimum as follows

We claim that $I_x= (l_x, r_x)\subset U$ and for $x,y \in U, I_x = I_y$ or $I_x \cap I_y =\emptyset.$

Suppose that $(l_x, r_x) \not\subset U.$ Without loss of generality, assume that $[x, r_x) \not\subset U,$ i.e. $r_x - \delta \not\in U$ with $x<r_x-\delta$ for some $\delta>0.$ Since $r_x = \sup A$, there exists $t_0\in A$ such that $r_x-\delta < t_0 \leq r_x,$ i.e. $[x, r_x-\delta) \subset [x, t_0) \subset U$.

It contradicts to the assumption that $r_x -\delta \not\in U$.

$\therefore [x, r_x) \subset U.$ Similarly, we can show that $(l_x, x] \subset U.$

$\therefore I_x = (l_x, r_x) \subset U.$

Now we want to show that for $x,y\in U$, $I_x = I_y$ or $I_x \cap I_y =\emptyset.$ Without the loss of generality assume that $x<y$. Suppose that $I_x \neq I_y$ and $I_x \cap I_y \neq \emptyset.$ Since $I_x \cap I_y \neq \emptyset$ and $x<y$, $l_y < r_x.$ Since $(l_x, r_x) \subset U, (l_x, l_y)\subset U.$

\[\begin{align} \begin{split} C &:=\{q: q>y \text{ and } [y,q)\subset U \}\\ D &:= \{p: p<y \text{ and } [p,y) \subset U \}\\ r_y &:= \sup C \\ l_y &:= \inf D \end{split} \end{align}\]For any $k\in (l_x, l_y), k\in U.$ i.e. $k\in D.$ If $l_y = l_x$, it is trivial case.

Even if $l_x\neq l_y$, however, we can show that $\inf D = l_y = l_x.$ Suppose that $\inf D = l_y \neq l_x$. Then for any $q \in (l_x, l_y), \inf D \leq q < l_y$, which contradicts to the assumption that $l_y = \inf D$.

$\therefore l_x = l_y.$

Similarly, we can prove that $(r_x, r_y) \subset U$ and $r_x = r_y.$

For any $r \in (r_x, r_y)$, $r \in U.$ i.e. $r\in A.$ Even if $r_x \neq r_y$, we can show that $\sup A = r_x = r_y$.

Suppose that $\sup A = r_x \neq r_y.$ Then for any $r \in (r_x, r_y), \sup A \geq r$ because $r \in (r_x, r_y) \subset (r_x, r_y)\subset A$. But it implies that $\sup A \geq r > r_x$, which contradicts to the assumption that $\sup A = r_x$.

$\therefore r_x=r_y$.

$\therefore I_x = (l_x, r_x) = (l_y, r_y) =I_y$, which contradicts to the assumption that $I_x \neq I_y.$

$\therefore I_x = I_y \text{ or } I_x \cap I_y = \emptyset.$

Lastly, we want to show that there is one-to-one correspondence between \(\mathcal{L} :=\{I_x: x\in U \} \text{ and } \mathbb{Q}.\) For each $I \in \mathcal{L}$, choose $r_l \in \mathbb{Q}$ such that $r_l \in I.$ Such $r_l$ exists because the rational number is dense in $\mathbb{R}.$ Then for two distinct two intervals $I, J \in \mathcal{L}, r_l \neq r_J$ because $I\cap J = \emptyset.$ Thus the collection of $\mathcal{L}$ is at most countable. There fore $\mathcal{L}$ can be enumerated as \(\{I_j\}_{j\in A}\) where cardinality of $A$ is at most equal to the cardinality of natural number $\mathbb{N}.$ Clearly, $U = \cup_{j\in A} I_j$. \(\tag*{$\square$}\)

## Definition 3.1.14

$X$ is a subset of $\mathbb{R}$.

- $U \subset X$ is
*open in $X$*if $\forall p \in U, \exists \epsilon >0$ such that $N_\epsilon (p) \cap X \subset U.$ - $C \subset X$ is
*closed in $X$*if $X\setminus C$ is open in $X$.

## Example 3.1.15

$X=[0,\infty], U=[0,1]$. $U$ is open in $X$ because $(-1, 1) \cap X = U.$ However, $U$ is not open in $\mathbb{R}.$

## Theorem 3.1.16

(a) $U \subset X$ is open in $X$ $\iff U = X \cap O$ for some open set $O$ in $\mathbb{R}.$

(b) $C \subset X$ is closed in $X \iff C = X \cap F$ for some closed set $F$ in $\mathbb{R}.$

<*proof*>
(a) $\Rightarrow$ Since $U$ is open in $X$, $\forall p\in U, \exists\epsilon_p >0$ such that $N_{\epsilon_p}(p) \cap X \subset U.$ Define a set $O := \bigcup_{p \in U} N_{\epsilon_p}(p).$ Since $N_{\epsilon_p}(p)$ is open, arbitrary union of open set is open. Thus, $O$ is open. Since for every $p\in U, N_{\epsilon_p}(p) \cap X \subset U$, clearly $O\cap X \subset U.$ Conversely, for every $p \in U, p\in X$ and $p\in N_{\epsilon_p}(p)$. Thus, $p \in N_{\epsilon_p}(p) \cap X$, i.e. $U\subset N_{\epsilon_p}(p) \cap X.$

$\therefore U = O \cap X.$

$\Leftarrow$ Suppose that there is open set $O$ in $\mathbb{R}$ such that $U=O\cap X.$ Let $p\in O\cap X$ be given. Since $p\in O$ and $O$ is open, there is $\epsilon >0$ such that $N_\epsilon (p) \subset O.$ Then $N_\epsilon (p) \cap X \subset O\cap X = U.$

$\therefore U$ is open in $X$.

(b) Suppose that $C$ is closed in $X$. That is $X \setminus C$ is open in $X$. By (a), there is an open set $O$ in $\mathbb{R}$ such that $X\setminus C = X\cap O.$ Define $F := \mathbb{R} \setminus O = O^c.$ Clearly, $F$ is closed in $\mathbb{R}.$

\[\begin{align} \begin{split} X\cap F &= X \cap O^c \\ &= X\setminus (X\cap O)\\ &= X \cap (X\cap C^{c}) ^c\\ &= X \cap (X^c \cup C) \\ &=(X\cap X^c) \cup (X\cap C) \\ &= X\cap C \\ &= C \end{split} \end{align}\]$\Leftarrow$ Suppose that $C=X\cap F$ for some closed $F$ in $\mathbb{R}$. We want to show that $X\setminus C$ is open in $X$.

\(\begin{align} \begin{split} X\setminus C &= X \setminus (X\cap F)\\ &= X \cap (X\cap F)^c \\ &= X \cap (X^c \cup F^c)\\ &= (X\cap X^c) \cup (X\cap F^c) \\ &= X\cap O \end{split} \end{align}\) By (a) $X\setminus C$ is open in $X$.

$\therefore C$ is closed in $X$.

\[\tag*{$\square$}\]## Definition 3.1.17

1) Let $A,B\subset \mathbb{R}$ be given. If $A\cap \overline{B}=\emptyset$ and $\overline{A}\cap B=\emptyset$, we say $A$ and $B$ are separated.

2) If $E\subset\mathbb{R}$ is **connected** if there is no $A, B\subset \mathbb{R}$ such that $A$ and $B$ are separated and $E=A\cup B$. Otherwise $E$ is disconnected.

## Lemma 1

Let $A,B$ be subsets of $\mathbb{R}$. Then $\overline{A\cap B} \subset \overline{A}\cap \overline{B}$.

<*Proof*>

Let $p\in A\cap B$ be given. Since $p\in A, p\in \overline{A}=A\cup A^\prime.$ Similarly, since $p\in B, p\in \overline{B}$. Thus, $p\in \overline{A}\cap \overline{B}$. Then, let $p$ be a limit point of $A\cap B$. For every $\epsilon >0$, $N^\prime_\epsilon (p)\cap (A\cap B) \neq \emptyset$. In other words, $N^\prime_\epsilon (p)\cap A \neq \emptyset$ and $N^\prime_\epsilon (p) \cap B\neq \emptyset$. Thus, $p$ is a limit point of $A$ and $B$, i.e., $p \in \overline{A}\cap \overline{B}$.

$\therefore$ For any $p\in \overline{A\cap B}, p\in \overline{A}\cap \overline{B}$.

$\therefore \overline{A\cap B}\subset \overline{A}\cap \overline{B}.$

\(\tag*{$\square$}\)

## Theorem 3.1.18

$E \subset \mathbb{R}$ is connected $\iff$ for any $x,y\in E$ with $x<y, (x,y)\subset E.$

<*Proof*>

$\Rightarrow$ Suppose that $x,y\in E$ but there exists $z\in \mathbb{R}$ such that $x<z<y$ and $z\notin E$. Define sets

\(\begin{equation*} A_z = E\cap (-\infty, z), B_z=E\cap (z,+\infty). \end{equation*}\) Then $E= A_z\cup B_z$ but $A_z$ and $B_z$ are separated, which contradicts to the assumption that $E$ is connected.

$\Leftarrow$ Suppose that for any $x,y\in E$ with $x<y, (x,y)\subset E.$ Assume that $E$ is disconnected. Then there are $A,B\subset E$ such that $A\cap \overline{B}=\emptyset$, $\overline{A}\cap B=\emptyset$, and $E=A\cup B$. Pick $x\in A$ and $y\in B$ with $x<y$. Consider $z:=\sup\left(A\cap [x,y] \right)$. Since $A\cap [x,y]$ is bounded above, $z\in \overline{A\cap [x,y]}$. By Lemma 1, $z\in \overline{A}$. Since $\overline{A}\cap B=\emptyset$, $z\notin B$. Since $x\in A$ and $y\in B$,

\[\begin{equation*} x \leq z < y \end{equation*}\]If $z\notin A$, then $x<z<y$ and $z\notin A\cup B=E$, which contradicts to the assumption that $z\in (x,y)\subset E$.

If $z\in A$, then $z\notin \overline{B}$ since $A\cap \overline{B}=\emptyset$. Since $z$ is not a limit point of $B$, there exists $\epsilon_0>0$ such that $N^\prime_{\epsilon_0}(z)\cap B=\emptyset$. So we can pick $z_1\notin B$ such that $z<z_1 < y$. Then we get,

\[\begin{equation*} x < z_1 <y \end{equation*}\]and $z_1 \notin A$. We can show it by a contradiction. Suppose that $z_1 \in A$. Then $z_1 \in A\cap [x,y]$. Since $z=\sup A\cap [x,y]$, $z_1 \leq z$ which is a contradiction. Thus $z_1 \notin A$.

Since $z_1 \notin A$ and $z_1 \notin B$, $z_1\notin A\cup B=E$, which is a contradiction that $z\in (x,y)\subset E$.

## Definition 3.2.1

\(E \subset \mathbb{R}, \{O_\alpha\}_{\alpha \in \Lambda}\) is a family of open sets. \(\{O_\alpha\}_{\alpha \in \Lambda}\) is an *open cover of $E$* if $E \subset \cup_{\alpha \in \Lambda}O_\alpha.$

## Definition 3.2.3

$K \subset \mathbb{R}$ is *compact* if every open cover of $K$ has a finite subcover of $K$, i.e. \(\forall \{O_\alpha \}_{\alpha \in \Lambda}\) is an open cover of $K, \exists \alpha_1, \ldots, \alpha_k \in \Lambda$ such that $K \subset O_{\alpha_1} \cup \cdots \cup O_{\alpha_k}.$

## Example 3.2.4

(a) Every finite set is compact. Let \(E=\{p_1, \ldots, p_n\}\) be a subset of $\mathbb{R}$. Let \(\{O_\alpha\}_{\alpha \in \Lambda}\) be an open cover of E. For each $p_i \in $, we can find $O_{\alpha_i}$ such that $p_i \in O_{\alpha_i}$. Then $E \subset O_{\alpha_1} \cup \cdots \cup O_{\alpha_n}$.

$\therefore E$ is compact.

\[\tag*{$\square$}\](b) $(0,1)$ is not compact. $I_n := (0, 1-\frac{1}{n}).$ Then \(\{I_n\}_{n\geq2}\) is an open cover of $(0,1)$ because $(0,1) = \cup_{n\geq 2} I_n$. Suppose that there are $I_{n_1}, \ldots, I_{n_k}$ such that $(0,1) \subset \cup_{i=1}^k I_{n_i}.$ Put \(N:=\max\{n_1, \ldots, n_k\}.\) Then $\cup_{i=1}^k I_{n_i} = I_N = (0, 1-\frac{1}{N}) \subsetneq (0,1)$, which contradicts to the assumption.

$\therefore (0,1)$ is not compact.

## Reference

- Manfred Stoll,
**『**Introduction to Real Analysis**』**, Pearson