Compact set

Theorem 3.2.5

(a) $K$ is compact $\Rightarrow$ $K$ is closed and bounded

(b) $K$ is compact and $F$ is closed and subset of $K$, then $F$ is compact

<Proof> (a) Let $K \subset \mathbb{R}$ be compact. Construct an open cover as follows: \(\begin{align} O:=\{(-k,k): k \in \mathbb{N} \} \end{align}\) Then $O$ is an open cover of $\mathbb{R}$, i.e. $O$ is an open cover of $K\subset \mathbb{R}$. Since $K$ is compact, there are $k_1, \ldots, k_n \in \mathbb{R}$ such that $K \subset(-k_1, k_1) \cup \cdots \cup (-k_n, k_n)$. Then take $N$ as follows: \(\begin{align} N := \max \{k_1, \ldots, k_n \} \end{align}\) Then $K \subset [-N, N].$

$\therefore K$ is bounded

Now, we want to show that $K$ is closed. In other words, we want to show that $K^c$ is open. Let $p \in K^c$ be given. For every $q \in K$, take $\epsilon_q >0$ such that $N_{\epsilon_q}(q) \cap N_{\epsilon_q}(p) = \emptyset.$ Such $\epsilon_q$ does exist since we can take $\epsilon_q := \frac{|p-q|}{2}.$ Then

\[\begin{equation*} \{ N_{\epsilon_q}(q): q\in K\} \end{equation*}\]

is clearly an open cover of $K$. Since $K$ is compact, there are finite subcover such that $K \subset N_{\epsilon_{q_1}}(q_1) \cup \cdots \cup N_{\epsilon_{q_n}}(q_n).$ Then take $\epsilon >0$ as follows:

\[\begin{align} \epsilon := \min\{ \epsilon_{q_1},\ldots, \epsilon_{q_n}\} \end{align}\]

Then $N_\epsilon (p) \cap N_{\epsilon}(q_i) = \emptyset$ for all $i=1,\ldots, n$. Since $\bigcup_{i=1}^nN_{\epsilon_{q_i}}(q_i)$ is an open cover of $K, N_\epsilon (p) \subset K^c,$ i.e. $p \in \text{Int}(K^c)$.

$\therefore K^c$ is open, i.e., $K$ is closed.

\[\tag*{$\square$}\]

(b) Let $K$ be compact and let $F$ be a closed subset of $K$. Let

\[\begin{equation*} \{O_\alpha\}_{\alpha \in \Lambda} \end{equation*}\]

be an open cover of $F$.
Then

\[\begin{equation*} \{O_\alpha \}_{\alpha \in \Lambda} \cup \{F^c\} \end{equation*}\]

is an open cover of $K$ since $F^c$ is open. Since $K$ is compact, there are $\alpha_1, \ldots, \alpha_n \in \Lambda$ such that

\[\begin{align} K \subset O_{\alpha_1} \cup \cdots O_{\alpha_n} \cup F^c \end{align}\]

$\therefore F$ has a finite subcover

\[\begin{equation*} \{ O_{\alpha_1}, \ldots, O_{\alpha_n}\}. \end{equation*}\]

$\therefore F$ is compact

\[\tag*{$\square$}\]

Remark

One might think the condition that $F$ is closed is not necessary. But as shown in the previous proof, it is very crucial element of proof.

Suppose $E$ is an open cover of $F$. What we want to show is the existence of finite subcover of $E$. Given $E$, we add open sets $O_\alpha$ to E until it covers $K$. Since $K$ is compact, we can choose finite open sets $U_{\alpha_i}$ from the extended $E$ to cover $K$. Since $F$ is subset of $E$, $F$ is covered by the sets $U_{\alpha_i}$. But some $U_{\alpha_j}$ might not be in the original open cover $E$, we may choose $U_{\alpha_j}$ such that $U_{\alpha_j} \cap F$. But if $F$ is closed, we can choose finite subcover of $E$. Since $F^c$ is open, we can add $F^c$ to $E$ and never choose such $U_{\alpha_j}$. That is why the condition is necessary.

Theorem

Let $X$ be a subset of $\mathbb{R}$, $Y$ is a subset of $X$, and $K$ is a subset of $Y$. If $K$ is compact relative in $Y$ if and only if $K$ is compact relative in $X$.

<Proof>

$\Rightarrow$ Suppose that $K$ is compact in $Y$. Let \(\{ H_\alpha\}_{\alpha \in \Lambda}\) be an open cover of $K$ in $X$. Since $K\subset Y$ and $K\subset \bigcup_{\alpha\in\Lambda}H_\alpha$, $K\subset Y\cap (\bigcup_\alpha H_\alpha )=\bigcup_\alpha Y\cap H_\alpha$. Since $Y\cap H_\alpha$ is open in $Y$, \(\{Y\cap H_\alpha\}_{\alpha\in\Lambda}\) is open cover of $K$. Since $K$ is compact in $Y$, we can choose finite subcover of $K$, denoted as \(\{Y\cap H_{\alpha_i} \}_{i=1}^n\). Since $K\subset\bigcup_{i=1}^n Y\cap H_{\alpha_i} \subset \bigcup_{i=1}^nH_{\alpha_i}$, \(\{ H_{\alpha_i}\}_{i=1}^n\) is a finite open subcover of $K$.

$\therefore K$ is compact in $X$.

$\Leftarrow$ Suppose that $K$ is compact in $X$. Let \(\{U_\alpha\}_{\alpha\in\Lambda}\) is an open cover of $K$ in $Y$. Then for each $\alpha\in\Lambda$, $U_\alpha=Y\cap H_\alpha$ for some open $H_\alpha$ in $X$.

Since $K$ is compact in $X$ and $K\subset\bigcup_{\alpha\in\Lambda} H_\alpha$, we can choose finite subcover of $K$, denoted as \(\{H_{\alpha_i} \}_{i=1}^n\). Since $K\subset Y$ and \(\{ H_{\alpha_i}\}_{i=1}^n\) is a finite open subcover of $K$, $K\subset Y\cap \bigcup_{i=1}^nH_{\alpha_i}$. By de Morgan’s law, $Y\cap \bigcup_{i=1}^n H_{\alpha_i}=\bigcup_{i=1}^nY\cap H_{\alpha_i}$. Thus, $K\subset \bigcup_{i=1}^n Y\cap H_{\alpha_i}$.

$\therefore K$ is compact in $Y$.

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

Theorem 3.2.6 (Bolazno-Weierstrass property)

If S is an infinite subset of a compact set $K$, then $S$ has a limit point in $K$.

<proof>

If $S^\prime \cap K = \emptyset$, then for each $q\in K$, there exists $N_q$ of $q$ such that $N_q \cap S$ has at most one element.

Now, \(\{ N_q: q\in K\}\) is an open cover of $K$. Since $K$ is compact, there are $q_1, \ldots, q_n \in K$ such that $K \subset N_{q_1} \cup \cdots \cup N_{q_n}$. i.e. \(S \subset K \subset N_{q_1} \cup \cdots \cup N_{q_n}\) That is, $N_{q_i} \cap S$ is infinite for some $i$ because $S$ is infinite. However, it is a contradiction because

\[\begin{equation*} N_q \cap S = \{q \} \text{ or } \emptyset. \end{equation*}\]

$\therefore S^\prime \cap K \neq \emptyset.$

\[\tag*{$\square$}\]

Theorem

Let \(\{K_\alpha\}_{\alpha\in\Lambda}\) be a collection of compact sets. If every intersection of finite subcollection is not empty, then $\bigcap_{\alpha\in\Lambda}K_\alpha$ is not empty.

<proof> Proof by a contradiction. Suppose that $\bigcap_{\alpha\in\Lambda}K_\alpha=\emptyset$. In other words, $\bigcup_{\alpha\in\Lambda}K^c_\alpha=X$, where the set $X$ is universal set. Then choose a set from \(\{K_\alpha\}_{\alpha\in\Lambda}\), call it $K_{\alpha_0}$. Since $K_{\alpha_0}$is compact and \(\{K^c_\alpha\}_{\alpha\in\Lambda}\) is open cover of $K_{\alpha_0}$, we can choose finite subcover such that $K_{\alpha_0}\subset \bigcup_{i=1}^nK^c_{\alpha_i}$. That is $K_{\alpha_0}\subset (\bigcap_{i=1}^nK_{\alpha_i})^c$. Thus, $\bigcap_{i=0}^n K_{\alpha_i}=\emptyset$. It is a contradiction to the assumption that every intersection of finite subcollection is not empty.

$\therefore \bigcap_{\alpha\in\Lambda}K_\alpha \neq \emptyset$.

\[\tag*{$\square$}\]

Theorem 3.2.7

\(\{K_n \}\) is a sequence of nonempty compact subsets of $\mathbb{R}$ such that $K_n \supset K_{n+1}$. Then $K:= \bigcap_{n=1}^\infty K_n$ is non-empty compact set.

<proof> Let $O_n = K^c_n$. Then $O_n$ is open because $K_n$ is closed. So it is equivalent that

\[\begin{align} \bigcap_{n=1}^\infty K_n = \emptyset \iff \bigcup_{n=1}^\infty O_n = \mathbb{R} \end{align}\]

Suppose that $\bigcap_{n=1}^\infty K_n = \emptyset$. Then \(\{O_n \}_{i=1}^\infty\) is an open cover of $\mathbb{R}$ and is an open cover of $K_1$. Since $K_1$ is compact, there exists $n_1 <\cdots <n_k$ such that

\[K_1 \subset O_{n_1} \cup \cdots \cup O_{n_k}\]

Since $O_{n_i} = K_{n_i}^c$, $K_1 \cap K_{n_1} \cap \cdots \cap K_{n_k} = \emptyset$. But it implies that $K_{n_k}$ is an empty set, which contradicts to the assumption.

$\therefore \bigcap_{n=1}^\infty K_n \neq \emptyset$

Next, we want to show that $K$ is compact. By the theorem 3.2.5, every $K_n$ is closed. Since arbitrary intersection of closed sets is also closed, $K$ is also closed. Moreover, $K\subset K_1$ and $K_1$ is compact.

$\therefore K$ is compact

\[\tag*{$\square$}\]

Theorem 3.2.8 (Heine-Borel)

Every closed and bounded interval $[a,b]$ is compact.

<Proof> Define an open cover of $[a,b]$ and a set $E$ as follows:

\[\begin{align} \begin{split} \mathcal{U} & :=\{U_\alpha \}_{\alpha \in \Lambda} \\ E&:= \{r \in [a,b]: [a,r] \text{ is covered by a finite number of the sets }U_\alpha\} \end{split} \end{align}\]

Since $E$ is bounded above by $b$ and $E$ is a non empty set ($\because a \in E $), there is $c=\sup E$ by the least upper bound property. Then $c \leq b$.

Since $c\in [a,b]$, $c\in U_\beta$ for some $\beta \in \Lambda$.
Since $U_\beta$ is open, there is $\epsilon >0$ such that $N_\epsilon (c) \subset U_\beta.$

Now, $c-\epsilon$ is not an upper bound of $E$, so there exists $r\in E$ such that $c-\epsilon < r \leq c.$ So, there are $\alpha_1, \ldots, \alpha_k \in \Lambda$ such that $[a,r] \subset \bigcup_{i=1}^k U_{\alpha_i}$. That is $[a,c]$ is covered by the finite subcover as follows:

\[\begin{align} [a,c] \subset \bigcup_{i=1}^k U_{\alpha_i} \cup U_\beta \end{align}\]

$\therefore c \in E$

Now, we want to show that $c=b$. Suppose that $c
\lneq b$. Then we can choose $c<s<b$ such that $c<s<c+\epsilon$. Then, \(\begin{align} [a,s] \subset \bigcup_{i=1}^k U_{\alpha_i} \cup U_\beta \end{align}\) $\therefore s \in E$, which contradicts the assumption that $c=\sup E$.

$\therefore c=b$

$\therefore b \in E$.

$\therefore [a,b]$ is compact.

\[\tag*{$\square$}\]

Theorem 3.2.8-2 (Heine-Borel)

All $k$-cells, $I=\prod_{i=1}^k[a_i,b_i]$ are compact where $a_i, b_i \in \mathbb{R}$.

<Proof>

Suppose that there is an open cover

\[\begin{equation*} S=\{ G_\alpha\} \end{equation*}\]

such that there is no finite subcover of $S$ In that case, dyadically divide $I$ into $2^k$ $k$-cells. At least one cell of $I$ cannot be covered with finite subcover of $S$. Iterating, we obtain a nested sequences of $k$-cells $(I_n)_{n=1}^\infty$ such that

1. diam$(I_n)\leq 2^{-n}\delta, \delta=\sqrt{\sum_{i=1}^k(b_i-a_i)^2}$
2. $\exists x_0\in \bigcap_{n=1}^\infty I_n$

Since $S$ is an open cover of $I$, $x_0\in G_\alpha$ for some $\alpha$. Since $G_\alpha$ is open, there is $\epsilon>0$ such that $N_\epsilon (x_0)\subset G_\alpha$. By Archimedean property, we can choose big enough $n\in\mathbb{N}$ such that $2^{-n}\delta <\epsilon$. Then $I_n \subset G_\alpha$, which is a contradiction.

\[\tag*{$\square$}\]

Theorem 3.2.9 (Heine-Borel-Bolzano-Weierstrauss)

Let $K$ is a subset of $\mathbb{R}$. Then the followings are equivalent (TFAE)

(a) $K$ is closed and bounded

(b) $K$ is compact

(c) Every infinite subset of $K$ has a limit point in $K$.

<Proof>

(a) $\Rightarrow$ (b): Since $K$ is bounded, $K\subset [-M,M]$ for some $M>0$. Now, $[-M,M]$ is compact and $K$ is closed. By the theorem 3.2.5, $K$ is compact.

(b) $\Rightarrow$ (c): See the theorem 3.2.6

(c) $\Rightarrow$ (a) Suppose that $K$ is not bounded. Then for all $n \in \mathbb{N}$, there is $p_n \in K$ s.t. $p_n \neq p_m$ for $n \neq m$ and $|p_n| >n$.

Then \(\begin{align} S=\{p_n: n \in \mathbb{N} \} \end{align}\) is an infinite subset of $K$. But $S$ does not have limit point in $K$. Let $p\in K$ be given. Take $r \in \mathbb{N}$ such that $r>|p|+1$. Then there are $r-1$ points of $S$ in $N_r(0)$. Then take

\[\begin{equation*} \epsilon = \min \{ |p-p_n|: n=1,\ldots, r-1\} \end{equation*}\]

Then $N^\prime_\epsilon(p) \cap S=\emptyset.$ Thus $p$ is not a limit point of $E$. By our assumption, $E$ must have a limit point in $K$, which is a contradiction.

$\therefore K$ is bounded.

Let $p$ be a limit point of $K$.

Then for each $n\in \mathbb{N}$, there is $p_n \in K \text{ with } p_n \neq p \text{ such that } |p_n-p| < \frac{1}{n}$.

Then \(\begin{align} S := \{p_n: n\in \mathbb{N}\} \end{align}\) is an infinite subset of $K$. By our assumption, $S^\prime \cap K \neq \emptyset$.

Suppose that $q\in \mathbb{R}$ with $q\neq p$. Put
\(\begin{align} \epsilon := \frac{1}{2}|p-q| \end{align}\) Choose $N\in \mathbb{N}$ such that $\frac{1}{N} <\epsilon$. If $n \geq N$, then \(\begin{align} \begin{split} |p-q| &\leq |p_n-p| + |p_n-q| \\ &<|p_n -q| + \frac{1}{n} \\ &<|p_n -q| + \epsilon \end{split} \end{align}\) That is $|p_n-q| > |p-q| - \epsilon = \frac{1}{2}|p-q|$.

$\therefore q \notin S^\prime$

$\therefore p$ is a unique limit point of $S$.

$\therefore p\in S^\prime \cap K$

$\therefore p\in K$

$\therefore K$ is closed.

Another proof for closedness of $K$.

Suppose that $K$ is not closed. Then there is $x_0\in K^c$ such that $x_0$ is a limit point of $K$. Then, for every $n\in\mathbb{N}$, there is $a_n\in N_{1/n}(x_0)\cap K$. We want to say that $x_0$ is the unique limit point of

\[\begin{equation*} E=\{a_n\}. \end{equation*}\]

Let $\epsilon>0$ be given. By Archimedean property, we can take $n\in\mathbb{N}$ such that $\frac{1}{n}< \epsilon$. Then

\(\begin{equation*} 0<|a_n-x_0|<\frac{1}{n}<\epsilon. \end{equation*}\) since $x_0 \notin K$ and $a_n\in K$. In other words, $N^\prime_\epsilon(x_0)\cap E=\emptyset.$ Now let $y\in K$ be given. With triangular inequality, we get the following inequality with big enough $n\in\mathbb{N}$.

\[\begin{align*} |a_n-y| &\geq |x_0-y|-|x_n-a_n| \\ &>|x_0-y| - \frac{1}{n} \\ &>|x_0-y| - \frac{1}{2} |x_0-y| \\ &=\frac{1}{2}|x_0-y| \end{align*}\]

That is for large enough $n$, there are finite points of $E$ around $y$. Thus $y$ cannot be a limit point of $E$.

Now $x_0$ is the unique limit point of $E$ but $x_0\notin K$. But it contradicts to the assumption that every infinite subset of $K$ has a limit point in $K$.

$\therefore K$ is closed.

\[\tag*{$\square$}\]

Theorem

Let $E$ be a non-empty subset of $\mathbb{R}$. If the set $E$ is bounded above, $\sup E\in \overline{E}$.

<Proof>

If $\sup E \in E$, we are done. Suppose that $\sup E \notin E$. We want to show that $\sup E \in E^\prime$. Let $p=\sup E$. Suppose that $\sup E \notin E^\prime$. Then there is $\epsilon>0$ such that $N_\epsilon(p)\cap E=\emptyset$.

Since $p$ is the least upper bound, there is some $x\in E$ such that $p- \frac{\epsilon}{2} < x \leq p$. But it implies that $x\in N_\epsilon (p)$, which is a contradiction that $p$ is a isolated point. Thus $\sup E \in E^\prime.$

$\therefore \sup E\in \overline{E}$.

\[\tag*{$\square$}\]