Lecture 11

Recall

$K$ is compact if $\forall$ open cover $\{G_{\alpha}\}_{\alpha\in A}$ of $K$, $\exists$ a finite subcover $\{G_{\alpha_i}\}_{i=1}^n$ (We can only start proof from the cover of our desired set)

Let $(X,d)$ be a metric space. Consider the following statement: If $K$ is compact and $p\in X$, then $K\cup \{p\}$ is compact.

1. To give a proof of the statement, we start with “Suppose $K$ is compact and $p\in X$.” What MUST be the next step in the proof be?
   Let $\{G_{\alpha}\}_{\alpha\in A}$ be any open cover of $K\cup \{p\}$. (Since we want to show $K\cup \{p\}$ is compact)
2. Complete the proof of the statement.
   Since $K\subset K\cup \{p\} \subset \{G_{\alpha}\}_{\alpha\in A}$ and $K$ is compact, then $\exists$ a finite subcover $\{G_{\alpha_i}\}_{i=1}^n$ relative to $X$.
   And $p\in K\cup \{p\}$, $\exists \beta \in A$ such that $p\in\beta$, such that $C=\{G_{\alpha_i}\}_{i=1}^n+\beta$. And $C$ is a be So $K\cup \{p\}$ is compact.
   Any sets has an open cover
3. Suppose $F\subset K\subset X,K$ is compact, and $F$ is closed (in $X$). Prove that $F$ is compact. [Hint: the proof structure is similar to 2.]
   By Theorem 2.33

New Materials

Compact sets

Theorem 2.35

If $F\subset K\subset X$, and $F$ is closed (relative to $X$), and $K$ is compact, then $F$ is compact.

Proof:

Let $\{G_{\alpha}\}_{\alpha\in A}$ be an open cover of $F$. (Since we want to show $F$ is compact)

And $(\bigcup_{\alpha\in A} G_\alpha)\cup F^c=X\supset K$ is an open cover of $K$. Since $K$ is compact, then this open cover has a finite subcover $\Phi\subset (\bigcup_{\alpha\in A} G_\alpha)\cup F^c$ of $K$.

Since $F\subset K$, So $\Phi$ is a cover of $F$ then $\Phi\backslash \{F^c\}$ is a finite subcover of $\{G_\alpha\}_{\alpha \in A}$ of $F$.

QED

Corollary 2.35

If $F$ is closed and $K$ is compact, then $F\cap K$ is closed.

Corollary 2.36 From Theorem 2.36

If $K_1\supset K_2\supset K_3$ is sequence of nonempty compact sets, Then $\bigcap ^{\infty}_1 K_n$ is not empty.

Proof:

We proceed by contradiction. Suppose $\bigcap ^{\infty}_1 K_n=\phi$

Then $\bigcap^{\infty}_{n=1} K^c_n=(\bigcup^{\infty}_{n=1} K_n)^c=X\supset K_i$, Since $K_n$ is compact, $K_n$ is closed. $K_n^c$ is open.

So $\{K_n^c\}_{n\in \mathbb{N}}$ is an open cover of $K_n$. By compactness of $K_1$, $\exists$ a finite subcover $\{K^c_{n_1},...,K^c_{n_m}\}$ of $K_1$. So

Let $l=\max\{n_1,...,n_m\}$, then $K_l=\bigcap^{m}_{i=i} K_{n_i}$. So $K_1\subset K_l^c$, so $K_1\cap K_l=\phi$

which contradicts with $K_l\subset K_1$

QED

Theorem 2.37

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

Proof:

We’ll prove the following equivalent statement (contrapositive).

If $K$ is compact, $E\subset K$, and $E'\cap K=\phi$, then $E$ is finite

Suppose $K$ is compact, $E\subset K$ , $E'\cap K=\phi$.

For each $q\in K, q\notin E'$, so $\exists$ neighborhood $\forall q\in B_{r_q}(q)$ such that $V_q\cap E\backslash \{q\}=\phi$ (By $E'\cap K=\phi$).

Then $\{V_q\}_{q\in K}$ is an open cover of $K$ , so it has a finite subcover $\{V_{q_i}\}^n_{n=1}$. Then $E\subset K\subset \bigcup_{i=1}^n$, and $\forall i,V_{q_i}\cap E\subset \{q_i\}$, then $E\subset\{q_1,...,q_n\}$

QED

Theorem 2.38

If $I_1,I_2,...$ is a sequence of closed and bounded intervals and $I_1\subset I_2\subset I_3\subset ...$, then $\bigcap{}^{\infty}_{n=1} I_n\neq \phi$

Proof:

Let $E=\{a_n:n\in \mathbb{N}\}$, $E$ is non empty (Since $a_i\in E$)

We claim $\forall m\in \mathbb{N}$, $b_m$ is an upper bound of $E$, i.e. $\forall m\in \mathbb{N},\forall n\in \mathbb{N},a_n\leq b_m$.

To see this:

Let $x=sup(E)$, we claim $x\in \bigcap_{n=1}^{\infty} I_n$, i.e $\forall n\in \mathbb{N}, a_n\leq x\leq b_n$

Fix $n\in \mathbb{N}$.

Since $x$ is an upper bound of $E$ and $a_n\in E$, then $a_n\leq x$.

Since $x$ is the least upper bound of $E$, and $b_n$ is an upper bound of $E$, then $x\leq b_n$. $x\in E,E\neq \phi$

QED