NoteNextra-origin/pages/Math4111/Math4111_L11.md

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.

EOP

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

K_1\subset \bigcup^m_{i=1} K_{n_i}^c =\left( \bigcap^{m}_{i=i} K_{n_i} \right)^c

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

EOP

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\}

EOP

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:

a_{n}\leq a_{n+m}\leq b_{n+m}\leq b_{n}

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

EOP