NoteNextra-origin/content/Math4111/Math4111_L9.md

Lecture 9

Review

1. Let X=\mathbb{R} (and as usual, let d(x,y)=|x-y|). What is the set B_1(0) B_1(0)=(-1,1)
2. Let X=[0,5] (and let d be as usual). What is the set B_1(0). B_1(0)=[0,1)
3. Let X=\mathbb{R} and let E=[0,2). Is E open? No, 0 is not a interior point.
4. Let X=[0,5] and let E=[0,2). Is E open? Yes, 0 is a interior point, we can set radius to 1 and all the points of B_1(0)\subset E

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Metric space

Theorem 2.27

If X is a metric spae and E\subset X, then.

1. \bar{E} is closed.
2. E=\bar{E} if and only if E is closed. Proof: E=\bar{E}\iff E=E\cup E'\iff E'\subset E\iff E is closed.
3. \bar{E}\subset F for every closed set F\subset X such that E\subset F. F\subset X closed, and E'\subset F'\subset F, so \bar{E}=E\cup E'\subset F

Theorem 2.28

E\subset \mathbb{R} non-\phi and bounded above, \sup E\in \bar{E}

Proof:

Let y\sup E, To show y\in \bar{E}, we need to show \forall h>0, B_h(y)\cap E\neq \phi

Let h>0. Since y-h is not an upper bound of E, \exists x\in E such that x>y-h.

Since y is an upper bound of E, x\leq y. So x\in B_n(y)\cap E, so B_h(y)\cap E\neq \phi.

QED

Remark 2.29

Let (X,d) be a metric space, E\subset X. "E is open" is short for "E is open in $X$"/E is open relative to X.

This means \forall p\in E, \exists r>0\{q\in X:d(p,q)<r\}\subset E.

Suppose E\subset Y\subset X. \exists r>0\{q\in Y:d(p,q)<r\}\subset E.

Example: X=\mathbb{R}, Y=[0,5], E=[0,2), E is open in Y, E is not open on X.

Theorem 2.30

Let E\subset Y\subset X,then E is open in Y\iff \exists G open in X such that E=Y\cap G

Proof:

Observe. If p\in Y, r>0, then \{q\in Y:d(p,q)<r\}=\{q\in X:d(p,q)<r\} (ball in X and ball in Y).

\impliedby

Suppose \exists G open in X such that E=Y\cap G. Let p\in E. Then p\in G so \exists r>0 such that \{q\in X:d(p,q)<r\}\subset G intersect both sides with Y to get \{q\in Y:d(p,q)<r\}\subset G\cap Y=E

\implies

Suppose E is open in Y. Then \forall p\in E,\exists r_p>0 such that \{q\in Y:d(p,q)<r_p\}\subset E

Let V_p=\{q\in X:d(p,q)<r_p\}, G=\bigcup_{p\in E}V_p.

Each V_p is open in X so G is open in X. We need to show E=G\cap Y.

To show E\subset G\cap Y. p\in E\implies p\in V_p\implies p\in G. So E\subset G. Also, by assumption, E\subset Y.

To show G\cap Y\subset E.

G\cap Y=\left(\bigcup_{p\in E}V_p\right)\cap Y=\bigcup_{p\in E}(V_p\cap Y)\subset E

QED

Compact sets

Definition 2.31/2.32

Let (X,d) be a metric space K\subset X.

Let A be an index set. Suppose \forall \alpha\in A, G_\alpha is an open set (in X) and suppose K\subset \bigcup_{\alpha\in E}G_\alpha. Then we say \{G_\alpha:\alpha\in A\} is an open cover of K.

Let \{G_\alpha\}_{\alpha \in A} be an open cover of K. Suppose \exists \alpha_1,...,\alpha_n\in A such that K\subset\bigcup_{i=1}^n G_{\alpha_i}. Then we say \{G_{\alpha_i}\}_{i=1}^n is a finite subcover of \{G_\alpha\}_{\alpha\in A} of K.

Definition

We say K is compact if every open cover of K contains a finite subcover.

i.e. \forall (This is important) open cover \{G_\alpha\}_{\alpha\in A} of K, \exists finite subcover \{G_{\alpha_i}\}_{i=1}^n of \{G_\alpha\}_{\alpha\in A} of K.

Examples: X=\mathbb{R}.

1.$K=\mathbb{R}$. \mathbb{R} is not compact.

as we can build an infinite open cover \bigcup_{i\in Z} (i,i+2) and it does not have a finite subcover because:

Suppose we consider the sub-collection \{n_i,n_i+2:i=1,..,k\}, Then N+3 is not in the union, where N=max\{n_1,...,n_k\}.

QED