NoteNextra-origin/pages/Math4111/Math4111_L8.md

Lecture 8

Review

Let (X,d) be a metric space. Recall that B_r(x)=\{z\in X:d(x,z)<r\}.

Let x,y\in X and let r=\frac{1}{2}d(x,y). What do you think is true about B_r(x)\cap B_r(y)? Can you prove it?

It should be empty. Proof any point cannot be in two balls at the same time. (By triangle inequality or contradiction)

Metric space defs

1. p\in X,r>0, B_r(p)=\{q\in X:d(p,q)<0\}, also called neighborhood.
2. p is a limit point of E(p\in E') if \forall r>0, (B_s(p)\cap E)\backslash \{p\}\neq \phi
3. If p\in E and p is not a limit point of E, then p is called an isolated point of E.
4. E is closed if E'\subset E
5. p is a interior point of E(p\in E^{\circ}) if \exists r>0 such that B_r(p)\subset E.

New materials

Metric space

Theorem 2.20

p\in E'\implies \forall r>0,B_r(p)\cap E is infinite.

Proof:

We will prove the contrapositive.

want to prove \exists r>0 such that B_r(p)\cap E is finite \implies p\notin E' (\exists s>0 such that (B_s(p)\cap E)\backslash \{p\}=\phi)

Suppose \exists r>0 such that B_r(p)\cap E is finite

let B_s(p)\cap E)\backslash \{p\}={q_1,...,q_n}

• If n=0, then B_s(p)\cap E)\backslash \{p\}=\phi, so p\in E'

• If n\geq 1, then let s=min\{d(p,q_m):1\leq m\leq n\}

  Each d(p,q_m) is positive and the set is finite, so s>0.

Then (B_s(p)\cap E)\backslash \{p\}=\phi, so p\notin E

EOP

Theorem 2.22 De Morgan's law

\left(\bigcup_a E_a\right)^c=\bigcap_a(E^c_a)

E^c=X\backslash E

Proof:

x\in \cup_{a\in A} E_x\iff \exists a\in A such that x\in E_a

So x\in \left(\bigcup_a E_a\right)^c\iff \forall a\in A, x\notin E_a\iff \forall a\in A,x\in E_a^c\iff \bigcap_a(E^c_a)

Theorem 2.23

E is open \iff E^c is closed.

Warning: E is open \cancel{\iff} E is closed. E is closed \cancel{\iff} E is open.

Example:
\phi, \R is both open and closed. "clopen set"
[0,1) is not open and not closed. bad...

Proof:

\impliedby Suppose E^c is closed. Let x\in E, so x\notin E^c

E^c is closed and x\notin E^c\implies x\notin (E^c)'\implies \exists r >0 such that (B_r(x)\cap E^c)\backslash \{x\}=\phi

So \phi=(B_r(x)\cap E^c)\backslash \{x\}=B_r(x)\cap E^c

So B_r(x)\in E

\implies

Suppose E is open

\begin{aligned}
    x\in (E^c)'&\implies \forall r>0, (B_r(x)\cap E^c)\backslash \{x\}\neq \phi\\
    &\implies \forall r>0, (B_r(x)\cap E^c)\neq \phi\\
    &\implies \forall r>0, B-r(x)\notin E\\
    &\implies x\notin  E^{\circ}\\
    &\implies x\notin E\\
    &\implies x\in E^c
\end{aligned}

So (E^c)'\subset E^c

EOP

Theorem 2.24

An arbitrary union of open sets is open

Proof:

Suppose \forall \alpha, G_\alpha is open. Let x\in \bigcup _{\alpha} G_\alpha. Then \exists \alpha_0 such that x\in G_{\alpha_0}. Since G_{\alpha_0} is open, \exists r>0 such that B_r(x)\subset G_{\alpha_0} Then B_r(x)\subset G_{\alpha_0}\subset \bigcup_{\alpha} G_\alpha

EOP

A finite intersection of open set is open

Proof:

Suppose \forall i\in \{1,...,n\}, G_i is open.

Let x\in \bigcap^n_{i=1}G_i, then \forall i\in \{1,..,n\} and G_i is open, so \exists r_i>0, such that B_{r_i}(x)\subset G_i

Let r=min\{r_1,...,r_n\}. Then \forall i\in \{1,...,n\}. B_r(x)\subset B_{r_i}(x)\subset G_i. So B_r(x)\subset \bigcup_{i=1}^n G_i

EOP

The other two can be proved by Theorem 2.22,2.23

Definition 2.26

The closure \bar{E}=E\cup E'

Remark: Using the definition of E', we have, \bar{E}=\{p\in X,\forall r>0,B_r(p)\cap E\neq \phi\}

Definition 2.27

\bar {E} is closed.

Proof:

We will show \bar{E}^c is open.

Suppose p\in \bar{E}^c. Then by remark, \exists r>0 such that B_r(p)\cap E=\phi (a)

Furthermore,, we claim B_r(p)\cap E'=\phi (b)

Suppose for contradiction that \exists q\in B_r(p)\cap E' By Theorem 2.19, \exists s>0 such that B_s(q)\subset B_r(p)

Since q\in E',(B_s(q)\cap E)\backslash \{q\}\neq \phi. This implies B_r(p)\cap E=\phi, which contradicts with (a)

This proves (b)

So \bar{E}^c is open

EOP