# Math 4111 Exam 2 review

$E$ is open if $\forall x\in E$, $x\in E^\circ$ ($E\subset E^\circ$)

$E$ is closed if $E\supset E'$

Then $E$ closed $\iff E^c$ open $\iff \forall x\in E^\circ, \exists r>0$ such that $B_r(x)\subset E^c$

$\forall x\in E^c$, $\forall x\notin E$

$B_r(x)\subset E^c\iff B_r(x)\cap E=\phi$

## Past exam questions

$S,T$ is compact $\implies S\cup T$ is compact

Proof:

Suppose $S$ and $T$ are compact, let $\{G_\alpha\}_{\alpha\in A}$ be an open cover of $S\cup T$

(NOT) $\{G_\alpha\}$ is an open cover of $S$, $\{H_\beta\}$ is an open cover of $T$.

...

QED

## K-cells are compact


We'll prove the case $k=1$ and $I=[0,1]$ (This is to simplify notation. This same ideas are used in the general case)

Proof:

That $[0,1]$ is compact.

(Key idea, divide and conquer)

Suppose for contradiction that $\exists$ open cover $\{G_a\}_{\alpha\in A}$ of $[0,1]$ with no finite subcovers of $[0,1]$

**Step1.** Divide $[0,1]$ in half. $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$ and at least one of the subintervals cannot be covered by a finite subcollection of $\{G_\alpha\}_{\alpha\in A}$

(If both of them could be, combine the two finite subcollections to get a finite subcover of $[0,1]$)

Let $I_1$ be a subinterval without a finite subcover.

**Step2.** Divide $I_1$ in half. Let $I_2$ be one of these two subintervals of $I_1$ without a finite subcover.

**Step3.** etc.

We obtain a seg of intervals $I_1\subset I_2\subset \dots$ such that

(a) $[0,1]\supset I_1\supset I_2\supset \dots$  
(b) $\forall n\in \mathbb{N}$, $I_n$ is not covered by a finite subcollection of $\{G_\alpha\}_{\alpha\in A}$  
(c) The length of $I_n$ is $\frac{1}{2^n}$

By (a) and **Theorem 2.38**, $\exists x^*\in \bigcap^{\infty}_{n=1} I_n$.

Since $x^*\in [0,1]$, $\exists \alpha_0$ such that $x^*\in G_{\alpha_0}$

Since $G_{\alpha_0}$ is open, $\exist  r>0$ such that $B_r(x^*)\subset G_{\alpha_0}$

Let $n\in \mathbb{N}$ be such that $\frac{1}{2^n}<r$. Then by $(c)$, $I(n)\subset B_r(x^*)\subset G_{\alpha_0}$

Then $\{G_{\alpha_0}\}$ is a cover of $I_n$ which contradicts with (b)

QED

## Redundant subcover question

$M$ is compact and $\{G_\alpha\}_{\alpha\in A}$ is a "redundant" subcover of $M$.

$\exists \{G_{\alpha_i}\}_{i=1}^n$ is a finite subcover of $M$.

We define $S$ be the $x\in M$ that is only being covered once.

$$
S=M\backslash\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)
$$

We claim $S$ is a closed set.

$G_{\alpha_i}\cap G_{\alpha_j}$ is open.

$\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)$ is closed

$S=M\backslash\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)$ is closed.

So $S$ is compact, we found another finite subcover yeah!

