# Lecture 12

## Review Questions

For a metric space $(X,d)$, we say a subset $S\subset X$  si bounded if there exists $p\in X$ and $r>0$ such that $S\subset B_r(p)$.

Consider the following statement: If a set $S\subset X$ is compact, the its is bounded.

1. Will the proof of this statement involve an arbitrary open cover (one that you, the prover, do not get to choose) or a specific open cover (one that you can choose)?
    We should choose a specific cover so that we can construct cover that have a set that is a superset of $S$.
2. Give a proof of the statement. [Suggestion: If you prefer, you could try proving the contrapositive. Both a direct proof and a proof by contrapositive are roughly of the same difficulty.]

### Continue on compact sets

#### Lemma

If $S\in X$ is compact, then $S$ is bounded.

Proof:

Fix $p\in X$, then $\{B_n(p)\}_{n\in \mathbb{N}}$ (specific open cover) is an open cover of $S$ (Since $\bigcup_{n\in \mathbb{N}}=X$). Since $S$ is compact, then $\exists$ a finite subcover ${n\in \mathbb{N}}_{i=1}^k=S$, let $r=max(n_1,...n_k)$, Then $S\subset B_r(p)$

QED

#### Definition k-cell

A 2-cell is a set of the form $[a_1,b_1]\times[a_2,b_2]$

#### Theorem 2.39 (K-dimension of Theorem)

Theorem 2.38 replace with "closed and bounded intervals" to "k-cells".

Ideas of Proof:

Apply the Theorem to each dimension separately.

#### Theorem 2.40

Every k-cell is 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

#### Theorem 2.41

If a set $E$ in $\mathbb{R}^k$ (only in $\mathbb{R}^k$, not for general topological space or metric spaces.) has one of the following three properties, then it has the other two:

(a) $E$ is closed and bounded.  
(b) $E$ is compact.  
(c) Every infinite subset of $E$ has a limit point in $E$.

We'll prove $(a)\implies (b)\implies (c)\implies (a)$

Proof:

$(a)\implies (b)$

Suppose $(a)$ holds, then $\exists$ k-cell $I$ such that $E\in I$.By **Theorem 2.40**, $I$ is compact. By **Theorem 2.35**, $E$ is compact.

$(b)\implies (c)$

Follows from **Theorem 2.37**

$(c)\implies (a)$

We will proceed by contrapositive, which says that $\neg (a)\implies \neg (c)$

$\neg (a)$: $E$ is not closed or not bounded.

$\neg (c)$: $\exists$ infinite subcover $S\subset E$ such that $S'\cup E=\phi$

Suppose $(a)$ does not hold. There are two cases to consider

Case 1: $E$ is not bounded. Then $\forall v\in \mathbb{N},\exists x_n\in E$ such that $|x_n|\geq n$

Let $S=\{x_n,...,n\in\mathbb{N}\}$, then $S'=\phi$ (by **Theorem 2.20**)

Case 2: $E$ is not closed. Then $z\in E'\backslash E$.

Since $z\in E'$, $\forall n\in \mathbb{N},\exists x_n\in E$ such that $|x_n-z|<\frac{1}{n}$

Let $S=\{x_n:n\in \mathbb{N}\}$, we claim $S'\subset \{z\}$. (In fact $s'\in\{z\}$, but we don't need this in the proof)

We'll show if $y\neq z$, then $y\notin  S'$

$\forall w\in B_r(y)$

$$
\begin{aligned}
    d(w,z)&\geq d(y,z)-d(y,w)\\
    &>d(y,z)-r\\
    &=d(y,z)-\frac{1}{2}d(y,z)\\
    &=\frac{1}{2}d(y,z)
\end{aligned}
$$

So $B_r(y)\cap S$ is finite. By **Theorem 2.20**, $y\notin  S$, this proves the claim so $S'\cap E=\phi$

QED