4. Topology

Cantor Set:

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Let $C_{0} = [0, 1]$
Keep removing middle 1/3

\begin{aligned} C_{1} & = C_{0} ∖ (\frac{1}{3}, \frac{2}{3}) \\ C_{2} & = [0, \frac{1}{9}] \cup [\frac{2}{9}, \frac{1}{3}] \cup [\frac{2}{3}, \frac{7}{9}] \cup [\frac{8}{9}, 1] \\ ⋮ \end{aligned}

C = ⋂_{n = 0}^{\infty} C_{n}

Length of complement: $(\frac{1}{3} + \frac{2}{9} + \dots + \frac{2^{n - 1}}{3^{n}} + \dots) \to \frac{\frac{1}{3}}{1 - \frac{2}{3}} = 1$
So length( $C$ )=0 but $C$ is uncountable with cardinality equal to $R$
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Fractal Dimension

$K \subseteq R$ , Let $N (K, ε)$ number of intervals of size $ε$ it takes to cover $K$
Dimension $(K) = lim_{ε \to 0} \frac{\log (N, ε)}{\log (\frac{1}{ε})}$
$K$ must be bounded

Example
if $K$ is finite

lim_{ε \to 0} \frac{\log (2)}{\log (\frac{1}{ε})} = 0

$K = [0, a] ⟹ \dim K = a$

$\dim (C) = \frac{\log (2)}{\log (3)}$

Open Set

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A set $O \subseteq R$ is open if for all $a \in O$ there exists an $ε - neighbourhood$ $V_{ε} (a) \subseteq O$

Limit Point

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A point $x$ is called a limit point of a set $A \subseteq R$ if every $ε - neighbourhood$ $V_{ε} (x)$ intersects $A$ at some point other than $x$

V_{ε} (x) \cap (A ∖ x) \neq \emptyset

Thm Limits and Limit Points

$x$ is a limit point of set $A$ iff $x = lim a_{n}$ for some sequence $(a_{n})$ contained in A satisfying $a_{n} \neq x$ for all $n \in N$ .

$a_{n} \neq x$ in order to distinguish it from isolated points. It may not be in the set while an isolated point is always in the set.

Proof

$⟹$ assume $x$ is a limit point of $A$ .
Since it's a limit point for each $n \in N$ we can pick:

a_{n} \in V_{1 / n} (x) \cap A

with the stipulation that $a_{n} \neq x$ . Clearly $(a_{n}) \to x$ .
$⟸$ assume $lim a_{n} = x$ where $a_{n} \in A$ but $a_{n} \neq x$ . Let $V_{ε} (x)$ be arbitrary. By the definition of convergence there exists a term a $N$ in the sequence satisfying $a_{N} \in V ε () x$ ,and the proof is complete.

Isolated Point

A point $a \in A$ is an isolated point it is not a limit point of $A$

Closed Set

A set $F \subseteq R$ is closed if it contains all its limit points

Thm Cauchy and Closed

A set $F \subseteq R$ is closed iff every Cauchy sequence contained in $F$ has a limit in $F$

Proof

$⟹$
Let $F$ be closed. Assume that $(x_{n})$ is a Cauchy sequence such that $x_{n} \in F$ for all $n \in N$
Since every Cauchy sequence is convergent then $\forall ε > 0, \exists N \in N$ so that $n > N$ then $| x_{n} - l | < ε$ since $l$ is a limit point by definition then it must be in $F$ .
$⟸$
If every Cauchy sequence has a limit in $F$ then so does every convergent sequence

Thm Density and Limit Points

For every $y \in R$ there exists a sequence of rational numbers that converge to $y$

Closure of a set

Given a set $A \subseteq R$ let $L$ be the the set of all limit points of $A$

\bar{A} = A \cup L

For any $A \subseteq R$ , $\bar{A}$ is closed and $\bar{A}$ is the smallest closed set that contains $A$ .

Thm Open and Complements

A set $O$ is open iff $O^{c}$ is closed

Proof

$⟹$
Let $O$ be open
Let $x$ be a limit point of $O^{c}$
Every neighbourhood of $x$ contains some $O^{c}$
$x \notin O$ or $V_{ε} (x) \subseteq O$ for some $ε$ . So $x \in O^{c}$
$⟸$
Let $O^{c}$ be closed. Since $O^{c}$ is closed. Let $x \in O$ .
$x$ cannot be a limit point, $\exists ε > 0$ so that $V ε (x) \cap O^{c} = \emptyset ⟹ V ε (x) \subseteq O$

Union and Intersection Theorems

De Morgan's Law

For any collection of sets ${E_{λ} : λ \in Λ}$

(⋃_{λ \in Λ} E_{λ})^{c} = ⋂_{λ \in Λ} E_{λ}^{c} and (⋂_{λ \in Λ} E_{λ})^{c} = ⋃_{λ \in Λ} E_{λ}^{c}

1) The unions of an arbitrary collection of open sets is open
2) The intersection of finitely many open sets is open
3) The intersection of an arbitrary collection of open sets is open
4) The union of finitely many closed sets is closed

Proof

${0_{λ} | λ \in Λ}$
Let $O = ⋃_{λ \in Λ} O_{λ}$
If $a \in O$ then $a \in O_{λ}$ for some $λ$ there exists $V_{ε} (a) \subseteq O_{λ} \subseteq O$

Suppose ${O_{1}, \dots, O_{N}}$ are open
let $a \in ⋂_{n = 1}^{N} O_{n}$
For each $n \in {1, \dots, N}$ there exists $V_{ε_{n}} (a) \subseteq O_{n}$
Take $ε = min {ε_{1}, \dots, ε_{N}}$ , Then $V_{ε} \subseteq V ε_{n}$ for all $n$

Thm Open from Disjoint Open

Every open set can be written as a countable union of (disjoint) open intervals

Interior/Exterior/Boundary Points

Let $A \subseteq R$

$a$ is an interior point if there exists $ε > 0$ so that $V_{ε} (a) \subseteq A$ ( $V_{ε} (a) \cap A^{c} = \emptyset$ )

$a$ is an exterior point if there exists $ε > 0$ so that $V_{ε} (a) \subseteq A^{c}$ ( $V_{ε} (a) \cap A = \emptyset$ )

a boundary point is not interior or not exterior ( $\partial A$ ), limit point of $A$ and $A^{c}$ , ( $V_{ε} (a) \cap A^{c} \neq \emptyset$ and $V_{ε} (a) \cap A \neq \emptyset$ )

Example
$A = (a, b) \cup [c, d]$
int $A$ = $(a, b) \cup (c, d)$ Ext $A = (- \infty, a) \cup (b, c) \cup (d, \infty)$
boundary = ${a, b, c, d}$

Properties
Int $A \cup Ext A \cup B d A = R$
$Int A = Ext A^{c}$
$Bd A = Bd A^{c}$
$A$ is open iff $A = Int A$
If $a$ is isolated then $a \in Bd A$

Closure and Boundaries

$A \subset R$ is closed iff $Bd A \subset A$

Proof

$⟹$
Let $A$ be closed and $a \in Bd A$
If $a \notin A$ then

(V_{ε} (a) ∖ {a}) \cap A \neq \emptyset

for $ε > 0$ since $V ε (a) \cap A \neq \emptyset$ so $a$ is a limit point of $A$ which is a contradiction since $A$ is closed

$⟸$
Let $Bd A \subset A$ , let $a$ be a limit point
suppose that $a \notin A$
Since it's a limit point $\forall ε > 0, V_{ε} (a) \cap A = \emptyset$
Sine $a \in A^{c}$ $\forall ε > 0, V ε (a) \cap A^{c} = \emptyset$
So $a \in Bd A \subset A$

$\bar{A} = Bd A \cup A$

Compact Sets

$K \subseteq R$ is compact if every sequence in $K$ has a subsequence that converges to a limit in $K$ .

Thm. Sup and Inf in Compact

If $K$ is compact and nonempty then $sup K$ and $inf K$ both exist in $K$

Proof

Let $s = sup K$ , for any $ε_{n} = \frac{1}{n}$ we can pick an $x_{n}$ such that $s - ε_{n} < x_{n} < s$ , thus $(x_{n}) \to s$ . Therefore $s \in K$ . Same idea for $inf K$ .

Bounded Sets

Set $K$ is bounded iff there exists $M > 0$ s.t $| a | \leq M$

Thm Compact Closed Bounded

A $K \subseteq R$ is compact iff it is closed and bounded
Ex. Cantor set is compact

Proof

$⟹$
Let $K$ be compact. For contradiction, assume it is not bounded.

Then there exists $x_{1} \in K$ with $| x_{1} | > 1$ and there is an $x_{n} \in K$ with $| x_{n} | > n$ . Since $K$ is compact there exists a convergent subsequence of $(x_{n})$ , $(x_{n_{k}})$ , But $| x_{n_{k}} | > n_{k}$ , so $(x_{n k})$ is not bounded and therefore not convergent and has no convergent subsequence, so it can't be in $K$ .
$⟸$
Let $K$ be closed and bounded. Let $(x_{n})$ be a sequence in $K$ . Since $K$ is bounded $(x_{n})$ is bounded by BW, $(x_{n})$ has a convergent subsequence $(x_{n_{k}})$ . Since $K$ is closed $(x_{n_{k}})$ converges to an element of $K$ .

Thm Nested Compact Sets

K_{1} \supseteq K_{2} \supseteq K_{3} \supseteq K_{4} \supseteq \cdot \cdot

s a nested sequence of nonempty compact sets, then the intersection $⋂_{n = 1}^{\infty} K_{n} \neq \emptyset$

Open Cover

Let $A \subseteq R$ .
An open cover for $A$ is a (possibly inﬁnite) collection of open sets ${O_{λ} : λ \in Λ}$ whose union contains the set $A$ or:

A ⊆\bigcup_{λ∈Λ}O_{\lambda}.$$Given an open cover for $A$, a **ﬁnite subcover** is a ﬁnite sub-collection of open sets **from the original open cover** whose union still manages to completely contain $A$. --- ## Thm Heine-Borel Let $K$ be a subset of $\mathbb{R}$. All of the following statements are equivalent in the sense that any one of them implies the two others: (i) K iscompact. (ii) K isclosedandbounded. (from earlier) (iii) Every open cover for K has a ﬁnite subcover. ### Proof (iii)$\to$(i) Let $O_{x}=V_{1}(x)$ for every $x\in K$. Clearly $\cup_{x\in K} O_{x}$ cover $K$. So there exists a finite subcover. Since each open set is bounded the finite union of bounded sets is clearly also bounded. For contradiction assume that $K$ is not closed. Let $(y_{n})$ be Cauchy contained in $K$ such that $(y_{n})\to y\not\in K$. Therefore every $x\in K$ is a positive distance from $y$. If we take our open cover to be $O_{x}=V\varepsilon_{x}(x)$ where $\varepsilon_{x}=\frac{\lvert x-y \rvert}{2}$. Since we are assuming (iii) this must cover $K$ for a finite union using some $\{ x_{1},x_{2},\dots x_{n} \}$. We assume that $y\not\in K$ but since $(y_{n})\to y$ there exists an $N$ so that

\lvert y_{N}-y \rvert <\min\left{ \frac{\lvert x_{i}-y \rvert}{2}:1\leq i\leq n \right}

So it is even closer to $y$ and therefore not included in $\cup^{\infty}_{1=1}O_{x_{i}}$ ## Sigma-Closed and Delta-Open A set $A\subseteq\mathbb{R}$ is called an $F_{\sigma}$ set if it can be written as the countable union of closed sets. A set $A\subseteq \mathbb{R}$ is called a $G_{\delta}$ set if it can be written as the countable intersection of open sets. ## Dense $A$ is dense in $B$ if $\bar{A}=B$ ## Nowhere-Dense A set E is nowhere dense if $\bar{E}$ contains no non-empty open intervals Example: Cantor Set ## Baire's Theorem $\mathbb{R}$ cannot be written as the countable union of nowhere-dense sets.