# Random math: 04

Source of most content: Gunhee Cho

# Topology

Topology is useful to classify geometric objects.

Def. Given a set $X \ne \emptyset$, say $T \subseteq 2^X$ is a topology on $X$ if $T$ satisifes:

1. $\emptyset, X \in T$
2. If $\forall \alpha \in I, \ E_\alpha \in T$, then $\cup_{\alpha \in I}E_\alpha \in T$
3. If $E_1, \cdots, E_n \in T$, then $\cap^n_{k=1} E_k \in T$.

We call the element of $T$ open set.

NB; Union of any collection of open sets is open while infinite intersection of open sets does not need to be open.

Def. If a set $X$ is equipped with a topology $T$, we will call (X, T) the topological space (TS).

e.g.1; Discrete toplogy $T=2^X$ : the largest topology contains all subsets as open sets.
e.g.2; Indiscrete topology $T=\{\emptyset, X\}$: the smallest topology contains only $\emptyset, X$ as open sets.

Def. For a set $X$ and toplogies $T, T’$ on $X$, $T’$ is called finer (larger) than $T$ if $T \subseteq T’$ and coarser (smaller) than $T$ if $T \supseteq T’$.

## Distance function

Def. Given a set $X$, $d: X\times X \rightarrow \mathbb{R}$ is a distance function if $d$ satisfies:

1. $\forall x,y \in X, \ d(x,y) \geq 0$ where the equality holds iff $x=y$
2. $\forall x,y \in X, \ d(x,y) = d(y,x)$
3. $\forall x,y,z, \in X, \ d(x,z) \leq d(x,y) + d(y,z)$ (triangle inequality)

Def. Given $(X, d),$
(1) Given $\epsilon >0, \ p \in X, \ N_\epsilon (p) := \{x \in X \mid d(x,p) < \epsilon \}$ is called $\epsilon$-neighborhood of $p$.
(2) $U$ is called open in $X$ if $\forall p \in U, \ \exists \epsilon_p > 0$ s.t. $N_{\epsilon_p}(p) \subseteq U$.

## Example; metric topology

$\text{Given } (X, d), \ T_d := \{ U \subseteq X \mid \forall p \in U, \exists \epsilon_p >0 \text{ s.t. } N_{\epsilon_p}(p) \subseteq U \} \text{ is a topology on } X. \nonumber$

### Proof

1. $\emptyset, X \in T_d$ by def. of $T_d$.
2. Let $p \in \cup_{\alpha \in I} E_\alpha \Leftrightarrow x \in E_k$ for some $k\in I \leadsto \exists \epsilon_p >0$ s.t. $N_{\epsilon_p}(p) \subseteq E_k \subseteq \cup_{\alpha \in I} E_\alpha$.
3. Let $p \in \cap^n_{i=1}{E_i} \Leftrightarrow \forall i \in [1, n] \ x \in E_i \leadsto \exists \epsilon^i_p > 0$ s.t. $N_{\epsilon^i_p}(p) \subseteq E_i$. Taking $\epsilon_p = \min_i (\epsilon^i_p)$ leads to $N_{\epsilon_p} (p) \subseteq N_{\epsilon^i_p}(p) \subseteq E_i, \ \forall i \in [1, n]$.

NB; A seq. $\{a_n \}^\infty_{n=1} \in \mathbb{R}$ which converges to $\alpha \in \mathbb{R} \Leftrightarrow \forall \epsilon >0, \ \exists N=N(\epsilon)$ s.t. $\forall n \geq N(\epsilon), \ |a_n - \alpha | < \epsilon$
$\Leftrightarrow \forall \epsilon >0, \ \exists N=N(\epsilon)$ s.t. $\forall n \geq N(\epsilon), \ a_n \in N_\epsilon(\alpha), \ d(a_n, \alpha) < \epsilon$ with $d(x,y) = | x - y |$.

## Convergence

Def. Given a topological space $(X, T)$,
(1) A sequence $f : \mathbb{N} \rightarrow X$ is a function, which $n \mapsto f(n) =: a_n \leadsto (a_n)^\infty_{n=1} \subseteq X$.
(2) Let $\alpha \in X, (a_n)^\infty_{n=1}$ converges to $\alpha$ if $\forall V \in T$ s.t. $\alpha \in V$, $\exists N= N(V) \in \mathbb{N}$ s.t. $\forall n\geq N, \ a_n \in V$. And denote $\lim_n a_n = \alpha$ or $a_n \xrightarrow{n\rightarrow \infty} \alpha$.
(3) In $(\mathbb{R}, T_d), \ d(x,y) := |x-y|, a_n \rightarrow \alpha$ means for all neighborhood $V$ of $\alpha$, there exists natural number $N$ s.t. $\forall n \geq N, \ a_n \in V$ $\Leftrightarrow \forall N_\epsilon (\alpha), \ \exists N \in \mathbb{N}$ s.t. $\forall n \geq N, a_n \in N_\epsilon(\alpha)$ $\Leftrightarrow \forall \epsilon >0, \ \exists N \in \mathbb{N}$ s.t. $\forall n \geq N,| a_n - \alpha | < \epsilon$.

NB1; Open set $V$ containing $\alpha \equiv V$ is neighborhood of $\alpha$.
NB2; Convegence of sequence depends on the topology of TS.

$X=\mathbb{N}, T = T_C = \{ U \subseteq \mathbb{N} \mid U^c \text{ is finite}\} \cup \{\emptyset \}$. Define $a_n :=n, \ \forall n \in \mathbb{N}$. Then, we can show that $a_n$ converges to every points in $\mathbb{N}$. Take any $m\in \mathbb{N}$, and any neighborhood $V$ of $m$ s.t. $V^c$ is finite set. Next, take $N=\max V^c + 1$. Then, $\forall n \geq N, \ n \notin V^c$ i.e., $n \in V \leadsto \ a_n$ converges.

NB3; To avoid above situation, we have to assume axioms which $(X, T)$ must satisfy such as Hausdorff axiom or T2-axiom which guarantees the uniqueness of the limit point.

## Limit and continuity

Def. A function $f:X\rightarrow Y$ between two toplogical sapces $(X,T_X)$ and $(Y, T_Y)$ is continous function if for every open set $V \subseteq Y$, $f^{-1}(V) = \{ x \in X \mid f(x) \in V\}$ is open subset of $X$.

Def. Let $\emptyset \ne E \subseteq \mathbb{R}$ and $f:E \rightarrow \mathbb{R}$. Let $p$ is a limit point of E $(\Leftrightarrow \forall \text{ neighborhood } V \text{ of } p, (V-\{p\}\cap E \ne \emptyset))$. We say $f(x)$ converges to $L$ as $x \rightarrow p$, $\lim_{x\rightarrow p} f(x) = L$ if $\forall \epsilon >0,\ \exists \delta = \delta(\epsilon) >0$ s.t. if $x\in E, 0< | x - p | < \delta$ then $|f(x)-L| < \epsilon$.

Def. A function $f: E (\subseteq \mathbb{R}) \rightarrow \mathbb{R}$ is continous at $p\in E$ if $\forall \epsilon > 0, \ \exists f = f(\epsilon) > 0$ s.t. if $| x - p | < \delta$ then $|f(x) - f(p)| < \epsilon$.

In topology, we can rewrite as $\forall \epsilon >0, \ \exists f = f(\epsilon) >0$ s.t. if $x\in N_\delta (p)$ then $f(x) \in N_\epsilon (f (p))$.

NB; $\lim_{x\rightarrow p}f(x) \ne L \Leftrightarrow \exists \epsilon >0,\ \exists x \in E, \ 0 <|x-p|<\delta$ but $|f(x)-L|\geq \epsilon$.

### Proposition

$\text{Let } \emptyset \ne E\subseteq \mathbb{R}, \ f:E\rightarrow \mathbb{R}, \ p \text{ is a limit point of }E. \\ \text{ Then, } \lim_{x \rightarrow p}f(x) = L \text{ iff } \forall (p_n) \subseteq E - \{p \} \text{ with } p_n \xrightarrow{n\rightarrow \infty} p, \text{ then } \lim_{n\rightarrow \infty}f(p_n)=L. \nonumber$

NB; limit of function $\Leftrightarrow$ limit of sequence.

#### Proof

1. $(\Rightarrow)$: Suppose $\lim_{x\rightarrow p}f(x) = L$. Take any seq. $(p_n) \subseteq E - \{p\}$ with $p_n \rightarrow p$. For any $\epsilon >0$, by Asm., $\exists \delta = \delta(\epsilon) >0$ s.t. if $0<|x-p|<\delta$ then (a) $|f(x) - L | < \epsilon$ with some $f>0$. Since $\lim_n p_n = p$, $\exists N \in \mathbb{N}$ s.t. $\forall n \geq N, \ 0 < |p_n-p|<\delta$. By (a), $|f(p_n) - L |<\epsilon, \ \forall n \geq N$. Hence $\lim_{n\rightarrow \infty} f(p_n)=L$.
2. $(\Leftarrow)$: Suppose $\lim_{x\rightarrow p}f(x) \ne L$, which means $\exists \epsilon >0, \ \forall \delta >0, \ \exists x \in E$ with $0<|x-p|<\delta$ but $|f(x)-L| \geq \epsilon$. $\forall n \in \mathbb{N}$, take $\delta = \frac{1}{n} \leadsto \ \exists p_n \in E$ with $0<|p_n-p|< \frac{1}{n}$ but $|f(p_n) -L | \geq \epsilon$. Thus, there exists a sequence $(p_n)\subseteq E - \{p\}$ s.t. $0<|p_n -p | < \frac{1}{n}$ and $|f(p_n) - L | \geq \epsilon > 0, \ \forall n \in \mathbb{N}$. However, since $0<|p_n-p|< \frac{1}{n},$ $\lim p_n = p$, $\lim f(p_n) = L$ by Asm. Then, with some $\epsilon >0, \ \exists N \in \mathbb{N}$ s.t. $\forall n \geq N, \ |f(p_n)-L|<\epsilon$, which is a contradiction.