Assuming we are in $\mathbb{R}$ and $\mathbb{C}$

Let $S$ be $\mathbb{R}$ or $\mathbb{C}$ and $d(x,y)$ be the usual distance in those spaces

Properties:

Neighbourhood

$N\subset S$ is a neighbourhood of $x\in S$ if it contains a ball $B(x,\delta )$

Open Set

$A$ is open if $\mathrm{\forall }z\in A,$ $\mathrm{\exists }{B}_{\delta }(z)\in A$

Every ball $B_{\delta }(x)$ is open:

For $x,y\in B_{\delta }(x)$, let ${\delta }^{\prime }=\delta -d(x,y)$,

Take $z\in B_{{\delta }^{\prime }}(y)$,

Intersection of Finite Open Sets

Is open;

Let $O_1,\dots ,O_n$ be open
Take $z\in \bigcap _{i=1}^n{O}_i$, $⟹\mathrm{\forall }i,z\in O_i$
There exists ${\delta }_i$ such that $B_{{\delta }_i}(z)\subseteq O_i$
just take $min\{{\delta }_1,\dots ,{\delta }_n\}$

Union of Any Open Sets

Is open
$B_{{\delta }_i}(z)\in O_i\subset \bigcup O_i$

Closed Set

The complement of an open set

Union of Finite Amount of Closed Sets

Is closed

Take $x\in (\bigcup ^n{C}_i{)}^c$, therefore for any $\delta$, $B_{\delta }(x)⊈\bigcup ^n{C}_i$

$x\notin C_i\mathrm{\forall }i⟹x\in \bigcap _{i=1}^n{C}_i^c$ using De Morgans rule.
Each $C_i$ is closed $\Rightarrow C_i^c$is open.

Thus, for each $i$ there exists ${\delta }_i>0$ such that

$B_{{\delta }_i}(x)\subseteq C_i^c$

Set $\delta =min\{{\delta }_1,\dots ,{\delta }_n\}>0$

Intersection of any Closed Set

Is closed
Take $x\in (\bigcap C_i{)}^c$, therefore for any $\delta$,

$⟹x\in \bigcup ^n{C}_i^c$ where $C_i^c$ is open.

$x\in C_i^c⟹B_{{\delta }^{\prime }}(x)\in C_i^c\subset {(\bigcap C_i)}^c$ so its open

Terminology

Interior

The interior of $X$ ($\text{int}(X)\text{ or }{X}^{\circ }$) is the union of all open sets inside $X$

$X$ is open iff $X^{\circ }=X$

Closure

The closure of $X$ ($\text{cl}(X)$ or $\overline{X}$ ) is the intersection of all closed sets that contain $X$

$X$ is closed iff $\overline{X}=X$

Boundary

Is the closure of $X$ minus the interior of $X$. A bound is a boundary ($\text{Bd}(X)$ or $\partial X$) iff all of its neighbourhoods intersect both $X$ and $X^c$.

Exterior

The exterior of $X$ is the interior of $X^c=(X^c{)}^{\circ }$

Isolated point

$x\in X$ is an isolated point of $X$ if $x$ has a neighbourhood whose intersection with $X$ is only $x$

Accumulation Point / limit point

$x\in X$ is an accumulation point if any neighbourhood $x$ contains infinitely many points from $X$.

Take $A=\{\text{accumulation points of }X\}$

Take $y\in A^c$, $\mathrm{\exists }{B}_{\delta }(y)$ where $B_{\delta }(y)\cap X=\{a_1,a_2,\dots ,a_n\}$ is finite

let ${\delta }^{\prime }=\frac{1}{2}min\{d(a_1,y),\dots ,d(a_n,y)\}$

So its complement is open. Therefore $A$ is closed.

Connectedness

A subset of $K$ of $\mathbb{C}$ or $(\mathbb{R})$ is called connected if:

where $U_1$ and $U_2$ are open and $U_1\cap U_2=\mathrm{\varnothing }$ and $K\subseteq U_1$ or $K\subseteq U_2$.
($K$ cannot be subset of two disjoint open sets)

Families of Connected Sets

If $(K_{\alpha }{)}_{\alpha \in I}$ is a family of connected sets $\bigcap _{\alpha \in I}{K}_{\alpha }\ne \mathrm{\varnothing }$ then $\bigcup _{\alpha \in I}{K}_{\alpha }$ is connected.

Assume $A$ and $B$, are open and partition $\bigcup _{\alpha \in I}{K}_{\alpha }$
Take $K_i$ assume $K_i\subset A$. Take $K_j,$ $j\ne i$.

Assume $z\in K_i\cap K_j$, $z\in A$

$K_j\cap A$ and $K_j\cap B$ are disjoint but $K_j$ is connected. Since $z\in A$ $K_j\subset A$

Connected Set Decomposition

Any $K\subset \mathbb{C}$ can be uniquely decomposed by

Where each $U_{\alpha }$ is connected.

Connected Intermediate Value Theorem

If $K$ is connected and $f$ is continuous on $K$ then $f(K)$ is connected

Proof:

Lemma:
If $f:X\to Y$ is cts, if $A\subseteq Y$ is open, $f^{-1}(A)$ is also open

Take $b\in f^{-1}(A)⟹f(b)\in A$
$\mathrm{\exists }\epsilon >0$ such that $B_{\epsilon }(f(b))\in A$ since $A$ is open

If $f$ is continuous it means $\mathrm{\exists }\delta >0$ s.t $d(b,x)<\delta$ for any $d(f(b),f(x))<\epsilon$, for any $\epsilon$. Therefore $f(x)\in B_{\epsilon }(f(b))$ and $B_{\delta }(b)\subseteq f^{-1}(A)$

IVT
Assume $f(K)=A\cup B$ which are disjoint non-empty open sets and $K$ is connected

Which contradicts $K$ being connected

Path-Connectedness

Arc/Path

A continuous $\gamma :[a,b]\to \mathbb{C}$ is called an arc or a path from $\gamma (a)$ to $\gamma (b)$

Path-Connect

$K\subset \mathbb{C}$ is path-connected if $z_1,z_2\in K$ then there exists $\gamma :[a,b]\to K$ so that $\gamma (t)$ is in $K$ for any $t\in [a,b]$

Theorems

Let $U\subset \mathbb{C}$ be open, then $U$ is connected iff it is path connected

A path-connected subset of $\mathbb{C}$ is always connected, but not all connected subsets of $\mathbb{C}$ are path connected

If $U\subset \mathbb{C}$ is open then each of its connected components is open

Regions

A region is an open connected subset of $\mathbb{C}$
Analytic functions on a region $D$ is represented by $A(D)$

Constant Functions on Regions

If $f\in A(D)$ and $f^{\prime }(z)=0$ for any $z\in D$ then $f$ is constant in $D$

Analytic Definition

$f$ is analytic on $K\subset \mathbb{C}$ if there exists an open set $U$, $K\subset U$,

A complex-valued function $f(z)$ defined on a region $\mathrm{\Omega }$ is to be analytic in $\mathrm{\Omega }$ if it has a derivative at each point of $\mathrm{\Omega }$

Compactness

If $K\subset \bigcup _{\alpha \in I}{U}_{\alpha },$ all $U_{\alpha }$ are open, then there is a finite number of $U_{\alpha },$ such that $K\subset U_1\cup U_2\cup \cdots \cup U_n$ (finite open-cover)

All are equivalent:

1. K is closed and bounded subset of $\mathbb{C}$
2. If $\{z_n\}\subset K$ then there exists a subsequence $z_{n_k}$ st $\underset{k\to \mathrm{\infty }}{lim}{z}_{n_k}=z\in K$
3. If $f:K\to \mathbb{R}$ is continuous then $f(K)$ is closed and bounded. The values $maxf(z)$ and $minf(z)$ are finite and they are achieved at some point on $K$
4. If $f:K\to \mathbb{C}$ is continuous then $f(K)$ is bounded and closed
5. If $f:K\to \mathbb{R}(\text{or }\mathbb{C})$ is continuous, then $f$ is uniformly continuous. $\mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta >0$ such that if $|z_1-z_2|<\delta ⟹|f(z_1)-f(z_2)|<\epsilon$

If for every $n$, $K_n\ne \mathrm{\varnothing }$ is compact and $K_n\supset K_{n+1}$ then $\bigcap _1^{\mathrm{\infty }}{K}_n\ne \mathrm{\varnothing }$