An $\varepsilon$ *neighbourhood*, also called
*open ball* or *open disk*, of a complex number $z_0$ consists of all points
$z$ lying inside but not on a circle centred at $z_0$ and with radius
$\varepsilon>0$ and is expressed by
\begin{eqnarray}
B_{\varepsilon}(z_0)=\{z:|z-z_0| \lt \varepsilon\}
\end{eqnarray}
The *closed* $\varepsilon$ *neighbourhood*
of $z_0$ is expressed by
\begin{eqnarray}
\overline{B}_{\varepsilon}(z_0)=\{z:|z-z_0|\leq \varepsilon\}
\end{eqnarray}
And finally, a *deleted* $\varepsilon$ *neighbourhood*
of $z_0$, also called *punctured balls or disks*, is expressed by
\begin{eqnarray}
B_{\varepsilon}(z_0)\setminus \{z_0\}=\{z:0 \lt |z-z_0| \lt \varepsilon\}
\end{eqnarray}
Figure 1 shows the geometrical representation of the following examples:

- $B_{1}(0)=\{z:|z|\lt 1\}$
- $\overline{B}_{\frac{7}{8}}\left(-1-\sqrt{2}i\right)=\left\{z:|z-\left(-1-\sqrt{2}i\right)|\leq \frac{7}{8}\right\}$
- $\overline{B}_{\frac{1}{2}}\left(2+\sqrt{3}i\right)\setminus \left\{2+\sqrt{3}i\right\}=\left\{z:0 \lt |z-\left(2+\sqrt{3}i\right)|\leq\frac{1}{2}\right\}$

A point $z_0$ is said to be an *interior point*
of a set $S\subset \mathbb C$ whenever there is some neighbourhood
of $z_0$ that contains only points of $S$; it is called an
*exterior point* of $S$ when
there exists a neighbourhood of it containing no
points of $S$. If $z_0$ is neither of these, it is a
*boundary point* of $S$. A
boundary point is, therefore, a point all of whose
neighbourhoods contain at least one point in $S$ and at
least one point not in $S$. The totality of all boundary points is called the
*boundary* of $S$.

In this text we will use the following notation:

- $\text{Int } S=\left\{z: z \text{ is an interior point of } S \right\}$
- $\text{Ext } S=\left\{z: z \text{ is an exterior point of } S \right\}$
- $\partial S=\left\{z: z \text{ is a boundary point of } S \right\}$

Considering the previous examples of neighbourhoods, let $$ S_1=B_1(0),\; S_2=\overline{B}_{\frac{7}{8}}\left(-1-\sqrt{2}i\right) \text{ and } S_3=\overline{B}_{\frac{1}{2}}\left(2+\sqrt{3}i\right)\setminus \left\{2+\sqrt{3}i\right\}. $$ Thus, for $S_1$ we have:

- $\text{Int } S_1=B_1(0)$
- $\text{Ext } S_1=\{z: |z|>1 \}$
- $\partial S_1=\{z: |z|=1 \}$

For $S_2$ we have:

- $\text{Int } S_2=B_{\frac{7}{8}}\left(-1-\sqrt{2}\right)$
- $\text{Ext } S_2=\left\{z:|z-\left(-1-\sqrt{2}i\right)|>\frac{7}{8}\right\}$
- $\partial S_2=\left\{z:| z-\left(-1-\sqrt{2}i\right)| =\frac{7}{8}\right\}$

And finally, for $S_3$ we have:

- $\text{Int } S_3=B_{\frac{1}{2}}\left(2+\sqrt{3}\right)$
- $\text{Ext } S_3=\left\{z:|z-\left(2+\sqrt{3}i\right)|>\frac{1}{2}\right\}$
- $\partial S_3=\left\{z:|z-\left(2+\sqrt{3}i\right)|=\frac{1}{2}\right\}\cup\{2+\sqrt{3}i\}$

A set $S$ is *open* if for every $z\in S$, exits
$\varepsilon>0$ such that
$$B_{\varepsilon}(z)\subset S.$$
That is, $\text{Int } S= S$.
A set $S$ is *closed* if it contains all of its
boundary points, that is
$$\partial S\subseteq S.$$
The set $\mathbb C$ is both open and closed since it has no boundary points.

The set $\mathbb C$, together with the collection $\tau =\{S\subseteq \mathbb C: S \text{ is open} \}$ is a topological space, and this is expressed by the pair $\left(\mathbb C, \tau\right)$. The topological space $\left(\mathbb C, \tau\right)$ satisfies the following:

- $\emptyset$ and $\mathbb C$ are open.
- Whenever two or more sets are open, then so is their union.
- Whenever sets $S_1$ and $S_2$ are open, then so is $S_1\bigcap S_2$.

**Remark:** The technical definition of topological space
is a bit unintuitive, particularly
if you haven't studied topology. In essence, it states that the geometric
properties of subsets of $\mathbb C$ will be preserved when continuous
transformations (functions or mappings) are applied.

The *closure* of a set $S$ is the closed set
consisting of all points in $S$ together
with the boundary of $S.$ In other words
$$\overline{S}=S\cup \partial S.$$

In general, an set is connected if it cannot be expressed as the union of
two disjoint nonempty open sets.
An open set $S$ is *pollygonally connected* if each pair of
points $z_1$ and $z_2$ in it can be
joined by a polygonal line, consisting of a finite number of
line segments joined end to end,
that lies entirely in $S.$
A open set $S$ is connected if and only if
$S$ is pollygonally connected.

Notice for example that the open set $|z| \lt 1$ is connected.
The *annulus* $1 \lt |z| \lt 2$ is open and
also connected, see Figures 4 and 5.

A nonempty open set that is
connected is called a
*domain*. In this context, any neighbourhood is a
domain. A domain together with some,
none, or all of its boundary points is referred to as a
*region*. In other
words, a set whose interior is a domain is called a *region*.
A set $S$ is bounded if there is
$R>0$ such that
$$S\subset B_R(0)=\{z\in \mathbb C: |z|\lt R\}.$$

**Exercise:** Sketch the set $S$ of points in
the complex plane satisfying the
given inequality.
Determine whether the set is (*a*) open, (*b*) closed,
(*c*) a domain, (*d*) bounded, or (*e*) connected.

- $\text{Im}(z)\lt 0$
- $-1\lt \text{Re(z)} \lt 1$
- $|z|>1$
- $2\le |z - 3+4i |\le 5$

NEXT: Complex Functions