Domains in $ \mathbb{C}$ .

Definition 2.1.1   Take $ z_0 \in \mathbb{C}$ and let $ r$ be a positive real number. The open ball with center $ z_0$ and radius $ r$ is $ B(z_0,r)= \{ z
\in \mathbb{C}\; ; \; \vert z-z_0\vert<r \}$ .

In Figure 1, we display the open unit disk, which is $ B(0,1)$ .

Figure 1: The unit disk.
\begin{figure}\mbox{\epsfig{file=UnitDisk.eps,height=4cm}}\end{figure}

Definition 2.1.2   An open neighborhood of $ z_0$ is a subset of $ \mathbb{C}$ containing an open ball centered at $ z_0$ . The most used neighborhoods are the open balls themselves.

Example 2.1.3       

Definition 2.1.4   A punctured neighborhood of $ z_0$ contains all the points of a neighborhood of $ z_0$ , excepted $ z_0$ itself.

Example 2.1.5   $ \mathbb{C}-\{ 0 \}$ is a punctured neighborhood of 0.

Definition 2.1.6   Let $ V$ be a subset of $ \mathbb{C}$ . It is an open subset of $ \mathbb{C}$ if, for each point $ z_0 \in V$ , there exists an open ball centered at $ z_0$ and included in $ V$ . .

In other words, $ V$ is an open susbet of $ \mathbb{C}$ if, and only if, the following condition holds:

$\displaystyle \forall z_0 \in V, \; \exists R>0 \; \vert \; B(z_0,R) \subset V.$    

Example 2.1.7       

Figure 3: Domains in $ \mathbb{C}$
\begin{figure}\mbox{
\subfigure[]{\epsfig{file=OpenHalfPlane.eps,height=4.5cm}}...
...quad
\subfigure[]{\epsfig{file=OpenUnitDisk.eps,height=4.5cm}}
}\end{figure}

Proposition 2.1.8   Let $ V_1$ and $ V_2$ be two open subsets of $ \mathbb{C}$ . Then the following hold:
1.
$ V_1 \cup V_2$ is open.
2.
$ V_1 \cap V_2$ is open.

Proof.     
1.
Let $ z_0 \in V_1 \cup V_2$ . Suppose that $ z_0 \in V_1$ ; as $ V_1$ is open, there exists R>0 such that $ B(z_0,R) \subset V_1$ , whence $ B(z_0,R) \subset V_1 \cup V_2$ . If $ z_0 \in V_2$ , a similar argument works, whence the result.
2.
Let $ z_0 \in V_1 \cap V_2$ . As $ V_1$ and $ V_2$ are open, there exist two positive numbers $ R_1$ and $ R_2$ such that $ B(z_0,R_1) \subset V_1$ and $ B(z_0,R_2) \subset V_2$ . Take $ R= \min (R_1,R_2)$ . Thus $ B(z_0,R) \subset V_1 \cap V_2$ and we are done.
$ \qedsymbol$

Figure 4: Intersection of two open subsets of $ \mathbb{C}$ .
\begin{figure}\centering
\mbox{
\epsfig{file=Intersection-OpenSets.eps,height=5cm}
}\end{figure}

Remark 2.1.9       

1.
Proposition 1.8 can be generalized to the union and the intersection of any finite number of open subsets of $ \mathbb{C}$ . The proof (by induction) is left to the reader.
2.
For an infinite family $ V_k, \; k \in \mathbb{N}$ of open subsets of $ \mathbb{C}$ , the intersection $ \underset{k \in \mathbb{N}}{\bigcap} V_k$ cannot be open. For example. take $ V_k=B(z_0,1/(k+1)$ , for any complex number $ z_0$ . Then $ \underset{k \in \mathbb{N}}{\bigcap} V_k = \{ z_0 \}$ , and it is easy to show that a set which contains only a single point is not open.

Definition 2.1.10   A closed set is the complement of an open set.

Example 2.1.11   The closed unit-disk $ \overline{\mathcal{D}}=\{z \in \mathbb{C}; \; \vert z\vert \leq 1 \}$ is a closed set, as its complementary set is open: it is $ \{ z \in \mathbb{C} \; \vert \; \vert z\vert > 1 \}$ (see Figure 5). For any $ z_0 \not\in \overline{\mathcal{D}}$ , i.e. for any $ z_0$ such that $ \vert z_0 \vert>1$ , let $ \varepsilon = \frac 12 \vert z_0 -1\vert$ ; then we have $ B(z_0,\varepsilon) \cap \overline{\mathcal{D}} =
\empty$ .

Figure 5: A closed set.
\begin{figure}\mbox{\epsfig{file=ClosedUnitDisk.eps,height=4cm}}\end{figure}

The following proposition will be very useful throughout our study of analytic functions (chapter chapter analytic functions) and further.

Proposition 2.1.12   Let $ A=\{ z_k, \; k=1, \dots, n \}$ be a finite set of points in $ \mathbb{C}$ . Then $ \mathbb{C} - A$ is open.

Proof. Denote $ V=\mathbb{C} - A$ and take any $ z_0 \in V$ . Now denote $ R_k=\vert z_0-z_k\vert$ (i.e. $ R_k$ is the distance between the images in the palne of the complex numbers $ z_0$ and $ z_k$ ). The set $ \{R_k, \; k=1, \dots, n \}$ is a finite set of positive real numbers, thus it has a minimal element, say $ R$ . Using the triangle inequality, it is easy to show that $ B(z_0,R) \subset V$ . $ \qedsymbol$

Proposition 2.1.13   Let $ V_1$ and $ V_2$ be two closed subsets of $ \mathbb{C}$ . Then the following hold:
1.
$ V_1 \cup V_2$ is closed.
2.
$ V_1 \cap V_2$ is closed.

The proof is left to the reader, using De Morgan laws.

Remark 2.1.14   There exist subsets of the complex plane which are neither open nor closed. For example, take the set $ T= \{ z
\in \mathbb{C} \; ; \; 1 \leq$   Re$ (z) < 3 \}$ (Figure 6).

If $ z_0=1+ib, \; b \in \mathbb{R}$ , then $ z_0 \in T$ but every open ball centered at $ z_0$ contains points which do not belong to $ T$ . Thus $ T$ is not open.

Now consider the complementary subset $ T^C$ of $ T$ in the $ \mathbb{C}$ and take $ z_1=3+ib, \; b \in \mathbb{R}$ . Then $ z_1 \in T^C$ , but every open ball centered at $ z_1$ contains points which do not belong to $ T^C$ . Thus $ T^C$ is not open, whence $ T$ is not closed.

Figure 6: A set which is neither open nor closed.
\begin{figure}\mbox{\epsfig{file=NonOpen-NonClosed.eps,height=5cm}}\end{figure}

Definition 2.1.15   A connected set is a subset of $ \mathbb{C}$ such that any two points in the set can be connected by a path of straight segments totally contained in the set. For example,
  1. The punctured plane $ \mathbb{C}-\{ 0 \}$ is connected and $ \mathbb{C}$ without a whole straight line is not connected.
  2. The subset $ D=\{ z \in \mathbb{C} \; \vert \; Re\; (z) \geq 1 \}$ is not connected. See Figure 7: it is impossible to connect the points $ A$ and $ B$ with a finite sequence of segments totally included in $ D$ .

    Figure 7: A non-connected subset of the plane.
    \begin{figure}\mbox{\epsfig{file=non-connected-subset.eps,height=4cm}}\end{figure}

A domain is an open connected set.

Example 2.1.16           

Figure 8: Domains in the complex plane.
\begin{figure}\mbox{
\subfigure[]{\epsfig{file=UnitDisk.eps,height=4.5cm}}
\qquad \qquad
\subfigure[]{\epsfig{file=HalfPlane.eps,height=4.5cm}}
}\end{figure}

Definition 2.1.17   A boundary point of a set $ A$ is a point $ z_0$ in $ \mathbb{C}$ such that every ball centered at $ z_0$ contains at least one point of $ A$ and at least one point not in $ A$ . The set of all the boundary points of $ A$ is called the boundary of $ A$ .

Example 2.1.18   Let $ E= \{ z \in \mathbb{C} \; \vert \; \operatorname{Re}\nolimits (z) > 1 \}$ . Then every point on the line whose equation is $ x=1$ is a boundary point.

Figure 9: Boundary, interior and exterior.
\begin{figure}\mbox{
\subfigure[Boundary points]{\epsfig{file=Boundary1.eps,hei...
...nts and exterior points]{\epsfig{file=IntExt.eps,height=5.5cm}}
}\end{figure}

Let $ z_0=1+yi$ , where $ y \in \mathbb{R}$ . Then, for any $ \varepsilon >0$ , the point $ z_0-\frac 12 \varepsilon$ is in the open ball $ B(z_0, \varepsilon)$ , but not in $ E$ . Thus $ z_0$ is a boundary point of $ E$ .

Definition 2.1.19   An interior point of a set $ A$ is a point $ z_0$ such that there exists an open ball centered at $ z_0$ and totally contained in $ A$ . An exterior point of a set $ A$ is a point $ z_0$ such that there exists an open ball centered at $ z_0$ and all of whose points are out of $ A$ (see Fig. 9(b)).

Example 2.1.20   Let $ E= \{ z \in \mathbb{C} \; \vert \; 1 \leq \vert z\vert < 2 \}$ (see Figure 10). A point $ z_0$ such that $ \vert z_0\vert<1$ or such that $ \vert z_0\vert>2$ is a point exterior to $ E$ :

A point $ z_0$ such that $ 1<\vert z_0\vert<2$ is an interior point; take $ \varepsilon = \frac 12$   min$ (\vert z_0\vert-1, 2-\vert z_0\vert))$ , then $ B(z_0,\varepsilon) \subset E$ .

Figure 10: Boundary points.
\begin{figure}\mbox{
\epsfig{file=IntExtAnnulus.eps,height=5.5cm}
}
\end{figure}

Definition 2.1.21   A bounded set $ A$ is a set for which there exists a positive number $ R$ such that $ \forall z \in A, \vert; \vert Z\vert< R$ (i.e. it is composed only of interior points of a certain circle centered at 0).

Example 2.1.22       

Figure 11: Domains in the complex plane.
\begin{figure}\mbox{
\subfigure[Bounded]{\epsfig{file=BoundedAnnulus.eps,height...
...ure[Non bounded]{\epsfig{file=NonBoundedDomain.eps,height=5cm}}
}\end{figure}

Remark 2.1.23   There is no connection between the notions of a closed set (definition 1.10) and of a bounded set (defintion 1.21). We mean a set can have either both properties, or only one of them, or none of them. For example:
(i)
The set $ \{ z \in \mathbb{C} \; \vert \; \vert z\vert \leq 1 \}$ is closed and bounded.
(ii)
The unit ball $ B(0,1)$ is bounded nut not closed.
(iii)
The set $ \{ z \in \mathbb{C} \; \vert \; 1 \leq \operatorname{Re}\nolimits (z) \leq 3 \}$ is closed but unbounded.
(iv)
The set $ \mathbb{C}^* = \mathbb{C}-{0}$ is neither closed nor bounded.

Noah Dana-Picard 2007-12-24