topology of the complex planeTopologyOfTheComplexPlane2003-05-25 06:16:152009-04-28 21:31:44Definitioncomplexopen diskaccumulation pointinterior pointopenclosedboundedcompact% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\sR}[0]{\mathbb{R}}
\newcommand{\sC}[0]{\mathbb{C}}
\newcommand{\sN}[0]{\mathbb{N}}
\newcommand{\sZ}[0]{\mathbb{Z}}The usual topology for the complex plane $\sC$
is the topology induced by the metric
$$d(x,\,y) := |x\!-\!y|$$
for\, $x,\,y \in \sC$.
Here, $|\cdot|$ is the \PMlinkname{complex modulus}{ModulusOfComplexNumber}.
If we identify $\sR^2$ and $\sC$, it is clear that the above
topology coincides with topology induced by the Euclidean metric on $\sR^2$.
Some basic topological concepts for $\sC$:
\begin{enumerate}
\item The open balls
$$B_r(\zeta) \;=\; \{z\in\sC\,\vdots\; |z\!-\!\zeta| < r\}$$
are often called \emph{open disks}.
\item A point $\zeta$ is an \emph{accumulation point} of a subset $A$ of $\sC$, if any open disk $B_r(\zeta)$ contains at least one point of $A$ distinct from $\zeta$.
\item A point $\zeta$ is an \emph{interior point} of the set $A$, if there exists an open disk $B_r(\zeta)$ which is contained in $A$.
\item A set $A$ is \emph{open}, if each of its points is an interior point of $A$.
\item A set $A$ is \emph{closed}, if all its accumulation points belong to $A$.
\item A set $A$ is \emph{bounded}, if there is an open disk $B_r(\zeta)$ containing $A$.
\item A set $A$ is \emph{compact}, if it is closed and bounded.
\end{enumerate}