A description/proof of that nonzero multiplicative translation from complex numbers Euclidean topological space onto complex numbers Euclidean topological space is homeomorphism
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of \(\mathbb{C}\) topological space.
- The reader knows a definition of multiplicative translation.
- The reader knows a definition of homeomorphism.
Target Context
- The reader will have a description and a proof of the proposition that any nonzero multiplicative translation from the \(\mathbb{C}\) Euclidean topological space onto the \(\mathbb{C}\) Euclidean topological space is a homeomorphism.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For the \(\mathbb{C}\) Euclidean topological space, the nonzero multiplicative translation, \(f: \mathbb{C} \rightarrow \mathbb{C}, c \mapsto c_0 c\), is a homeomorphism.
2: Proof
\(f\) is bijective, because for \(c_1, c_2 \in \mathbb{C}\) such that \(c_1 \neq c_2\), \(c_0 c_1 \neq c_0 c_2\), and for any \(c_1 \in \mathbb{C}\), there is the \(c_2 = c_1 / c_0\) such that \(c_0 c_2 = c_1\).
\(c = r e^{\theta i}\) and \(c_0 = r_0 e^{\theta_0 i}\). \(c_0 c = r_0 r e^{(\theta_0 + \theta) i}\). On \(\mathbb{R}^2\), \(\begin{pmatrix} r cos \theta \\ r sin \theta \end{pmatrix} \mapsto \begin{pmatrix} r_0 r cos (\theta_0 + \theta) \\ r_0 r sin (\theta_0 + \theta) \end{pmatrix} = \begin{pmatrix} r_0 cos \theta_0 r cos \theta - r_0 sin \theta_0 r sin \theta \\ r_0 sin \theta_0 r cos \theta + r_0 cos \theta_0 r sin \theta \end{pmatrix}\). That is \(\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \mapsto \begin{pmatrix} r_0 cos \theta_0 x_1 - r_0 sin \theta_0 x_2 \\ r_0 sin \theta_0 x_1 + r_0 cos \theta_0 x_2 \end{pmatrix}\), which is continuous. So, \(f\) is continuous by the definition of the \(\mathbb{C}\) topological space.
The inverse, \(f^{-1}: \mathbb{C} \rightarrow \mathbb{C}\), is \(c \mapsto c_0^{-1} c\), which is continuous because it is a nonzero multiplicative translation.
3: Note
The \(\mathbb{C}\) Euclidean topological space is the \(\mathbb{R}^2\) Euclidean topological space where each \(c = r_1 + r_2 i\in \mathbb{C}\) is mapped to \((r_1, t_2) \in \mathbb{R}^2\), which means that any set of complex numbers is open if and only if the mapped image on \(\mathbb{R}^2\) is open.
