2023-08-20

348: Conjugation from Complex Numbers Euclidean Topological Space onto Complex Numbers Euclidean Topological Space Is Homeomorphism

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A description/proof of that conjugation 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



Target Context


  • The reader will have a description and a proof of the proposition that the conjugation 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 conjugation, \(f: \mathbb{C} \rightarrow \mathbb{C}, c \mapsto \overline{c}\), is a homeomorphism.


2: Proof


\(f\) is bijective, because for \(c_1, c_2 \in \mathbb{C}\) such that \(c_1 \neq c_2\), \(\overline{c_1} \neq \overline{c_2}\), and for any \(c_1 \in \mathbb{C}\), there is the \(c_2 = \overline{c_1}\) such that \(\overline{c_2} = c_1\).

\(c = r e^{\theta i}\). \(\overline{c} = r e^{- \theta i}\). On \(\mathbb{R}^2\), \(\begin{pmatrix} r cos \theta \\ r sin \theta \end{pmatrix} \mapsto \begin{pmatrix} r cos \theta \\ - r sin \theta \end{pmatrix}\). That is \(\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \mapsto \begin{pmatrix} x_1 \\ - x_2 \end{pmatrix}\), which is continuous. So, \(f\) is continuous by the definition of the \(\mathbb{C}\) Euclidean topological space.

The inverse, \(f^{-1}: \mathbb{C} \rightarrow \mathbb{C}\), is \(c \mapsto \overline{c}\), which is continuous because it is the conjugation.


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.


References


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