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

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of $\mathbb{C}$ Euclidean topological space.
• The reader knows a definition of homeomorphism.

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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.

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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.

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Main Body

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1: Description

For the $\mathbb{C}$ Euclidean topological space, the conjugation, $f:\mathbb{C}\to \mathbb{C},c\mapsto \bar{c}$, is a homeomorphism.

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2: Proof

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

$c=r{e}^{\theta i}$. $\bar{c}=r{e}^{-\theta i}$. On ${\mathbb{R}}^2$, $\left(\begin{array}{c}rcos\theta \\ rsin\theta \end{array}\right)\mapsto \left(\begin{array}{c}rcos\theta \\ -rsin\theta \end{array}\right)$. That is $\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)\mapsto \left(\begin{array}{c}{x}_1\\ -x_2\end{array}\right)$, which is continuous. So, $f$ is continuous by the definition of the $\mathbb{C}$ Euclidean topological space.

The inverse, $f^{-1}:\mathbb{C}\to \mathbb{C}$, is $c\mapsto \bar{c}$, which is continuous because it is the conjugation.

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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.

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References

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