Does my theorem capture the idea that "For topological questions, the sets $\mathbb C^m$ and $\mathbb R^{2m}$ can be identified"? My textbook Analysis I by Amann/Escher states that



I manage to reformulate this statement into the below theorem.

Let $\mathbb K$ denote $\mathbb R$ or $\mathbb C$, and $m \in \mathbb{N}^+$. We consider $\mathbb K^m$ to be a space endowed with the Euclidean inner product $\langle \cdot, \cdot \rangle: \mathbb K^m \times \mathbb K^m \to \mathbb K$ and the induced norm $\| \cdot \|: \mathbb K^m \to \mathbb R$ where for all $x=(x_{1}, \ldots, x_{m})$ and $y=(y_{1}, \ldots, y_{m})$ in $\mathbb{K}^{m}$ $$\langle x,y \rangle :=\sum_{j=1}^{m} x_{j} \overline{y_j}, \quad \| x\| := \sqrt{\langle x,x \rangle}$$
We define a bijection $\mathcal F: \mathbb C^m \to \mathbb R^{2m}$ by $$\mathcal F (x_{1}+i y_{1}, \ldots, x_{m}+i y_{m}) = (x_{1}, y_{1}, \ldots, x_{m}, y_{m})$$

Theorem: $U \subseteq \mathbb C^m$ is a neighborhood of $z:=(x_{1}+i y_{1}, \ldots, x_{m}+i y_{m}) \in \mathbb C^m$ if and only if $\mathcal F [U] \subseteq \mathbb R^{2m}$ is a neighborhood of $\mathcal F (z) \in \mathbb R^{2m}$.


My questions:


*

*Does my theorem successfully capture the idea of the authors' statement?

*Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!

My attempt:
We have $U$ is a neighborhood of $z$ if and only there is an open ball $\mathbb B(z,r) \subseteq U$ for some $r>0$. To prove that $\mathcal F[U]$ is a is a neighborhood of $\mathcal F (z)$, it suffices to show that $\mathcal F [\mathbb B(z,r)]$ is an open ball centered at $\mathcal F (z)$ with radius $r$.
$$\begin{aligned} z' \in \mathbb B(z,r) & \iff \| z'-z \| < r \iff \| z'-z \|^2 < r^2 \\
& \iff \sum_{j=1}^{m} (z_j'-z_j) \overline{z_j'-z_j} < r^2 \\
& \iff \sum_{j=1}^{m} |z_j'-z_j|^2 < r^2 \\
& \iff \sum_{j=1}^{m} |(x'_{j}+i y'_{j}) - (x_{j}+i y_{j})|^2 < r^2 \\
& \iff \sum_{j=1}^{m} |(x'_{j}-x_{j})+i (y'_{j} - y_{j})|^2 < r^2 \\
& \iff \sum_{j=1}^{m} \left ((x'_{j}-x_{j})^2 + (y'_{j} - y_{j})^2 \right ) < r^2 \\
& \iff \|\mathcal F(z') - \mathcal F (z)\|^2 < r^2 \\
& \iff \|\mathcal F(z') - \mathcal F (z)\| < r \\
& \iff \mathcal F(z') \in \mathbb B(\mathcal F(z),r)\end{aligned}$$
This completes the proof.
 A: Everything you said from "Let $\Bbb K$ be", up to but not including "We define a bijection" has little to do with topology, and is more algebra and analysis related. If you keep only the bijection and the theorem that comes after, then I believe that is what the authors' statement is meant to reflect.
Of course, you're using that machinery during your proof later, so it's not unimportant. But it's not part of the authors' sentiment the way I read it.
The proof looks good as far as I can see. All the arrows go in both directions, although you could probably make the notation a bit clearer with regards to which $\Bbb B$ is complex and which $\Bbb B$ is real. What you have really shown is that the open balls are preserved by the bijection (both ways). This makes the bijection a homeomorphism, meaning for topological purposes the spaces are the same.
A: It's all correct, but your argument has more steps than necessary, which buries the important idea in a pile of notation. The core of your argument is that $F$ preserves the norm, meaning $\|F(x)\|=\|x\|$. So you should say that! Don't just hide it in step 6.
