$\mathbb{C}$ similar to $\mathbb{R^2}$ for some properties

Question

I have a doubt that I would like to clear, it concerns the similarities between $\mathbb{R^2}$ and $\mathbb{C}$.

In particular, I am investigating the properties about the compactness of the domain of a function and I have found the following proposition:

Let $D \subset\mathbb{R^n}$ be a closed set, let $f: D \rightarrow \mathbb{R^p}$ be continuous and let $M \subset D$ be compact. Then $f[M] = \{f(x) | x \in M \}$ is compact.

Can it be used also for a continuous function f moving from:

$$ f: [0,1] \rightarrow \mathbb{C} $$

I know that when we are working at the level of rings and fields, most of the differences come out, but it seems that for this preposition we are just working at the set level, so it seems to be reasonable enough to assume that the extension is possible. If not, why?

Answer 1

As a set and as a metric space (if we are dealing with the distance $d\bigl(x_1,x_2),(y_1,y_2)\bigr)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$ and $d(z_1,z_2=|z_1-z_2|^2$), then $\mathbb{R}^2$ and $\mathbb C$ are the same thing. Therefore, the answer is affirmative.

Answer 2

The continuous image of a compact set is compact, this is true for any topological space. Let $f: X \to Y$ continuous and $\mathscr{U}$ an open cover of $Y$. Then for any $U \in \mathscr{U}$, $f^{-1}(U)$ is open by continuity and the set $\{f^{-1}(U) | U \in \mathscr{U}\}$ is an open cover of $X$. By compactness, pass to a finite subcover, $f^{-1}(U_1),...,f^{-1}(U_n)$. Then $U_1,..,U_n$ are a finite subcover of $Y$.