If $f \colon X \to \mathbb{C}$ is continuous and $X \subset \mathbb{C}$ is closed and bounded then $f(X)$ is closed and bounded.

Question

Let $X \subset \mathbb{C}$. I want to forget all others definitions or equivalent characterizations of compact subsets of $\mathbb{C}$ and only work with “A subset of the complex numbers is compact if is closed and bounded”. With this definition I want prove the following:

Let $X \subset \mathbb{C}$ be a compact subset and let $f \colon X \to \mathbb{C}$ a continuous function, then $f(X)$ is compact.

I have tried to take an accumulation point of $f(X)$ and prove that this point is in $f(X)$ but without success.

Could you give me some idea or help?

Thank you.

Answer 1

Let $a$ be an accumulation point of $f(X)$. Then there is a sequence $f(x_i)$ converging to $a$. The sequence $x_i$ in $X$ has a convergent subsequence because $X$ is compact. Let $x$ be the limit. What can you say about $f(x)$?