Is the preimage of a bounded set also bounded?

Question

I need to prove the following statement:

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ a continuous function and $B \subseteq \mathbb{C}$ bounded, implies, that the set $A=f^{-1}(B)$ to be bounded.

I do know that the statement is true for a continuous function on closed sets.

Is it possible, that this statement is also true for just bounded sets; and if not, what is the argumentation?

Answer 1

Consider constant function $f(z)=c$ for all $z$, then $\{c\}$ is bounded, but $f^{-1}(\{c\})=\mathbb{C}$.

Answer 2

For a nonconstant (in fact injective) counterexample, since you're not requiring $f$ to be differentiable, you can just do something like $$ f(z) = \frac{z}{1+|z|} $$ which is a continuous function on a closed set (namely, all of $\mathbb C$). The preimage of the open unit disc is then all of $\mathbb C$.