Why is the map $f(x)=e^{i2\pi x}$ from $[0, 1)$ to the unit circle continuous?

Question

This seems to be a really silly question, I just couldn't think it straight.

The definition of a continuous map:

$f: X \to Y$ is continuous if for any open set $U$ in $Y$ , $f^{-1}(U)$ is open in $X$.

So for the map defined by $f(x)=e^{i2\pi x}$ from $[0, 1)$ to the unit circle on $\mathbb C$, the preimage of an open arc containing $1+0i$ is of form $[0, a)\cup(b,1)$ in the half open interval $[0,1)$, which is not open. So why is this map continuous?

Thanks.

Answer 1

Since $ [0,a) = (-2,a) \cap [0,1)$, $[0,a)$ is open in the subspace topology on $[0,1)$.