# Open and closed mapping are not necessarily continuous

it says

"To every point on the unit circle we can associate the angle of the positive x-axis with the ray connecting the point with the origin. This function from the unit circle to the half-open interval $[0,2π)$ is bijective, open, and closed, but not continuous."

So is $(1,0)$ the only point on the circle that isn't continuous?

Also, the Wikipedia article says the floor function is both open and closed. I don't see how it can be open since the image of the set $(0,1)$ under the floor function is $\{0\}$, which is definitely not open.

-

1) Yes, $(1,0)$ is the only point on the circle where this function is discontinuous.

2) You left out something: it says "the floor function from R to Z is open and closed, but not continuous." $\{0\}$ is open as a subset of the discrete space $\mathbb Z$.

-

You're correct that the only point where the map $$f: S^1 \longrightarrow [0, 2\pi),$$ $$f(\cos \theta, \sin \theta) = \theta$$ is discontinuous is $(1, 0)$.

As for the floor map, they are considering it as a map $$\lfloor \cdot \rfloor : \Bbb R \longrightarrow \Bbb Z,$$ where $\Bbb Z$ has the discrete topology. Hence every subset of $\Bbb Z$ is both open and closed. Of course $\lfloor \cdot \rfloor$ is not continuous since, for example, the inverse image of the open set $\{0\}$ is $[0, 1) \subset \Bbb R$, which is not open.

-

The wiki article talks about the floor function mapping $\mathbb{R}$ to $\mathbb{Z}$ where $\mathbb{Z}$ has the discrete topology. This means by definition that every subset of $\mathbb{Z}$ (like $\{0\}$) is open, and thus also closed because the set $\mathbb{Z} \setminus {0}$ is also a subset of $\mathbb{Z}$ and thus open, therefore it's complement which is $\{0\}$ must be closed.

-