Why is this subset not open? I have a function, $f:[0, 1) \rightarrow \mathbb{S}^1$ given by $f(x) = (\cos2\pi x, \sin2\pi x)$. I have to show that $f$ is bijective and continuous and that $f^{-1}$ is not continuous. 
I have proved that $f$ is bijective and continuous but struggle to prove $f^{-1}$ is not.
The solution says that by taking the open set $[0, \frac{1}{2}) \subset [0, 1)$ we can see that f maps this to a non open subset of $\mathbb{S}^1$. 
My question is why is this subset of $\mathbb{S}^1$ non open? I can't see how this is correct and any explanation would be greatly appreciated.
Thanks
 A: First of all two basic notions:
I assume that $S^1$ is equipped with the Subspace Topology. (the most natural topology in this case)
So the open sets $V$ in $S^1$ are precisely of the form $V = S^1 \cap \Omega$ where $\Omega$ is open in $\mathbb{R}^2$. It's very easy to see that in this case ($\mathbb{R}^2$ equipped with the usual topology - generated by the open balls-) we have a base for our topology in $S^1$ (explained in the link). In this case the base is the family of "open" circular arcs ( the starting and final point of the arc doesn't belong to the arc). So if a set in $S^1$ can be written as union of this arcs it's open and viceversa.
The problem:
Let's figure out what is $f([0,\frac{1}{2}))$. The image of $f$ trace a path beginning from the point $(1,0) = (\cos(2\pi 0),\sin(2\pi 0)$ in anti clockwise direction along the circumference and stop at the point $(\cos(2\pi \frac{1}{2} ),\sin(2\pi \frac{1}{2})=(-1,0)$ but it didn't reach it, it's arbitrarily near it.
The informal explanation:
The problem is that the image (:= $f([0,\frac{1}{2}))$) we traced isn't open in the topology of $S^1$, because the point $(1,0)$ has not a neighborhood entirely contained in such image. (By the definition of basis, the neighborhood can be thought of the form of an open arc centered in $(1,0)$ - it's immediate to notice that half of this arc lies outside our image, even if the open arc is arbitrarily small but still contains $(1,0)$). Please note that this reasoning doesn't work if we consider a point near $(-1,0)$. Call $y$ such point, $y$ has a open neighborhood contained in the image because we can go arbitrarily near $-1,0)$ so there is always space inside the image for a little arc which contains $y$.
The formalization:
Is a boring reasoning with arbitrarily small ball (centered in $(1,0)$ and $y$) intersected with $S^1$.
A: Hint: Take an open set $V$ containing $(1,0)$ and show that $f^{-1}$ contains both $0$ and points arbitraily close to $1$.
An alternate way to prove it is to note that $\mathbb{S}^1$ is compact and continuous functions map compact sets onto compact sets.
Edit: I didn't read the question carefully enough.
The reason that your set is not open is that $f([0,1))$ is the part of the circle that lies in the right lower quadrant of the plane together with $(1,0)$. We can of course find points on the circle in the right upper quadrant of the plane that are arbitrarily close to $(1,0)$
