Find the smallest topology on the unit circle $S^1$ which makes $\alpha$ continuous I need to find the smallest topology on $S^1$ which makes the following function $\alpha:S^1\rightarrow\mathbb{R}:\alpha(x,y)=x$ continuous. This is what I have so far:
$(x,y)=(x,\pm\sqrt{1-x^2})\in S^1$. But because we need the inverse here I'm guessing I just should take the plus sign. For the inverse $\alpha^{-1}:\Bbb R \to S^1$ we have $\alpha^{-1}(x) = (x,\sqrt{1-x^2})$. Now we define the set
$$
S := \{U \subset S^1: \exists U\in\tau: U=\alpha^{-1}(U) \}.
$$
Because $x\in[0,1], S=[0,1]\times[0,1]$. Now the book says that the smallest topology on $S^1$ that makes $\alpha$ continuous is the topology generated by $S$. So
$$
\langle S \rangle := \hskip{-2em} \bigcap_{\tau\text{-topology on $S^1$ containing $S$}}\hskip{-3em} \tau
$$
However I have no idea how to find this. Thinking about $S=[0,1] \times [0,1]$, I was thinking about the Euclidean Topology, but this is not possible because I also have to show that it is not Hausdorff, and any metric space is Hausdorff. Can I get some push in the right direction, because I think i'm overcomplicating the problem.
 A: The topology will be generated by inverse images of open sets in $[-1,1]$. Open intervals in the interior of this interval will have two open arcs as preimage. The inverse image of a neighborhood of an endpoint will be a symmetric open arc around the point $(1,0)$ or $(-1,0)$. In general, the topology that these preimages generate consists of standard open sets in $S^1$ with the additional property that they are symmetric in the $x$-axis. This is not a Hausdorff topology since no open sets can separate a point from its reflection in the $x$ axis.
A: You cannot ignore the fact that most points of $[-1,1]$ have two preimages under $\alpha$.
Suppose that $-1\le a<b\le 1$; then 
$$\alpha^{-1}[(a,b)]=\{\langle x,y\rangle\in S^1:x\in(a,b)\}=S^1\cap\Big((a,b)\times\Bbb R\Big)\;.$$
This is the union of two open arcs on $S^1$, one in the upper semicircle and the other in the lower semicircle. If $\tau$ is the desired topology, then all sets of this type must belong to $\tau$. So must all sets of the form
$$\alpha^{-1}[[-1,b)]=\{\langle x,y\rangle\in S^1:x\in[-1,b)\}=S^1\cap\Big([-1,b)\times\Bbb R\Big)\;,$$
which are symmetric open arcs of $S^1$ with centres at $\langle -1,0\rangle$, and all sets of the form
$$\alpha^{-1}[(a,1]]=\{\langle x,y\rangle\in S^1:x\in(a,1]\}=S^1\cap\Big((a,1]\times\Bbb R\Big)\;,$$
which are symmetric open arcs of $S^1$ with centres at $\langle 1,0\rangle$.
Since the sets $[-1,b),(a,b)$, and $(a,1]$ with $-1\le a<b\le 1$ are a base for the topology of $[-1,1]$, their inverses under $\alpha$ are a base for $\tau$.
To show that $\tau$ is not Hausdorff, just show that $\langle 0,1\rangle$ and $\langle 0,-1\rangle$ cannot be separated by disjoint members of $\tau$. You can actually show something stronger: every $\tau$-open set containing $\langle 0,1\rangle$ also contains $\langle 0,-1\rangle$, and vice versa.
