Separating Sets In The Radial Plane Topology I'm attempting to show that given the radial plane topology and sets $A$ and $B$ defined by: $$A=\{(\cos(t), \sin(t)): t \in [0,2\pi) \cap \mathbb{Q}\}$$ and $$B=\{(\cos(t), \sin(t)): t \in [0,2\pi) \cap \mathbb{I}\}$$ That $A$ and $B$ cannot be separated by disjoint open sets.
My idea was to suppose by contradiction that we do have such a separation, say $\{U,V\}$ with $A \subseteq U$ and $B \subseteq V$, and then consider the point $(0,1) \in A$. 
Then $(0,1) \in U$. So there must be line segments through $(0,1)$ in every direction which are back in $U$. One of these line segments must hit $B$ in some way which is a contradiction. I'm just not sure about how to say this formally.  
 A: Let's construct a point in $U\cap V$. With $u,l$ we denote the up and left direction.
We take a (suitable) point $x_0\in A$ (on the lower right quadrant) such that $[x_0,x_0+\epsilon_0 u]\subset U$ is inside the unit circle. Draw a line a $45^°$ degree back to the circle and choose $y_0\in B$ such that $y_0$ is on the arc between $x_0$ and the intersection. Now we repeat the step for $[y_0,y_0+\delta_0 l]\subset V$ and choose a point $x_1\in A$ on the arc between the (new) intersection an $y_0$. (A small drawing would be good)
Repeating it we get two convergent sequences $z:=\lim x_n = lim y_n$ with $z\in A \lor z\in B$. By construction the line $[z, z + \epsilon(u+l)]$ - i.e. a line at $45^°$ degree - intersects at least one $[x_n,x_n+\epsilon_n l]$ and $[y_n,y_n+\delta_n l]$ for all $\epsilon > 0$. Because of $z\in A \lor z\in B$ we have  $[z, z + \epsilon(u+l)]\subset U \lor [z, z + \epsilon(u+l)]\subset V$ for some $\epsilon$. Therefore we find a point $z_0$ on $[z, z + \epsilon(u+l)]$ such that $z_0\in U\cap V$.
A: This is a job for the Baire category theorem.
I will just use that $A$ and $B$ are a partition of the unit circle with $A$ and $B$ dense in the usual topology on the unit circle.
Suppose for contradiction that there are disjoint sets $U,V$ open in the radial plane topology, with $A\subseteq U$ and $B\subseteq V.$
For all positive integers $n,$ define:


*

*$U_n$ to be the set of $t\in[0,2\pi)$ such that $(\cos t-\lambda\sin t,\sin t+\lambda\cos t)\subseteq U$ for all $|\lambda|\leq 1/n$ (i.e. the tangent line segment of length $1/n$ in each direction is contained in $U$)

*$V_n$ to be the set of $t\in[0,2\pi)$ such that $(\cos t-\lambda\sin t,\sin t+\lambda\cos t)\subseteq V$ for all $|\lambda|\leq 1/n$
Using the definition of the radial plane topology, every point in the unit circle is in some $U_n$ or $V_n.$
By the Baire category theorem for the usual topology on $[0,2\pi),$ there exists $n$ such that either $U_n$ or $V_n$ is dense in a non-degenerate open interval $I\subset [0,2\pi)$ in the usual topology.
Swapping $A$ and $B$ if necessary we can assume it's $U_n$ that's dense in $I.$ Since $\bigcup_{m\geq n} V_m$ is dense in $I,$ there exists $b\in I$ with $(\cos b,\sin b)\in V_m$ for some $m\geq n.$
Because $U_m\supseteq U_n$ is dense in $I,$ there also exists $a$ with $b<a<b+2\arctan(1/m)$ and $(\cos a,\sin a)\in U_m.$
The tangent lines to the points $P_1=(\cos a,\sin a)$ and $P_2=(\cos b,\sin b)$ meet at some third point $P_3.$ There is a right triangle $OP_1P_3$ where $O$ is the origin, the angle at $O$ is $(b-a)/2,$ and the adjacent side $OP_1$ has length $1,$ so the length of the opposite side $P_1P_3$ is $\tan(\tfrac{b-a}2).$ The length of $P_3P_2$ is the same by symmetry.
The choice of $a<b<a+2\arctan(1/m)$ ensures $\tan(\tfrac{b-a}2)\leq 1/m,$ which implies $P_3\in U_m\cap V_m,$ contradicting the assumption that $U$ and $V$ are disjoint.
