Subsets of complex plane with euclidean metric that are complete Here is a problem from a past topology exam I found and was trying to tackle:
Let $A \subset S^1 \subset\mathbb{C}$ and define $X(A) := \bigcup_{z\in A} \{ tz : t \in \mathbb{R}\}$:
a) Charachterize those sets $A$ for which $X(A)$ with euclidean metrix is complete
b) Show that if A is countable then there exists a complex $w \in S^1$ such that $\forall {z \in A}$ $\mathrm{arg}(wz)$ is irrational (here $\mathrm{arg}(x)$ is the argument of a complex number $x$).
ad. a)
If you think of $X(A)$ as a family of lines with direction vector $(a,b)$ where $a+bi \in A$ than I think the answer is all closed subsets of $A$ (because we need the limit point to be in the set $X(A)$ and the interior is complete anyway as a subset of $\mathbb{R}^2$) but is the answer full?
ad.b)
I intuitively see that I need to show that you can "rotate" a countable set of points on a circle in such a way that their argument is irrational, but I have no idea how to prove it formally.  
I'd appreciate any help, thanks.
 A: In a complete metric space, a subset is closed iff it is complete (with respect to the same metric). Note that $S^1=(\mathbb C - \{0\})/\sim$ where $z\sim w$ if there exists $\lambda\in\mathbb R$ such that $z =\lambda w$. With the quotient topology (which is equivalent to the subspace topology), $A\subset S^1$ is closed iff $X(A)-\{(0,0)\}$, iff $X(A) $ is closed.
For part b, suppose there existed a countable subset $A\subset S^1$ such that $\forall w\in\mathbb C$, $\arg(aw)\in\mathbb Q$ for some $a\in A$. Assume $\arg$ is to be taken with the branch $[0,2\pi)$. Let $Q$ be the subset $\{q+2\pi K: q\in\mathbb Q, K\in\mathbb Z\}$, note that $Q$ is countable. Since $\arg(aw) =\arg(a) \arg(w) \mod 2\pi$ and $\arg(w) \in [0,2\pi)$, we may express our assumption using $Q$ as follows: for any $ r\in (\frac 1{2\pi},\infty)$, there exists $q+2\pi K\in Q$ so that $\frac 1r = \frac1{q+2\pi K} \arg(c) $ for some $c\in A$. This statement is absurd because $(0,2\pi)$ is uncountable while $Q\times A$ is countable. 
