Topology on union of circumferences Let $$X = \bigcup^{\infty} C_{n}$$ be the subset of $\mathbb{R}^2$ union of the circumferences $ C_n $ having ray 1/n and center (1/n,0), n>0, endowed with the topology induced by that of $\mathbb{R}^2$.
I know the open sets are exactly the intersections of X with the open sets in $\mathbb{R}^2$, and that the circumferences are closed and not open.
1)I now have to prove that X is separable and metrizable, but I cannot see how. The only countable set that seems promising to me and could be dense is {(1/n,0), (0,0)}; if this is the case, I think I recognise it is not closed, so its closure must be bigger, but I fail to notice why it must be all of X.
I also have no idea why it is metrizable.
2) I would also need some hint on how to proceed to show it's (locally) path-connected.
Thanks in advance!
 A: $X$ is a subspace of $\mathbb{R}^2$. Subspace of metrizable is metrizable, and subspace of second-countable is second-countable. Being metrizable (and second-countable), it is also separable. So that part is accounted for.
Now, if $p \in X$ is not $(0,0)$, you can get arbitrarily small neighbourhoods $V$ of $p$ in the induced topology consisting of only a segment of the circle that $p$ belongs to. If $p$ is $(0,0)$, you can get arbitrarily small neighbourhoods which are similar to "star convex" domains. So it follows that it is locally path-connected.
A: Any circle has a countable dense subset. (say it is centered at $(0,0)$ and take intersections of it with rays $y=rx,$ where $r$ is rational.) Then your space has a countable union of countable sets dense in it.
For local path connectedness one way is to take two points in the union and "go along" each circle to $(0,0)$ and then connect the two paths.
EDIT The last paragraph doesn't work for local path connected, since the two arcs meeting at $(0,0)$ might go outside a neighborhood of the point $P.$ However if $P=(0,0)$ and $Q$ is near $P$ one can just go along the circle through $Q$ in the direction toward $P.$ And if $P \neq (0,0)$ then it is at some definite distance $d>0$ from $(0,0)$ and then all but finitely many of the circles will be inside the disc of radius $d$ centered at $(0,0).$ Then if one takes a small enough neighporhood of $P$ it will contain points only from the circle through $P$ and close to it, and such points connect to $P$ along the circle through $P.$
