Find connected components of a a set Y Let Y be the union of all the circles of center (1,0) radius 1-1/n in R^2.  Then, we have circles of increasing radius, finally reaching r=1 as n goes to infinity.  The connected components are the equivalence classes under the equivalence relation x~y if and only if there exists a path from x to y. I think each equivalence class consists of the points on each circle in the set, since we proved that each circle is path-connected.  Thus for any points a,b on C1 there exists a path from a to b. So [a]=points on C1.  I'm not sure if there are more points in each equivalence class.  We also must find if each connected component is open in Y.  Assuming that each connected component in each circle in Y, it seems that there is a ball around each of the points on one circle, as represented by the successive circle.  I'm not sure of the last circle, C((1,0),1), since it is the last circle in the set as n goes to infinity.  Thanks!
 A: There is no "last circle"; there are no points in $Y$ whose distance to $(1,0)$ is exactly equal to $1$. If $y\in Y$, then there exists $n\in\mathbb{N}$ such that $y$ lies in the circle of radius $1 - \frac{1}{n}$ centered at $(1,0)$; since these circles are pairwise disjoint, there is a unique $n$ with this property, so $d(y,(1,0)) = 1 - \frac{1}{n}\lt 1$. 
Indeed, each circle is path connected. So you just need to show that no connected component contains points from more than one circle.
To show this, it suffices for each $n\in\mathbb{N}$ to find a set $U$ which is open in $\mathbb{R}^2$, and such that $U\cap Y$ is exactly the circle with radius $1 - \frac{1}{n}$. This will also show that each circle is open in $Y$ (in the induced topology).
And for this, you might consider an annulus around the $n$th circle, whose width is sufficiently small that it does not include any points whose distance to $(1,0)$ is $1 - \frac{1}{n+1}$ or $1-\frac{1}{n-1}$. Say...
$$\left\{ (x,y)\in \mathbb{R}^2 \;\left|\; 1-\frac{1}{n}-\epsilon\lt \sqrt{(x-1)^2 + y^2} \lt 1-\frac{1}{n}+\epsilon\right.\right\}$$
where $\epsilon$ is chosen so that $1-\frac{1}{n}-\epsilon\gt 1-\frac{1}{n-1}$ and $1-\frac{1}{n}+\epsilon\lt 1-\frac{1}{n+1}$. 
