Is the set C closed? Let $C = \bigcup\limits_{n\geq 1} C_n$ where $$C_n = \left\{(x,y)\middle|x^2+y^2=\left(\frac{n}{n+1}\right)^2\right\} $$ 
Is C closed? 
I think that it is not closed since it doesn’t contain all of its limit points, but I don’t know how to prove it. 
 A: The set is a union of concentric circles of different radii.  Each radii is less than $1$ but $\frac n{n+1}\to 1$.
So the point $(0,1)$ is a limit point (as is any $(x,y)$ on a unit circle).  For every $\epsilon > 0$ there is an $n$ so that $0< \frac 1{n+1}<\frac 1n < \epsilon$ and so the point $(0, 1-\frac 1{n+1})\in B_\epsilon((0,1))$ and $0^2 + (1-{n+1})^2 = (\frac n{n+1})^2$ so $(0, 1-\frac 1{n+1}) \in C$.  
But $(0,1)\not \in C$ as $0^2 + 1^2 = 1$ and there is no $n$ so that $(\frac {n}{n+1})^2 = 1$.
So not closed.
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It might be worth noting that $\mathbb R \subset \mathbb R^2$ so if $C$ is closed/open in $\mathbb R^2$ then $C\cap \mathbb R$ would be closed/open in $\mathbb R$.
And $C\cap \mathbb R$ is $\{x|x = \pm\frac {n}{n+1}\}$ which should be one of the standard easily recognize non-closed sets.
So as $C\cap\mathbb R$ is not closed it's not possible for $C$ to be closed.
A: As long as you show is does not contain all of its limit points, then it isn't closed in the metric topology on $\mathbb{R}^{2}$. It sounds like you've already come up with a sequence with a limit point that isn't contained in one of these sets. 

 If not, then $\left(\frac{n}{n+1},0\right)_{n \geq 1}$ is one such example. 

