Closure and accumulation point in $\mathbb R^2$ I gave a topology defined on $\mathbb{R}^2$ by it basis  
$$O_m=\{(x,y)\in\mathbb{R}^2, \max(x,y)\leq m\}$$
where  $m\in \mathbb{N}$.
The question is to find the closure and the set of accumulation points of  
$$A=\{(x,y)\in \mathbb{R}^2, x^2+y^2-6x+5\leq 0\}$$
I found 
$$cl{(A)}= \mathbb{R}^2- O_0$$
and 
$$A'=cl(A)-\{(1,0)\}$$ 
is it right ? thank you 
 A: Note that I have made significant edits to this, years after the first answer. The answer had a few errors. To whomever upvoted the answer, I apologise for misleading you.
It is not hard to see that $A$ is the set of all points distance $d \le 2$ from
the point $(3,0)$.
That is, $A= \{ (x,y) | \|(x,y)-(3,0)\|_2 \le 2 \}$. Note that $(1,0) \in A$.
It is not hard to see that the open sets are $\emptyset, O_1, O_2,...$ and $\mathbb{R}^2$. Hence the closed sets are $\mathbb{R}^2, O_1^c,...$ and $\emptyset$. The only closed set containing $A$ is $\mathbb{R}^2$. In particular,
$\bar{A} = \mathbb{R}^2$.
I am presuming that $p$ is an accumulation point iff any open set that contains $p$ also contains an element of $A$ different from $x$.
Since the closure is the disjoint union of the accumulation points and the isolated points, it suffices to identify the isolated points.
Note that $A \cap O_n$ contains many points for $n>1$ and $A \cap O_1 = (1,0)$. Hence $(1,0)$ is the only isolated point.
Hence the accumulation points are $\{(1,0)\}^c$.
