Closed in $\mathbb{R}^{2}$ vs closed in $\mathbb{R}$ Define the following subsets of $\mathbb{R}$ and $\mathbb{R}^{2}$ respectively:
$A=\left\{\frac{1}{n}: n \in \mathbb{N}\right\}$
$B=\left\{(x,\frac{1}{x}): x\in\mathbb{R} \smallsetminus \left\{0\right\} \right\}$
I know the first subset, $A$, is neither open nor closed, and the second subset, $B$, is closed.
However, the set $B$ is exactly $A$ but "viewed" in $\mathbb{R}^{2}$, so, why $B$ is neither open nor closed in $\mathbb{R}^{2}$?   
EDIT: 
Redefine $B$ as:
$B=\left\{(x,\frac{1}{x}): x \in \mathbb{N}\right\}$. 
$B$ is closed in $\mathbb{R}^{2}$, my question is why intuition fails? ie, $A$ is the projection of $B$ on $\mathbb{R}$ and since $A$ is neither open nor closed in $\mathbb{R}$, I thought $B$ was equally neither open nor closed in $\mathbb{R}^{2}$.
 A: The set $$ B=\left\{(x,\frac{1}{x}): x\in\mathbb{R} \smallsetminus \left\{0\right\} \right\}$$ is closed because its complement is open.
The set$$ A=\left\{\frac{1}{n}: n \in \mathbb{N}\right\}$$ is not closed and it is not  the same as the set $B$ because it does not contain its limit point $0$
The two sets are not the same and obviously one is countable and the other is not.
A: If you change the definition of $B$ to $B=\left\{(x,\frac{1}{x}): x \in \mathbb{N}\right\}$ then you have a discrete set, and any pair of points are at least one unit of distance separated. So there is no limit point for $B$ that is outside of $B$.
This is not the case with $A$, where points in $A$ can get arbitrarily close to each other as they converge to $0$, and the point $0$ is a limit point that is not in $A$ itself.
Roughly, open/closed is based on distance for these spaces under the usual topology. And in $A$, the points are arranged so that their distances can be arbitrarily close. But in the redefined $B$, the points are arranged in a different way that spreads them out too much to avoid getting a closed set.
