Multiple choice question from general topology Let $X =\mathbb{N}\times \mathbb{Q}$ with the subspace topology of $\mathbb{R}^2$ and $P = \{(n,  \frac{1}{n}):  n\in \mathbb{N}\}$ .
Then in the space $X$
Pick out the true statements
1 $P$ is closed but not open
2 $P$ is open but not closed
3 $P$ is both open and closed
4 $P$ is neither open nor closed
what can we say about boundary of $P$ in $X$?
I always struggle to figure out subspace topology. Though i am aware of basic definition and theory of subspace topology. I need a bit explanation here about how to find out subspace topology of $P$? 
Thanks for care
 A: For example, 1 is true. You can see that $P$ is not open by looking at an $\varepsilon$-ball around any point $p = (n, \frac1n )$ in $P$. Then there will be a rational $q$ such that $(n,q)$ is inside the ball hence $P$ is not open. (because $\mathbb Q$ is dense in $\mathbb R$)
Also, it's closed: think about why its complement is open. (You can make an $\varepsilon$-ball around the point that doesn't intersect with $P$)
Now that we have established that 1 is true, we know that 2, 3 and 4 are false.
A: Draw a picture: $P$ is a set of points on the positive half of the graph of the hyperbola $y=\frac1x$. Let’s look at one of those points, say $\left\langle 4,\frac14\right\rangle$. Is $\{4\}$ an open set in $\Bbb N$? Is $\Bbb Q$ an open set in $\Bbb Q$? If the answers to these questions are both yes, then $$\{4\}\times\Bbb Q=\{\langle 4,y\rangle:y\in\Bbb Q\}$$ is an open set in $X$. Call this set $U$; what is $U\cap P$?
This should help in deciding whether $P$ is open in $X$. To decide whether $P$ is closed in $X$, you need to consider a point $\langle n,q\rangle\in X\setminus P$ and ask whether this point can possibly be a limit point of $P$. It will help to realize that sets of the form $\{n\}\times(q-\epsilon,q+\epsilon)$ are open in $P$, because $\{n\}$ is open in $\Bbb N$, and $(q-\epsilon,q+\epsilon)$ is open in $\Bbb Q$. (Here I’m taking the interval in $\Bbb Q$, not in $\Bbb R$.)
A: P is closed but not open because epsillion ball around p=(n, 1/n) in P Then there will be rational q s.t (n, p) is inside the ball hence p is not open 
