Boundary of the set $\{(n,{1\over n}):n\in \mathbb N\}$ in $\mathbb N\times \mathbb Q$ $X=\mathbb N\times\mathbb Q$ is the topological space with induced topology from $\mathbb R\times \mathbb R.$ Consider the set $$P=\left\{\left(n,{1\over n}\right):n\in \mathbb N\right\}.$$
$1.$ Closedness: This set $P$ is not closed.Because the points $\left\{(n,0):n\in \mathbb N\right\}$ are all limit points of $P$ but not in $P.$
$2.$ Openness: This set $P$ is not open either.Let's take an open neighbourhood $U$ of a fixed point $\left(n,{1\over n}\right)$ from $\ \mathbb R\times \mathbb R\ $ in such a way that it stays completely in the first co-ordinate.Because of subspace topology , we intersect $U$ with $X.$ But $U\cap X\notin P$ because obviously $U\cap X$ will have points of form $(n,q)$ where $q\in \mathbb Q$ and $q\neq {1\over n}\forall n\in \mathbb N.$
$3.$ Boundary set of $P$: A point $p\in X$ is a boundary point of $A\subset X$ if for every open nbd $V$  of $p$, $V\backslash \{p\}$ contains points from both $A$ and $A^c.$ Following this definnition , I see the whole of $\{(n,0):n\in \mathbb N\}$ is  the boundary of $P.$ That is cardinality of the boundary set is countably infinite.
Are my assessments correct ? This was from a multiple choice question. Now the neither closed nor open option was there but boundary is countably infinite option wasn't. That's confusing me if I made a mistake. For boundary, the question assumed the cardinality to be either $0$ or finite. Typos?
Thanks.
 A: $P$ is closed and not open, and its boundary is $P$ itself. It is also discrete.


*

*$P$ is closed: If $\langle n,q\rangle\notin P$, then $q\ne\frac1n$. Let $U=\Bbb Q\setminus\left\{\frac1n\right\}$; $U$ is an open subset of $\Bbb Q$, and $\{n\}$ is an open subset of $\Bbb N$, so $\{n\}\times U$ is an open set in $\Bbb N\times\Bbb Q$. Clearly $\langle n,q\rangle\in\{n\}\times U$. If $\left\langle m,\frac1m\right\rangle\in P$, then either $m\ne n$, in which case $\left\langle m,\frac1m\right\rangle\notin\{n\}\times U$ because $m\notin\{n\}$, or $m=n$, in which case $\left\langle m,\frac1m\right\rangle=\left\langle n,\frac1n\right\rangle\notin\{n\}\times U$ because $\frac1n\notin U$. In short, $P\cap(\{n\}\times U)=\varnothing$, so $\{n\}\times U$ is an open nbhd of $\langle n,q\rangle$ disjoint from $P$. Thus, each point of $(\Bbb N\times\Bbb Q)\setminus P$ has an open nbhd disjoint from $P$, so $(\Bbb N\times\Bbb Q)\setminus P$ is open, and $P$ is therefore closed.

*$P$ is not open: In fact the interior of $P$ is empty: $P$ does not contain any non-empty open subset of $\Bbb N\times\Bbb Q$. Let $\mathscr{B}=\{(p,q)\cap\Bbb Q:p,q\in\Bbb Q\text{ and }p<q\}$, the set of open intervals in $\Bbb Q$ with rational endpoints; $\mathscr{B}$ is a base for the topology of $\Bbb Q$. The sets $\{n\}$ for $n\in\Bbb N$ form a base for the topology of $\Bbb N$. Thus, $\{\{n\}\times(p,q):n\in\Bbb N\text{ and }(p,q)\in\mathscr{B}\}$ is a base for the product topology on $\Bbb N\times\Bbb Q$. Let $\left\langle n,\frac1n\right\rangle\in P$. Every open nbhd of $\left\langle n,\frac1n\right\rangle$ contains a basic open nbhd of $\left\langle n,\frac1n\right\rangle$, i.e., a set of the form $\{n\}\times(p,q)$, where $p<\frac1n<q$, and $\left\langle n,\frac1n\right\rangle$ is the only point of $P$ in this open nbhd: the other points of $P$ all have first coordinates different from $n$. Thus, $\{n\}\times(p,q)\nsubseteq P$, and $P$ does not contain any non-empty open set.

*The boundary of $P$ is $P$: $P$ is closed, so every point of $(\Bbb N\times\Bbb Q)\setminus P$ has an open nbhd disjoint from $P$ and therefore cannot be a boundary point of $P$. If $\left\langle n,\frac1n\right\rangle\in P$, then of course every open nbhd of $\left\langle n,\frac1n\right\rangle$ contains a point of $P$, and we just saw in the previous point that no open nbhd of $\left\langle n,\frac1n\right\rangle$ is a subset of $P$. This means that every open nbhd of $\left\langle n,\frac1n\right\rangle$ contains points of $(\Bbb N\times\Bbb Q)\setminus P$, so every point of $P$ is a boundary point of $P$. Thus, $\operatorname{bdry}P=P$.

*$P$ is discrete: For each $\left\langle n,\frac1n\right\rangle\in P$, $\{n\}\times\Bbb Q$ is an open nbhd of $\left\langle n,\frac1n\right\rangle$ whose intersection with $P$ is $\left\{\left\langle n,\frac1n\right\rangle\right\}$.
