Difficulty in finding open sets and boundary of set in subspace topology

Question

Question: let $X=\mathbb{N}×\mathbb{Q}$ with subspace topology of the usual topology on $\mathbb{R^2}$ and $P=\{(n,\frac{1}{n}): n∈\mathbb{N}\}$ then

(1) In the space $X$, $P$ is open or closed or both or neither?

(2) what is boundary of $P$ in $X$

i known open sets in $(X, τ)$ is of the form, $\mathbb{N}×\mathbb{Q}∩S$ where $S$ is open set in $\mathbb{R^2}$ but by the representation of $P$, I am unable to determine whether $P$ is of this form or not and further, upto yet I had only find boundary of sets in $\mathbb{R}$ with usual topology, But they are asking to find boundary of set in the subspace topology. please help me, stuck on it from hours..... :-(

Answer 1

$P$ is closed in $X$. Consider $(n,q) \in X \setminus P$. Then $q \ne \frac1n$ so notice that $$B_X\left((n,q), \min\left\{\left|\frac1n - q\right|, 1\right\}\right) \subseteq X \setminus P$$

so $X \setminus P$ is open, therefore $P$ is closed.

$P$ is not open, for any $r > 0$ we have $$\left(1, 1+q\right) \in B_X(\underbrace{(1,1)}_{\in P}, r) \setminus P$$

where $q$ is a rational number in $\left\langle 0, \frac{r}2\right\rangle$.

The boundary $\partial P$ has to be a subset of $P$ since $P$ is closed. On the other hand, for every $\left(n, \frac1n\right) \in P$ and $r > 0$ we have that

$$\left(n, \frac1n+q\right) \in B_X\left(\left(n, \frac1n\right), r\right) \setminus P$$

where $q$ is a rational number in $\left\langle 0, \frac{r}2\right\rangle$. We conclude that $\partial P = P$.

Answer 2

The answer is closed but not open and the boundary is empty.

Because any open ball $B(x,\delta)$ for a point $x\in X$ is of the form {n}$\times(\mathbb{Q}\cap(a,b))$, contains countably many points. But any point $x\in P, x=(n,\frac{1}{n})$ the $B(x,\delta)$ contains only one point of P, the center itself, x is not an interior point of P and has no limit point. So P is closed but not open.

P has no interior, the boundary is P itself.