What is the closure of the set $B=\{(\frac{1}{n}, 0) \in\mathbb{R}^2 \ \colon \ n\in\mathbb{N}\}$

what is the closure of the following set

$B=\{(\frac{1}{n}, 0) \in\mathbb{R}^2 \ \colon \ n\in\mathbb{N}\}$

I get the following;

$cl(B)=int(B)\ \cup \ \partial(B)= B \ \cup \{0,1\} = [1,0]$.

However in the solutions it states

$cl(B)=int(B)\ \cup \ \partial(B)= B \ \cup \{0\} = (1,0]$.

Where have i gone wrong?

Image of question and solution

Question 2(ii)(b)

Answer 2(ii)(b)

Solution 1:

If our set $B$ is the set of points of $\mathbb{R}^2$ of the form $(\frac{1}{n},0)$ s.t. $n\in\mathbb{N}$ then this is obviously a subset of $\mathbb{R}^2$ but more specific, its a subset of the line $y=0$ in the real plane. So the closure of $B$ is ( i think obviously ) the set $\bar{B}=B\cup (0,0)$.