Closure of a subset in ordered square I want to find the closure of these sets on the ordered square:
$$B=\left\{(1-\frac{1}{n})\times \frac{1}{2}|\ n \in Z_{+})\right\}$$
Consider $I^2 = [0,1] \times [0,1]$ in the dictionary order topology.
Our claim is $\overline{B} = B \cup \{1 \times 0\}$. At first, we will show $1 \times 0$ is a limit point of $B$. Every neighborhood $U$ of $1 \times 0$  is of form  $U: = (a \times b, 1 \times c)$ for some $a <1$ and $c > 0$. There exists $n \in \mathbb{N}$ such that $a < 1 - \frac{1}{n} <1$. Now $U$ intersects $B$ at points $1-\frac{1}{n} \times  \frac{1}{2}$, hence $1 \times 0 \in \overline{B}$.
Now we will see that no no other points of $I^2$  except $1 \times 0$ is a limit point.
Case 1: The point $0 \times y$ such that $y \in [0, \frac{1}{2})$ is not a limit point of $A$. Clearly $(0 \times 0, 0 \times \frac{1}{2})$ is a neighborhood of the point $0 \times y$, which does not intersect $B$, hence $0 \times y \in \overline{B}$.
Case 2: Let $x \times y \not\in B$, and $x <1$, then there exists $n \in \mathbb{N}$ such that $1-\frac{1}{n} <x < 1-\frac{1}{n+1}$. Then $((1-\frac{1}{n}) \times \frac{1}{2},(1-\frac{1}{n+1}) \times \frac{1}{2}) $ is a neighborhood which does not intersect $B$, hence $x \times y \not\in \overline{B}$.
$\textbf{UPDATE: }$  Let $x \times y \not\in B$, and $x <1$, then there exists $n \in \mathbb{N}$ such that $1-\frac{1}{n} \leq x < 1-\frac{1}{n+1}$. Now we have two cases:

*

*If $x = 1-\frac{1}{n}$ and $y <2$, then $((1-\frac{1}{n}) \times \frac{1}{2},(1-\frac{1}{n+1}) \times \frac{1}{2}) $ is a neighborhood of $x \times y$ which does not intersect $B$, hence $x \times y \not\in \overline{B}$.

*Otherwise, $((1-\frac{1}{n}) \times \frac{1}{2},(1-\frac{1}{n+1}) \times \frac{1}{2}) $ is a neighborhood of $x \times y$ which does not intersect $B$, hence $x \times y \not\in \overline{B}$.

Case 3: The point $1 \times y$ such that $y \in (0,1]$ is not a limit point of $B$. Clearly $(1 \times 1, 1\times \frac{1}{2})$ is a neighborhood which does not intersect $B$, hence $1 \times y \not\in \overline{B}$.
$\textbf{UPDATE:}$  The point $1 \times y$ such that $y \in (0,\frac{1}{2}]$ is not a limit point. Clearly $1 \times 0, 1 \times 1)$ is a neighborhood of $1 \times y$ which does not intersect $B$.
I am not sure about my solution. Is it correct? Please check.

Close.
Case 2 is missing when $x=1-1/n$. Case 3: $1\times 1> 1\times 1/2$, so your interval doesn’t make much sense.
A simpler way of doing this is showing that the complement of your proposed set is open by showing it’s a union of open intervals. Letting the $i$th element of $B$ be $b_i$, you could consider the union of $(b_i,b_{i+1})$ and $(1\times 0,1\times1]$.
