Closure question 
Determine the closures of the following subsets of the ordered square:
$$A = \{(1/n) \times 0 : n \in Z_+ \}$$

The key says
$\bar{A} = A \cup \{ 0 \times 1 \}$
I thought it was $\bar{A} = A \cup \{ (0,0) \}$
why is it  $\bar{A} = A \cup \{ 0 \times 1 \}$?
 A: You have to notice that the topology is the lexicographic order topology on the square $[0,1]\times[0,1]$: Define a linear order on the square by $(a,b)≤(c,d)$ if $a≤c$, or $a=c$ and $b≤d$. Sometimes referred to as the "dictionary" ordering for obvious reasons. Now you just give this set the order topology induced by ≤. An open set looks like a union of vertical open (in $[0,1]$) intervals.To picture this linear order: go from $(0,0)$ up to $(0,1)$, then start back at $(\epsilon ,0)$ and go up to $(\epsilon,1)$, back to $(2\epsilon,0)$, etc. As an exercise to test your understanding, prove that $\omega \times\mathbb [0,1)$, in the lexicographic order topology, is homeomorphic to $[0,\infty)$.
Now to to the problem. $(0,0)$ is NOT a limit point of $A$ in this topology: for instance, $0\times[0,1)$ is an open set containing $(0,0)$ but no points of $A$. 
$(0,1)$, on the other hand, is a limit of $A$ in this topology:  An open set containing $(0,1)$ contains a set of the form $0\times (y,1] \cup (0,x)\times [0,1]$, which contains an entire tail of the sequence that makes up $A$.
The ordered square is Hausdorff, which implies uniqueness of sequential limits. Thus $\overline {A}=A\cup ${$(0,1)$}.
