Finding the closure of some subsets of the ordered square I need to find the closure of these sets on the ordered square:
$$A = \{(1/n)\times 0 | n\in \mathbb{Z}_+\}$$
$$B = \{(1-1/n)\times \frac{1}{2}| n\in \mathbb{Z}_+\}$$
$$C = \{x\times 0 | 0<x<1\}$$
$$D = \{x\times \frac{1}{2}| 0<x<1\}$$
$$E = \{\frac{1}{2}\times y | 0<y<1\}$$
I know that the closure is just the set itself united with its limit points. For $A$, the set of limit points, I imagine, are $\{<1,0>\}$, that is, the $x$ coordinate is $1$ and the $y$ is $0$. But here is says the reverse. What am I getting wrong?
Following the same reasoning, $B' = \{<1,\frac{1}{2}>\}$
$C' = \{<0,0>,<0,1>\}$
What am I doing wrong?
 A: Let $X=[0,1]\times[0,1]$ with the lexicographic order topology, and let $\preceq$ denote the lexicographic order on $X$. The points of $A$ are the points $\left\langle\frac1n,0\right\rangle$ with $n\in\Bbb Z^+$:
$$A=\left\{\langle 1,0\rangle,\left\langle\frac12,0\right\rangle,\left\langle\frac13,0\right\rangle,\left\langle\frac14,0\right\rangle,\ldots\right\}\;.$$
Make a sketch. The set $\left(\frac34,1\right]\times[0,1]$ is an open nbhd of $\langle 1,0\rangle$ that contains no point of $A$ except $\langle 1,0\rangle$, so $\langle 1,0\rangle$ is not a limit point of $A$. In the usual topology on $X$ these points would have $\langle 0,0\rangle$ as a limit point, but it has an open nbhd in $X$ that is disjoint from $A$: $\{0\}\times[0,1)$. (Again, make a sketch.) However, every open nbhd of $\langle 0,1\rangle$ does intersect $A$.
To see this, let $p=\langle 0,1\rangle$, and suppose that $U$ is an open nbhd of $p$. Then there are $u,v\in X$ such that $p\in(u,v)\subseteq U$, where $(u,v)$ is the open interval between $u$ and $v$ in the order topology. This means that $u\prec p\prec v$. If $u=\langle a,b\rangle$ and $v=\langle c,d\rangle$, we have
$$\langle a,b\rangle\prec\langle 0,1\rangle\prec\langle c,d\rangle\;,$$
which implies that $a=0,b<1$, and $c>0$. If you make a sketch, it should be clear that
$$(u,v)=\big(\{0\}\times(b,1]\big)\cup\big((0,c)\times[0,1]\big)\cup\big(\{c\}\times[0,d)\big)\;.$$
If $n>\frac1c$, then $\frac1n<c$, and $\left\langle\frac1n,0\right\rangle\in A\cap(u,v)\subseteq A\cap U$. Thus, every open nbhd of $p=\langle 0,1\rangle$ intersects $A$, and $\langle 0,1\rangle$ is in the closure of $A$.
To complete the argument that $\operatorname{cl}A=A\cup\{\langle 0,1\rangle\}$ you should show that no other point of $X$ besides $\langle 0,1\rangle$ is a limit point of $A$.
Now look at the set $B$: it lies on the line $y=\frac12$, and in the usual topology on $X$ it would converge to the point $\left\langle 1,\frac12\right\rangle$. But $\{1\}\times(0,1)$ is an open nbhd of $\left\langle 1,\frac12\right\rangle$ disjoint from $B$, so $\left\langle 1,\frac12\right\rangle$ clearly isn’t a limit point of $B$. (Specifically, $\{1\}\times(0,1)$ is the open interval from $\langle 1,0\rangle$ to $\langle 1,1\rangle$ in $X$.) If $\langle a,b\rangle\in X$ with $a<1$, it’s easy to find an open nbhd of $\langle a,b\rangle$ that contains no point of $B$ (except $\langle a,b\rangle$ itself, if $\langle a,b\rangle\in B$), so for limit points of $B$ we need to look at points $\langle 1,b\rangle$. The open interval $\{1\}\times(0,1)$ that shows that $\left\langle 1,\frac12\right\rangle$ isn’t a limit point of $B$ works just as well for every point $\langle 1,b\rangle$ with $b>0$, so the only remaining candidate is the point $\langle 1,0\rangle$. Showing that $\langle 1,0\rangle$ is a limit point of $B$ is very much like showing that $\langle 0,1\rangle$ is a limit point of $A$: show that every open interval in $X$ containing $\langle 1,0\rangle$ contains a set of the form $(a,1)\times[0,1]$ for some $a<1$ and therefore contains infinitely many points of $B$.
What you’re missing, I think, is a feel for the lexicographic order. The first point in that order is $\langle 0,0\rangle$. After that the order runs up the segment $\{0\}\times[0,1]$, and $\langle 0,1\rangle$ is the last point on that vertical segment: it comes after every other point on the segment, and before every point of $X$ with a positive $x$-coordinate. Similarly, $\langle 1,1\rangle$ is the last point in the order; if you start there and run through the order backwards, you run down the segment $\{1\}\times[0,1]$, and the last point on that segment that you hit is $\langle 1,0\rangle$: it’s the point of that segment that is ‘closest’ (in the lexicographic order) to any point with an $x$-coordinate less than $1$.
Because of that, you seem to be looking at the product topology rather than the order topology. Take $C$: $\langle 0,0\rangle$ would indeed be a limit point in the usual (product) topology, but in the order topology the set $\{0\}\times[0,1)$ is an open nbhd of $\langle 0,0\rangle$ disjoint from $C$. On the other hand, $A\subseteq C$, so $\operatorname{cl}A\subseteq\operatorname{cl}C$, and we know that $\langle 0,1\rangle$ is a limit point of $A$, so it must be a limit point of $C$ as well. It is true that $\langle 1,0\rangle$ is a limit point of $C$, but probably not for the reason that you had in mind: the argument is similar to the proof that it’s a limit point of $B$. But it also turns out that every point $\langle a,1\rangle$ with $0\le a<1$ is a limit point of $C$:
$$\operatorname{cl}C=\big((0,1]\times\{0\}\big)\cup\big([0,1)\times\{1\}\big)\;$$
all of the bottom edge of the square except $\langle 0,0\rangle$ and all of the top edge except $\langle 1,1\rangle$. To prove this, let $p=\langle a,1\rangle$, where $0\le a<1$, and show that every open nbhd of $p$ contains a set of the form $(a,b)\times[0,1]$ for some $b$ such that $a<b\le 1$; since every such set certainly contains points of $C$, it will follow that $p$ is a limit point of $C$. It still remains to show that no other points are in the closure of $C$, but that’s easy: the complement of the set
$$\big((0,1]\times\{0\}\big)\cup\big([0,1)\times\{1\}\big)$$
is
$$\big(\{0\}\times[0,1)\big)\cup\big((0,1)\times(0,1)\big)\cup\big(\{1\}\times(0,1]\big)\;,$$
which is easily shown to be open in $X$.
I’ll leave the other two for you to think about further. You should find that $\operatorname{cl}D=D\cup\operatorname{cl}C$ and that $\operatorname{cl}E$ is the same as it would be in the usual topology, $\left\{\frac12\right\}\times[0,1]$.
