Closures in ordered squares Prove that the following closures of the following subsets of the $[0,1]\times [0,1]$ are true:
(a) $B=\left\{(1-\frac{1}{n})\times \frac{1}{2}|\ n \in Z_{+})\right\}$ has $Cl(B)= B\cup \left\{1\times 0\right\}$
(b) $C=\left\{x\times 0\ |\ 0<x<1\right\}$ has $Cl(C)= C\cup [0,1)\times 1$
(c) $D = \left\{x\times \frac{1}{2}\ |\ 0<x<1\right\}$ has $Cl(D) = D\cup (0,1]\times 0\cup [0,1)\times 1$
(d) $E = \left\{\frac{1}{2}\times y\ |\ 0<y<1\right\}$ has $Cl(E) = E\cup \left\{\frac{1}{2}\times 0, \frac{1}{2}\times 1\right\}$
I have thought of the proof for a while, but I couldn't get how we can obtain the extra points in those closure formulas above. Can someone please help me with this part?
 A: Assuming that $[0,1]\times[0,1]$ has the lexicographic order topology and that $x\times y$ is your notation for the ordered pair $\langle x,y\rangle$, then (a), (c), and (d) are correct. However, (b) is not quite right: you need to add the point $1\times 0$ to what you already have.
Added: I’ll use $\prec$ and $\preceq$ for the lexicographic order. To see why $1\times 0\in\operatorname{cl}B$, let $U$ be an open nbhd of $1\times 0$. Then there are $x_0\times y_0$ and $x_1\times y_1$ such that $1\times 0\in(x_0\times y_0,x_1\times y_1)\subseteq U$. There’s nothing ‘below’ $1\times 0$ on the line $x=1$, so the only way to have $x_0\times y_0\prec 1\times 0$ is to have $x_0<1$: if $x_0$ were $1$, we’d have $x_0\times y_0=1\times 0$ if $y_0=0$, and $1\times 0\prec x_0\times y_0$ if $y_0>0$, so no matter what $y_0$ is, we’d have $1\times 0\preceq x_0\times y_0$ instead of $x_0\times y_0\prec 1\times 0$. Now choose $n\in\Bbb Z^+$ large enough so that $x_0<1-\frac1n$; then $x_0\times y_0\prec\left(1-\frac1n\right)\times\frac12\prec 1\times 0$, so 
$$\left(1-\frac1n\right)\times\frac12\in U\cap B\;.$$
$U$ was an arbitrary open nbhd of $1\times 0$, so we’ve shown that every open nbhd of $1\times 0$ contains a point of $B$ and hence that $1\times 0\in\operatorname{cl}B$.
The arguments for the extra points in (b) are similar to those for (c), so I’ll skip them. In (c) consider the point $p=x\times 0$, where $0<x\le 1$. If $U$ is an open nbhd of $p$, there must be an $x_0<x$ and a $y_0>0$ such that $p\in(x_0\times 1,x\times y_0)\subseteq U$. 

You may have to think about this a little: you know that there must be some open interval $(q,r)\subseteq U$ such that $q\prec p\prec r$, and with a bit of thought you should be able to convince yourself that we can always shrink the interval $(q,r)$ if necessary so that $r$ lies ‘above’ $p$, and $q$ is at the ‘top’ of its vertical segment, i.e., has second coordinate $1$.

Now choose any $x_1$ such that $x_0<x_1<x$; then $x_1\times\frac12\in U\cap D$. Thus, every open nbhd of $p$ contains points of $D$, and $p\in\operatorname{cl}D$. The argument that $p=x\times 1\in\operatorname{cl}D$ whenever $0\le x<1$ is similar, except that you work to the right of $p$ instead of to the left.
Finally, (d) is the easiest, since the set $E$ is just the vertical open interval between $\frac12\times 0$ and $\frac12\times 1$. Any open interval containing $\frac12\times 0$ must ‘reach up’ into this interval, and any open interval containing $\frac12\times 1$ must ‘reach down’ into it, so both endpoints of the interval must be in its closure.
In each part of the problem you should also try to see why no other points are in the closure of the given set: around each point not in the closure you should be able to find an open interval disjoint from the given set.
A: Any open set of the lexicographic order that contains $1$ x $0$ is an open 'interval' of the form:
 $U: = (a,1)$ x $[0,1] \cup \{a\}$ x $(b,1] \cup \{1\}$ x $[0,c)$ where $a < 1$ and $b,c \in [0,1]$.
Thus  $U \cap C \not= \emptyset$
