Determine the closures of the following subsets of second order:
$(a)$ $E = \left\{\frac{1}{2}\times y\ |\ 0<y<1\right\}$
$(b)$ $D = \left\{x\times \frac{1}{2}\ |\ 0<x<1\right\}$
My Attampt: For $(a)$, since $E$ it 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. Hence, $$Cl(E) = E\cup \left\{\frac{1}{2}\times 0, \frac{1}{2}\times 1\right\}$$
For $(b)$, since the set $D$ is the horizontal open interval between $0\times\frac12$ and $1\times\frac12\ $, any open interval containing $0 \times \frac12$ must ‘reach left’ to this interval and any open interval containing $1\times\frac12$ must ‘reach right’ to it. Thus, both endpoints of the interval must be in its closure:
$$Cl(D) = D\cup \left\{0\times \frac{1}{2}, 1\times \frac {1}{2}\right\}$$
Is this correct. If not, where have I gone wrong?
First $E = \{\frac{1}{2}\} \times (0,1)$. A short check on cases shows that indeed all basic neighbourhoods of $(\frac{1}{2},0)$ and $(\frac{1}{2}, 1)$ have points in $E$, and so they lie in $\overline{E}$. And as $\{\frac{1}{2}\} \times [0,1]$ is closed (as a closed interval in an ordered space) we see that $E$ plus these two points is the whole closure.
As to $D = (0,1) \times \{\frac{1}{2}\}$, the seeming endpoints $(1, \frac{1}{2})$ are and $(0, \frac{1}{2})$ are not in the closure: $((0,0), (0,1))$ is an open neighbourhood of $(0,\frac{1}{2})$ (an open interval) that misses $D$ and similarly for $((1, 0), (1,1))$ and $(1, \frac{1}{2})$.
But consider $(a,0)$ for $a \in (0,1]$:a basic neighbourhood of that point is an open interval (it’s not one of the two endpoints), with $(b,c) < (a,0) < (d,e)$ as endpoints. We cannot have $b=a$ as then we’d have $c <0$ which cannot be, so $b < a$. Pick some $b’$ with $b < b’ < a$. Then $(b,c) < (b’, \frac{1}{2}) < (a,0)$ (all determined by the first coordinate), and so this open interval intersects $D$. This shows that $(0,1] \times \{0\} \subseteq \overline{D}$ as well.
A similar argument can be made symmetrically at the top edge to show that $[0,1) \times {1} \subseteq \overline{D}$ as well.
I claim that $\overline{D} = C:= D \cup (0,1]\times \{0\} \cup [0,1) \times \{1\}$ and one inclusion we have seen and the other follows from the fact that $$[0,1] \times [0,1]\setminus C = ([(0,0),(0,1)) \cup \bigcup_{x \in (0,1)} [((x,0), (x,\frac{1}{2})) \cup ((x,\frac{1}{2}),(x,1))] \cup ((1,0),(1,1)]$$ which is a union of basic open sets of $[0,1]\times [0,1]$ and so $C$ is closed.