Munkres Topology Chapter 1.3 Exercise 15 Does $[0,1] × [0,1]$ in the dictionary order have the least upper bound property? What about $[0,1] × [0,1)$ and $[0,1) × [0,1]$?
Since both $[0,1]$ and $[0,1)$ have the least upper bound property, I conclude that all three of the above sets have it too. However, some online solution states that $[0,1] × [0,1)$ doesn't have the property because the subset $\{0\} × [0,1)$ does not have the least upper bound. I fail to see why. 
 A: $[0,1] \times [0,1)$ indeed does not have the lub-property.
$A=\{0\} \times [0,1)$ is bounded above, e.g. $(\frac12, 0)$ is an upperbound, because if $(x,y) \in A$ then $x=0 < \frac12$ and so $(x,y)<_l (\frac12,0)$.
Suppose $(a,b)$ is some upperbound (in $[0,1]\times[0,1)$!) for $A$. 
We cannot have that $a=0$ because then $b < 1$ and if $b < b' < 1$, $(0,b')\in A$ but $(0,b') >_l (a,b)$ and $(a,b)$ is not an upperbound for $A$ at all.
So $a >0$ but then, picking some $a'$ with $0 < a' < a$, say $\frac{a}{2}$ for definiteness, we have that $(a',0) <_l (a,b)$ and the fact that $a'>0$ implies that $(a',0)$ is an upperbound for $A$.
So in short: for every upperbound of $A$ we can find a smaller one. So the lub of $A$ does not exist.
It is fairly easy to see (and proved on this site several times) that the full lexicographically ordered square $[0,1]\times [0,1]$ does have the lub property (is a linear continuum in fact, I believe Munkres shows this in the text somewhere). The set $[0,1) \times [0,1]$ is a closed order convex (for any two points in it, it also contains the whole interval between them) subset of it and from that it also has the lub property. 
