Assuming that the real line has the least upper bound property, does [0,1] x [0,1] in dictionary order have the least upper bound property? I am not sure if I am in the right direction I want to say [0,1] x [0,1] has the least upper bound property because
[0,1] x [0,1] has the dictionary order of the form (x,y). That is
(0,0),(0,1/2)...(0,1),(1/2,0),(1/2,1/2)...,(1/2,1)...(1,0)(1,1/2)...(1,1).
The least upper bound in this dictionary order is in the form (x,0).Therefore, the least upper bound would have to exist between 0 and 1(that is in the interval [0,1]). Someone please guide me and tell me if I am in the right direction.
 A: HINT: Let $X=[0,1]\times[0,1]$, and let $\preceq$ be the lexicographic (dictionary) order on $X$. Let $A$ be a non-empty subset of $X$. It is not necessarily true that $\sup_\preceq A$, the least upper bound of $A$ with respect to $\preceq$, is of the form $\langle x,0\rangle$. For instance, if $A=\{0\}\times\left[0,\frac12\right)$, then $\sup_\preceq A=\left\langle 0,\frac12\right\rangle$.
To find $\sup A$, let
$$x_0=\sup\{x\in[0,1]:\langle x,y\rangle\in A\text{ for some }y\in[0,1]\}\,,$$
where the supremum (least upper bound) is with respect to the usual order on $\Bbb R$, and consider two cases:

*

*$A\cap\big(\{x_0\}\times[0,1]=\varnothing$, and

*$A\cap\big(\{x_0\}\times[0,1]\ne\varnothing$.

In each of these cases it is possible to specify a $y_0$ such that $\sup_\preceq A=\langle x_0,y_0\rangle$.
A: Let $(X, <_X)$ be a linear order with the lub-property, and let $(Y,<_Y)$ be a linear order that has the lub property and a minimum $\min(Y)$ and maximum $\max(Y)$.
Then $(X \times Y$ in the lexiographic product order $<_l$ defined by
$$(x,y) <_l (x',y') \iff (x < x') \lor ((x=x') \land y < y') $$ also has the lub property.
Proof: If $A \subseteq X \times Y$ has an upper bound $(p,q)$ then $p$ is an upper bound for $\pi_X[A]$ (if $a \in \pi_X[A]$ then for some $y \in Y$, $(a,y) \in A$ and $(a,y) <_l (p,q)$ which implies $a \le p$). So $p_0:= \sup_X \pi_X[A]$ exists as it is upper bounded and $X$ has the lub-property.
Now there are two cases:

*

*$p_0 \in \pi_X[A]$ so for some $y \in Y$ we have $(p_0, y) \in A$ so $\pi_Y[(\{x_0\} \times Y) \cap A]$ is non-empty and upper bounded by $\max(Y)$ so has a supremum $q_0$ in $Y$. Then $(p_0, q_0) = \sup_{<_l} A$.


*If $p_0 \notin \pi_X[A]$, show that $(p_0, \min(Y)) = \sup_{<_l} A$.
In either case the sup exists and the lexicographic product has the lub-property.
Note that $[0,1] \times [0,1)$ does not have this property ($\{0\} \times [0,1)$ has upper bounds but no sup) and neither has $[0,1] \times (0,1]$ (no sup for $[0,\frac12) \times \{\frac12\}$ e.g.) so it seems we need the bounds in the second factor.
