Dictionary order and least upper bound for linear continuum I am trying to figure out why does the set $L=\mathbb{Z_+} \times [0,1)$ with dictionary order has the least upper bound property? I say this is the counter example: Suppose $[5,1)<[9,1)$ that is we look at the two sets $[a,1)$ and $[b,1)$ such that $a,b \in \mathbb{Z_+}$, we can never find a least upper bound for those sets because the set of positive integers is not bounded.
Is my reasoning correct and the definition of dictionary order doesn't say anything if the second co-ordinates are equal
 A: I’ll write $\preceq$ for the lexicographic order on $L$. Suppose that $A$ is a non-empty subset of $L$ that is bounded above by some $\langle m,y\rangle\in L$. Let $N=\{n\in\Bbb Z_+:\langle n,0\rangle\text{ is an upper bound for }A\}$; $\langle m,y\rangle\prec\langle m+1,0\rangle$, so $\langle m+1,0\rangle$ is an upper bound for $A$, $m+1\in N$, and therefore $N\ne\varnothing$. Thus, $N$ is a non-empty set of positive integers and has a least element $n_0$.
If $\langle n_0,0\rangle\in A$, then clearly $\langle n_0,0\rangle=\max A$, so $\langle n_0,0\rangle$ is the least upper bound for $A$. If $\langle n_0,0\rangle\notin A$, let $$S=\{x\in[0,1):\langle n_0-1,x\rangle\in A\}\;;$$ $n_0-1\notin N$, so $\langle n_0-1,0\rangle$ is not an upper bound for $A$, and there is therefore an $x\in(0,1)$ such that $\langle n_0-1,x\rangle\in A$, so $S\ne\varnothing$. $S$ is a bounded subset of $\Bbb R$, so it has a least upper bound $s_0$. To finish the argument, just show that $\langle n_0-1,s_0\rangle$ is the least upper bound of $A$ in $L$.
