Supremum in a dictionary ordered set.

Question

Take double arrow space $[0,1]\times \{1,2\}$

Where topology is dictionary topology defined by linear order as $(a,b)<(c,d)$ iff $a<c$; or ;$a=c \;\&\; b<d$

open sets are in the form $$((a, b] \times\{1\}) \cup([a, b) \times\{2\})$$

I want to know that is this topology has least upper bound property?

I know that $\mathbb R$ has this axiom, by using this can I show the above topology has least upper bound property?

By doing this I want to show the above topology is compact. I have seen https://dantopology.wordpress.com/2009/10/07/the-lexicographic-order-and-the-double-arrow-space/ proves it but it seems a bit advanced and less informative.

Answer 1

Let $X$ be the double arrow space. Your description of the order topology on $X$ isn’t quite right. The set that you describe is the open interval in $X$ between $\langle a,1\rangle$ and $\langle b,2\rangle$, but that is not the only kind of open interval in $X$; for instance, the open interval between $\langle a,2\rangle$ and $\langle b,1\rangle$ is $(a,b)\times\{1,2\}$. And of course all unions of open intervals are open as well.

Let $\preceq$ be the lexicographic order on $X$; the linear order $\langle X,\preceq\rangle$ does have the least upper bound property. To prove this, let $\varnothing\ne A\subseteq X$. Let

$$P=\{x\in[0,1]:\exists i\in\{1,2\}\,(\langle x,i\rangle\in A)\}\,;$$

$[0,1]$ has the least upper bound property, so let $y=\sup P$, where the supremum is with respect to the usual order on $[0,1]$.

• Show that if $y\notin P$, then $\langle y,1\rangle=\sup_\preceq A$.
• Show that if $y\in P$, then $$\sup_\preceq A=\begin{cases}\langle y,1\rangle,&\text{if }\langle y,2\rangle\notin A\\\langle y,2\rangle,&\text{if }\langle y,2\rangle\in A\,.\end{cases}$$

Thus, in all cases $\sup_\preceq A$ exists.

Answer 2

Let $S\subseteq [0,1]\times\{1,2\}$ be a nonempty set. Let $$T=\{x\in [0,1]\mid \exists t\in\{1,2\}\text{ such that }(x,t)\in S\}.$$ Then $T$ is a nonempty subset of $[0,1]$, and hence has a least upper bound, $s$.

If $(s,2)\in S$, then $(s,2)$ is the maximum of $S$ and hence the least upper bound of $S$: indeed, if $(x,t)\in S$, then $x\in T$, so $x\leq s$; and since $t\leq 2$, we must have $(x,t)\leq (s,2)$.

If $(s,1)\in S$ and $(s,2)\notin S$, then $(s,1)$ is the maximum of $S$: if $(x,t)\in S$, then $x\leq s$; if $x\lt s$ then $(x,t)\lt (s,1)$, and if $x=s$, then it must be the case that $t=1$, so $(x,t)=(s,1)$. Thus, $(s,1)$ is the maximum of $S$, and hence the least upper bound of $S$.

Assume then that $(s,1),(s,2)\notin S$.

We know that for all $\epsilon\gt 0$ there exists $x\in T$ such that $s-\epsilon\lt x\leq s$ (in fact, $s-\epsilon\lt x\lt s$, since $x\in T$ but $s\notin T$).

Claim. $(s,1)$ is the least upper bound of $S$.

First, note that it is an upper bound for $S$: if $(x,t)\in S$, then $x\leq s$; since $(s,1)\notin S$ and $(s,2)\notin S$, then $x\lt s$, so we are done: $(x,t)\lt (s,1)$. Thus, $(s,1$ is an upper bound for $S$.

Now, let $(x,t)\lt (s,1)$. Since by assumption $x\lt s$, then there exists $x_1\in T$ such that $x\lt x_1\leq s$, and there exists $t_1\in\{1,2\}$ such that $(x_1,t_1)\in S$, so $(x,t)\lt (x_1,t_1)\leq (s,1)$. Thus, $(x,t)$ is not an upper bound for $S$. Thus, $(s,1)$ is the least upper bound for $S$.

Thus, $S$ has a least upper bound in $[0,1]\times\{1,2\}$, ordered lexicographically.