Example 2, Sec. 24, in Munkres' TOPOLOGY, 2nd ed: If $X$ is a well ordered set, then $X \times [0, 1)$ is a linear continuum

Solution 1:

I fully agree with the proof that $X \times [0,1)$ is a dense order (item (2)). Note that we only use two properties of $[0,1)$ there: if $r < s$ in $[0,1)$, we have $r',s' \in [0,1)$ such that $r < r' < s < s' < 1$. So we use that $[0,1)$ is a dense order without a maximum.

Indeed if $a_0 = \sup \pi_X[S] \notin \pi_X[S]$, we can see that $(a_0, 0) = \sup S$, because it's easily seen to be an upperbound and if $(x,r) \prec (a_0,0)$, $x \prec_X a_0$ (where we use that $0= \min [0,1)$) and $x$ cannot be an upperbound for $\pi_X[S]$ (as $a_0$ is the smallest one by definition) so we have some $(x',s) \in S$ with $x \prec_X x'$ (and hence $(x,r) \prec (x',s)$) but then $(x',s) \nprec (x,r)$, so the smaller $(x,r)$ is not an upperbound of $S$ any more.

The interesting case is indeed when $a_0 = \sup \pi_X[S] \in \pi_X[S]$, so $a_0$ is the maximum of $\pi_X[S]$.

Define $S' = \{(t \in [0,1) \mid (a_0, t) \in S\} = \pi_{[0,1)}[(\{a_0\} \times [0,1)) \cap S]$, which by definition is non-empty. If $S'$ has an upperbound $u \in [0,1)$, $S'$ has a least upperbound $\sup S' \in [0,1)$ (well-known property of $[0,1)$ which follows from the one for the reals) and then $(a_0, \sup S')$ is $\sup S$ (that it is an upperbound is clear and if $(x,r) \prec (a_0 ,\sup S')$ then either $x \prec_X a_0$ and for any $s \in S'$, $(x,r) \nprec (a_0,s)$ or $x=a_0$ and so $r < \sup S'$, so $r$ is not an upperbound for $S'$ witnessed by some $(a_0, r')$ with $r < r', r' \in S'$ and then also $(x,r) \nprec (a_0, r')$, so smaller elements are not an upperbound).

But if it has not, then as $S$ is bounded above, there must be some $a_1 > a_0$, $a_1 \in X$. (If $(p,q)$ is an upperbound of $S$, pick $t \in S'$ with $t > q$ and note that $(a_0, t) \prec (p,q)$ must hold and $a_0 = p$ cannot be as $t > q$ so $a_0 \prec_X p$, properly. In that case $a_0 + 1 := \min\{x \in X: x > a_0\}$ exists and we then show that $(a_0+1, 0) = \sup S$ (It's clearly an upperbound by the first coordinate alone and when $(x,r) \prec (a_0 +1 ,0)$ we must have $x \preceq a_0$. Then picking $t \in S$ with $t > r$ we have that $(a_0, t)\nprec (x,r)$, so a smaller element is not an upperbound). Note that here we really use the well-order on $X$ (before just the lub property of $X$ implied by it, and we again use the minimum of $[0,1)$).

This follows your basic argument with some minor details filled in (why we really have the $\sup S$ in every case e.g.)

Solution 2:

It is easier to prove the equivalent greatest bound property.
Let A be a not empty, bounded below subset of the simlply ordered set under consideration.
P1, the first projection of A is a subset of a well ordered set.
Thus it is well ordered. Let b be the least element.
So there is a not empty subset K. of [0,1)
for which {b}×K is a subset of P1
Let k be the glb of K. It is easy to show that (b,k) is the glb of A.