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

Question

Let $X$ be a well ordered set. Then how to show that the set $X \times [0, 1)$ is a linear continuum in the dictionary order? Here $$ [0, 1) \colon= \{ \ x \in \mathbb{R} \ \colon \ 0 \leq x < 1 \ \}.$$

Definition of Linear Continuum:

A simply ordered set $L$ having more than one element is called a linear continuum if the following hold:

(1) $L$ has the least upper bound property.

(2) If $x < y$, there exists $z$ such that $x < z < y$.

My Attempt:

In what follows, $x \times r$ would denote an order pair, and $r < s$ would denote the usual less than relation between real numbers $r$ and $s$.

Suppose $x_1 \times r_1$ and $x_2 \times r_2$ are two arbitrary elements of $X \times [0, 1)$ such that $x_1 \times r_1 \prec x_2 \times r_2$. This means that either $x_1 \prec_X x_2$, or $x_1 = x_2$ and $r_1 < r_2$.

Case 1. Suppose $x_1 \prec_X x_2$. Then $x_1 \times \frac{r_1+1}{2}$ is an element of $X \times [0, 1)$ such that $$ x_1 \times r_1 \prec x_1 \times \frac{r_1+1}{2} \prec x_2 \times r_2.$$

Case 2. Suppose $x_1 = x_2$ and $r_1 < r_2$. Then $x_1 \times \frac{r_1+r_2}{2}$ is an element of $X \times [0, 1)$ such that $$ x_1 \times r_1 \prec x_1 \times \frac{r_1+r_2}{2} \prec x_2\times r_2.$$

Let $\pi_1 \colon X \times [0, 1) \to X$ and $\pi_2 \colon X \times [0, 1) \to [0, 1)$ be the maps $x \times r \mapsto x$ and $x \times r \mapsto r$, respectively.

Now let $S$ be a non-empty subset of $X \times [0, 1)$ such that $S$ is bounded from above in $X \times [0, 1)$. Let $a \times u$ be an upper bound in $X \times [0, 1)$ of set $S$. Then, for every element $x \times r \in S$, we have $$ x \times r \preceq a \times u, $$ which is tantamount to saying that, either $x \prec_X a$, or $x = a$ and $r \leq u$. In either case, we have $x \prec_X a$ or $x = a$ for every element $x \in \pi_1(S)$. Thus the set $\pi_1(S)$ is bounded from above in $X$ (by the element $a \in X$, for example). Let $a_0$ be the supremum in $X$ of set $\pi_1(S)$. [The existence of this supremum follows from the fact that the set of all the upper bound in $X$ of the set $\pi_1(S)$ is non-empty and thus has a smallest element, because $X$ is well ordered.]

Case 1. If $a_0 \not\in \pi_1(S)$, then, for every element $x \times r \in S$, we have $x \prec_X a_0$ and hence $x \times r \prec a_0 \times 0$. Moreover, if there were some element $a \times u \in X \times [0, 1)$ such that $a \times u \prec a_0 \times 0$, then $a \prec_X a_0$ and so $a$ could not be an upper bound for $\pi_1(S)$, which would imply the existence of an element $x \times r \in S$ for which $a \prec_X x = \pi_1(x \times r) $. Thus the element $a_0 \times 0 \in X \times [0, 1)$ is the supremum of set $S$.

Case 2. If $a_0 \in \pi_1(S)$, then there is some $r \in [0, 1)$ for which $a_0 \times r \in S$. Thus the set $S \cap \left( a_0 \times [0, 1) \right)$ is non-empty.

Is what I have done so far correct? If so, then what next? If not, then where have I gone wrong?

PS:

The set $a_0 \times [0, 1)$ has the order type of $[0, 1)$ and so has the least upper bound property.

If $S \cap \left( a_0 \times [0, 1) \right)$ has an upper bound in $a_0 \times [0, 1)$, then the set $\pi_2 \left( \ S \cap \left( a_0 \times [0, 1) \right) \ \right)$ is bounded above in $[0, 1)$; let $r_0$ be the supremum of this set. Then $a_0 \times r_0$ is the supremum of $S$ in $X \times [0, 1)$.

So we suppose that $S \cap \left( a_0 \times [0, 1) \right)$ has no upper bound in $a_0 \times [0, 1)$. But as the set $S$ is bounded above in $X \times [0, 1)$, so $a_0$ cannot be the largest element of $X$, because otherwise the set $S \cap \left( a_0 \times [0, 1) \right)$ and hence the set $S$ would be unbounded in $X \times [0, 1)$.

And, as $a_0$ is not the largest element of $X$, so the set $$ \{ \ x \in X \ \colon \ a_0 \prec_X x \ \}$$ is a non-empty subset of $X$ and so has a smallest element, say $a_1$. Then the open interval $$\left( a_0, a_1 \right) \colon= \left\{ \ x \in X \ \colon \ a_0 \prec_X x \prec_X a_1 \ \right\}$$ is an empty subset of $X$. Therefore, the element $a_1 \times 0$ is the supremum of $S$.

Is my logic correct? If so, then does this PS part complete the proof satisfactorily enough?

Answer 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.)

Answer 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.

Answer 3

As the set $X×[0,1)$ can be thought of as having been constructed by "fitting in" a set of order type of $(0,1)$ immediately following each element of $X$, it suffices to show that $X$ has the least upper bound property.

Since $X$ is well-ordered, every nonempty subset of $X$ has a smallest element.

If $A$ is a nonempty subset of $X$ which is bounded above, then $B=$ the set of upper bounds of $A$ is a nonempty subset of a well-ordered set $X$; hence, $B$ has the smallest element say $s$. Then, $s=lub A$. Hence, $A$ has a least upper bound

Thus, $X$ has the least upper bound property.