# Example 2, Sec. 24 in Munkres' TOPOLOGY, 2nd ed: How to show this set to be a linear continuum?

Let $X$ be a well-ordered set, let $[0,1)$ denote the half-open interval (open from the right) on the real line, and let $X \times [0,1)$ have the dictionary order.

Then how to show that $X \times [0,1)$ is a linear continuum?

My effort:

Let $x_1 \times r_1$, $x_2 \times r_2$ be any two elements of $X \times [0,1)$ such that $$x_1 \times r_1 <_{X \times [0,1)} x_2 \times r_2.$$

Then either $$x_1 <_X x_2,$$ or $$x_1 = x_2 \ \ \ \mbox{ and } \ \ \ r_1 <_{\mathbb{R}} r_2.$$

If $x_1 = x_2$, then $$r_1 <_{\mathbb{R}} r_2;$$ so $$r_1 <_{\mathbb{R}} \frac{r_1+r_2}{2} <_{\mathbb{R}} r_2,$$ and hence $$x_1 \times r_1 <_{X \times [0,1)} x_1 \times \frac{r_1+r_2}{2} <_{X \times [0,1)} x_2 \times r_2.$$ Am I right?

What if $x_1 <_X x_2$? In this case, how to show the existence of an element in $X \times [0,1)$ that lies between $x_1 \times r_1$ and $x_2 \times r_2$?

Let $A$ be a non-empty subset of $X \times [0,1)$ such that $A$ is bounded above, say, by an element $y \times s$. Then $$a \times r \leq_{X \times [0,1)} y \times s \ \ \ \mbox{ for all } \ a \times r \in A.$$ Thus, $$a <_X y \ \ \ \mbox{ for all } \ a \in X.$$ Let $\pi_1 \colon X \times [0,1) \to X$ be the projection onto the first coordinate; that is, let $$\pi_1(x \times r) \colon= x \ \ \ \mbox{ for all } \ x \times r \in X \times [0,1).$$

Then $y$ is an upper bound of the image set $\pi_1[A]$. So the set of all the upper bounds in $X$ of $\pi_1[A]$ is a non-empty subset of the well-ordered set $X$ and hence has a smallest element, say, $z$.

Now there are two cases:

If $z \in \pi_1[A]$, then $\{z \} \times [0,1)$ intersects $A$. Now $\{z \} \times [0,1)$ has the ordered type of $[0,1)$; so $A \cap \left[ \{z \} \times [0,1) \right]$ has a least upper bound, say, $t$. Then $z \times t$ is the least upper bound of $A$. Am I right?

If $z \not\in \pi_1[A]$, then $z \times 0$ is the least upper bound of $A$. Am I right?

I’ll use $\preceq$ for the order on $X\times[0,1)$. Your first argument is correct. If $x_1<_Xx_2$, let $r=\frac12(r_1+1)$; then $x_1\times r_1\prec x_1\times r\prec x_2\times r_2$. (Note that this is possible, since $r_1<1$.) Thus, $\preceq$ is a dense linear order.
Your proof of the least upper bound property goes a little bit astray right at the beginning: it’s not necessarily true that $a<_Xy$ for all $a\times r\in A$: you can guarantee only that $a\le_Xy$. (And note that $a$ is in $X$, not in $A$.) However, it is true that $y$ is an upper bound in $X$ for $\pi_1[A]$, so that you can indeed pick $z$ as you did, and the rest of your argument is fine.
• @Saaqib: You don’t want to say that $a<_Xy$ for all $a\in X$: you just want it for the ones that are in $\pi_1[A]$. The easiest solution, I think, would be to omit that sentence altogether, go on and define $\pi_1$, and then say that $a\le_Xy$ for all $a\in\pi_1[A]$. – Brian M. Scott May 5 '15 at 22:13