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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?

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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.

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  • $\begingroup$ Thank you so much, @Brian M. Scott. I've just editted my post in accordance with the error you've pointed out. By the way, I'd already managed to figure out an answer to my question as to the denseness of the continuum part. $\endgroup$ – Saaqib Mahmood May 5 '15 at 7:13
  • $\begingroup$ @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 $a\in\pi_1[A]$. $\endgroup$ – Brian M. Scott May 5 '15 at 22:13

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