Is $X \times [0, 1)$ a linear continuum? What about $X \times (0, 1]$?

Question

I'm reading from Topology by Munkres and in example 2 of section 24, 'Connected Subspaces of the Real Line', the author discusses that $X \times [0, 1)$ is a linear continuum in the dictionary order where $X$ is a well ordered set.

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 argument was as follows:
$(1)$ Let $L \ \subset \ X \times [0, 1)$ and $b = \text{sup}(\ \pi_1(L)\ )$. If $b \in L$, then since $b \times [0, 1)$ is homeomorphic to $[0,1)$, there exists a $c = \text{sup}(\ \pi_2(\ b \times [0, 1)\ )\ )$. Then $b\times c$ is the least upper bound of $L$. If $b \notin L$, then $b \times 0$ will be the least upper bound of $L$ since if $b^{'} \times c$ with $b^{'} < b$ was an upper bound of $L$, then $b^{'}$ would be an upper bound for $\pi_1(A)$ which contradicts our assumption.

(2) Let $(\alpha, x) < (\beta, y)$. Then $\alpha < \beta$ or $\alpha = \beta, \ x <y$. If $\alpha < \beta$ then $(\alpha, x) < (\alpha, \frac{x+1}{2}) < (\alpha, y)$. Otherwise $(\alpha, x) < (\alpha, \frac{x+y}{2}) < (\alpha, y)$.

Hence $X \times [0, 1)$ is a linear continuum.

Is this argument valid?
Where do we use the fact that $X$ is well ordered?
Also instead if we consider $X \times (0, 1]$ will that be a linear continuum?
My intuition is that it'll not be since $b \times 0$ will not exists which we used in the proof of (1)

Answer 1

"Also" question. No.

Let $X = \omega+1 = \{0,1,\dots,\omega\}$. Then $X \times (0,1]$ is not order complete. The subset $\omega\times(0,1]$ is bounded above, but has no least upper bound. [All points $(\omega,t), t\in (0,1]$ are upper bounds, but among them there is no least one.]

Answer 2

First, let me address a notational issue with your proof. In the definition of "linear continuum", $L$ is the set that you are trying to show is a linear continuum. In this case, $L = X \times [0,1)$. In your proof, you use $L$ as an arbitrary (bounded) subset of $X \times [0,1)$, which you are trying to show has a least upper bound. This introduction of $L$ in one role, then immediate use of it in a different role without explanation is very confusing. The reader is led to believe the original role, and discovers the actual role only by recognizing your statements are inconsistent with that interpretation, forcing them to re-evaluate what they've already read. There are many letters in the alphabet. Use a different one for this different job.

Is this argument valid?

Overall yes, but your proof needs several corrections and additions. Some appear to just be typos, but you make too many of them.

$(1)$ Let $L \ \subset \ X \times [0, 1)$ and $b = \text{sup}(\ \pi_1(L)\ )$.

There is no reason that $b$ has to exist. You need to assume $L$ is bounded above. For example, $X = \Bbb N$ is a well-ordered set, and $L$ could be the set of even numbers times $[0,1)$. That set has no upper bound at all, much less a least one. And even when $L$ is bounded, you need to show that $\pi_1(L)$ must also be bounded.

there exists a $c = \text{sup}(\ \pi_2(\ b \times [0, 1)\ )\ )$.

I assume you mean $c = \sup\bigg(\pi_2\big((b \times [0, 1))\cap L\big)\bigg)$, as your set is just $[0,1)$ and has no supremum. And again you must show by the boundedness of $L$ that $\pi_2\big((b \times [0, 1))\cap L\big)$ is bounded before you can demand the existence of $c$.

If $b \in L$, then since $b \times [0, 1)$ is homeomorphic to $[0,1)$, there exists a $c = \text{sup}(\ \pi_2(\ b \times [0, 1)\ )\ )$. Then $b\times c$ is the least upper bound of $L$. If $b \notin L$, then $b \times 0$ will be the least upper bound of $L$ since if $b^{'} \times c$ with $b^{'} < b$ was an upper bound of $L$, then $b^{'}$ would be an upper bound for $\pi_1(A)$ which contradicts our assumption.

You mean "If $b \notin \pi_1(L)$". Similarly later $\pi_1(A)$ should be $\pi_1(L)$ (or far better, all $L$ outside the definition should be "$A$").

But this only argues that $b \times 0$ is an upper bound, not the least upper bound. Also, an argument should be made that $b\times c$ is the least upper bound of $L$ when $b \in \pi_1(L)$. Though that argument is trivial, I think it wiser for you not to skip it at this time. Only later when you have gotten your feet under you with these proofs should you omit it as obvious.

If $\alpha < \beta$ then $(\alpha, x) < (\alpha, \frac{x+1}{2}) < (\alpha, y)$.

$(\alpha, \frac{x+1}{2}) < (\beta, y)$. You have no idea how $y$ is related to $x$ here, much less $\frac{x+1}2$.

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Where do we use the fact that $X$ is well ordered?

In assuming that $X$ satisfies the least upper bound property, which you did in defining $b$. What you've actually proven is that any set with the least upper bound property times $[0,1)$ is a linear continuum under the dictionary order. If it has not already been shown that well-ordered sets have the least upper bound property, then you need to add that argument to your proof.