Exercise 24.6 Munkres: $X \times [0,1)$ is a linear continuum

The problem is as follows:

Show that if $X$ is a well-ordered set, then $X \times [0,1)$ in the dictionary order is a linear continuum.

Let $X \times [0,1)=L$. Now, by the definition of a linear continuum, I sought to prove that

1. $L$ has the least upper bound property, and that
2. if $x<y$ for any two elements $x,y \in L$ there exists $z \in L$ such that $x < z < y$.

However, since we are given the information that $X$is a well-ordered set, I was wondering if proving 1. was equivalent to proving that $L$ has a greatest lower bound? Which is easy. Is this equivalent? If yes, why is it so?

Hope you can help me clarify this issue. Thanks!

I think having the least upper bound property is always equivalent to having the greatest lower bound property.

Indeed, suppose $(X,<)$ is a totally ordered set with the the l.u.b. property. Take a nonempty set $S \subset X$ which possesses a lower bound. Therefore, $L = \{ x \in X : x \leq S\}$ is nonempty. Also, $L$ posses an upper bound (any element of the nonempty set $S$ is an example). Since the l.u.b. property holds, $L$ has a least upper bound $y$. It is straightforward to check that $y$ is also the greatest lower bound of $S$.

Be careful, we still have the second criterion and must use the definition of dictionary order.

That is given an order relation $<_A$ on the set $A$ and an order relation $<_A$ on the set $B$, we define the order relation $<$ on $A\times B$ such that: $$a_1\times b_1 < a_2 \times b_2$$ if $a_1<_Aa_2$, or if $a_1=a_2$ and $b_1<_Bb_2$. The relation $<$ is called the dictionary order relation on $A\times B$.

This is the catch that we are missing. Suppose that $X$ is not a linear continuum. Then either axiom is false. Suppose that the second axiom is not true for $X$. Then we see that there exists at least two elements $a,b\in X$ with $a<_Xb$ such that there is no $d\in X$ such that $a<_Xd<_Xb$. This implies that through the definition of the dictionary order that there are no $d\times z$ such that $a\times x < d\times z < b\times y$. We cannot resort to the second part of the definition because by the assumption of this case, $a<_Xb$.

If the first is false instead, then we still have a problem in the dictionary order. This is equivalent to saying $X$ is unbounded above. If $X$ is unbounded above, this will imply through the use of the dictionary order that there is no $a\in X$ such that $b\leq_X a$ for any $b\in X$. This implies that there is no $a\times x$ in $X\times[0,1)$ such that $b\times y < a\times x$ for any $b\times y$ in $X\times[0,1)$.

The claim the problem asks you to prove is in fact false by these definitions.