Suslin Line and completeness

Question

I am studying the chapter about Suslin hypothesis and diamond principle in the book Set Theory for the Working Mathematician, from Ciesielsky and I have the following problem:

A Suslin line is a ccc linearly ordered space $\langle X,\leq \rangle$ that has neither a first or a last element and non-separable.

The existence of a Suslin line implies the existence of a complete Suslin line

Hint: Take $X$ a Suslin line and consider $X^*$ the family of all proper nonempty initial segments of X without last elements.

I have shown that $X^*$ is also a Suslin line and for the sake of completeness, I was trying to show that, for an upper bounded subset $S\subset X^*$, would follow that $\bigcup S\in X^*$.

However, I am not sure if this will work and I am not able to use the ccc and non-separable properties to continue the proof.

EDIT: In part of my proof that $X^*$ is indeed a Suslin line, I was ignoring the fact that $X$ is dense in itself and some of my arguments became rather complicated. I will post my solution to this question.

Answer 1

Topologically this is quite straightforward. He is sketching the construction of the order completion $X^\ast$ of $X$. This is just like constructing $\Bbb R$ from $\Bbb Q$, in many ways.

Suppose that $X^\ast$ were separable, show that $X$ is too ($X$ order-embeds into $X^\ast$, and in the order topology, if a space is separable, so are all its subspaces).

And as $X$ is (topologically and order-)dense in $X^\ast$ and $X$ is ccc, so is $X^\ast$.

Answer 2

To prove the completeness of $X^*$, let $S\subset X^*$ a nonempty, bounded from above. Clearly, $\bigcup S$ is an initial segment of $X$ and notice that $\bigcup S \in X^*$:

Suppose that there exists a $x\in X$ such that $x$ is the last element of $\bigcup S$. Then there exists an $A\subset \bigcup S$ such that $x\in A$ and, since $A\in X^*$, there exists $x'\in A$, $x' > x$, a contradiction.

It is easy to check that $\bigcup S$ is the least upper bound of $S$.