Background:
I'm reading Kuratowski and Mostowski's set theory on my own.
A totally ordered set $A$ is said to be dense if for each pair $x,y \in A$ satisfying $x < y$ there exists an element $z \in A$ such that $x < z < y$ in $A$.
A totally ordered set $A$ which contains no infinite dense subset is said to be scattered.
The authors have proved the following two propositions.
Proposition 1: Each subset of a scattered set is scattered.
Proposition 2: The union of two scattered sets is scattered.
Let $V_x$ be the set of all $y \in A$ such that the set $[x,y]$ is scattered, where $[x,y]$ denotes the set of those $z \in A$ which equal either $x$ or $y$, or which satisfy one of the conditions $x < z < y$ or $y < z < x$.
It claims that each $V_x$ is scattered for all $x \in A$.
It says on the top of page 214 that
Suppose that on the contrary $C \subseteq V_x$ and that the set $C$ is infinite and dense. For any $c_1,c_2 \in C$ such that $c_1 < c_2$ we have $[c_1,c_2] \subseteq [c_1,x] \cup [c_2,x]$, thus the set $[c_1,c_2]$ is contained in the union of two scattered sets, by proposition 2, this union is scattered. But this is impossible, for between any two distinct elements of the set $C$ there always lies at least one other of $C$. Thus the assumption that $V_x$ is not scattered leads to a contradiction.
My attempt: I guess that $[c_1,c_2]$ is an interval in $A$ and the only two given elements of $C$ are $c_1$ and $c_2$, so $[c_1,c_2]$ is infinite and I have tried to show it is dense, I assume that there exist $c_1',c_2' \in [c_1,c_2]$ but I can't find a $c \in A$ such that $c_1' < c < c_2'$. I have been stuck in this problem for several weeks and I just can't find a contradiction.
My question: So how to show that each $V_x$ is scattered?Or is this statement really true?
Thanks in advance.
$C\cap[c_1,c_2]$ is infinite and dense, so $[c_1,c_2]$ is not scattered. There’s the contradiction.