Recently, I have read the proof of theorem 5.3 from textbook Introduction to Set Theory by Karel Hrbacek and Thomas Jech.
Although the authors explicitly use the fact that $S_c=S_{c^\ast}$ at the end without justification, I think this fact is not trivial at all and thus requires a proof.
Please help me check it out! Thank you for your help!
5.3 Theorem Let $(P, <)$ be a dense linearly ordered set without endpoints. Then there exists a complete linearly ordered set $(C,\prec)$ such that
(a) $P\subseteq C$.
(b) If $p,q\in P$ then $p < q$ if and only if $p \prec q$ ($<$ coincides with $\prec$ on $P$).
(c) $P$ is dense in $C$, i.e., for any $a,b\in C$ such that $a\prec b$, there is $p\in P$ with $a\prec p \prec b$.
(d) $C$ does not have endpoints.
Moreover, this complete linearly ordered set $(C,\prec)$ is unique up to isomorphism over $P$. In other words, if $(C^\ast,\prec^\ast)$ is a complete linearly ordered set which satisfies (a)-(d), then there is an isomorphism $h$ between $(C,\prec)$ and $(C^\ast,\prec^\ast)$ such that $h(x) = x$ for each $x \in P$. The linearly ordered set $(C,\prec)$ is called the completion of $(P, <)$.
As is usual in theorems of this type, the uniqueness part of the theorem is easier to prove. For that reason, we do it first.
Proof of Uniqueness of Completion. Let $(C,\prec)$ and $(C^\ast,\prec^\ast)$ be two complete linearly ordered sets satisfying (a)-(d). We show that there is an isomorphism $h$ of $C$ onto $C^\ast$ such that $h(x)=x$ for each $x\in P$.
If $c\in C$, let $S_c=\{p\in P\mid p \preccurlyeq c\}$. Similarly, let $S_{c^\ast}=\{p\in P\mid p\preccurlyeq^\ast c^\ast\}$ for $c^\ast\in C^\ast$. If $S$ is a nonempty subset of $P$ bounded from above, let $\sup S$ be the supremum of $S$ in $(C,\prec)$ and $\sup^\ast S$ be the supremum of $S$ in $(C^\ast,\prec^\ast)$. Notice that $\sup S_c=c$, $\sup^\ast S_{c^\ast}=c^\ast$.
We define the mapping $h$ as follows: $h(c)=\sup^\ast S_c$.
Clearl, $h$ is a mapping of $C$ into $C^\ast$; we have to show that the mapping is onto and that
(a) If $c\prec d$, then $h(c)\prec^\ast h(d)$.
(b) $h(x)=x$ for each $x\in P$.
To show that $h$ is onto, let $c^\ast\in C^\ast$ be arbitrary. Then $c^\ast=\sup^\ast S_{c^\ast}$, and if we let $c=\sup S_{c^\ast}$, then $S_c=S_{c^\ast}$ and $c^\ast=h(c)$.
My attempt:
For $c^\ast \in C^\ast$, $S_{c^\ast}=\{p\in P\mid p \preccurlyeq^\ast c^\ast\}$. Let $c=\sup S_{c^\ast}$, then $c\in C$. We have $S_c=\{p\in P\mid p \preccurlyeq c\}$. Notice that $S_{c^\ast} \subseteq P$ and $S_c \subseteq P$.
For any $p\in S_c$, $p \preccurlyeq c$ and thus $p \preccurlyeq \sup S_{c^\ast}$. Then $p \preccurlyeq p'$ for some $p' \in S_{c^\ast}$. If not, $p' \prec p$ for all $p' \in S_{c^\ast}$ and thus $p' \prec p \preccurlyeq \sup S_{c^\ast}$ for all $p' \in S_{c^\ast}$. Then $p=\sup S_{c^\ast}$ and thus $S_c$ has only one element. This clearly contradicts the fact that $P$ does not have a least element. Hence $p \preccurlyeq p'\preccurlyeq^\ast c^\ast$ for some $p' \in S_{c^\ast}$ and consequently $p \preccurlyeq^\ast c^\ast$. Thus $p\in S_{c^\ast}$. To sum up, $p\in S_c \implies p\in S_{c^\ast}$ and thus $S_c\subseteq S_{c^\ast}$.
For any $p\in S_{c^\ast}$, $p \preccurlyeq \sup S_{c^\ast}$ and thus $p \preccurlyeq c$. Then $p\in S_c$. Hence $S_{c^\ast} \subseteq S_c$.
As a result, $S_{c^\ast}=S_{c^\ast}$.