This is from textbook Introduction to Set Theory by Karel Hrbacek and Thomas Jech.
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.
The linearly ordered set $(C,\prec)$ is called the completion of $(P, <)$.
In proving this theorem, the authors define a new set $(P',<')$ which is isomorphic to $(P,<)$. Then they define $(C',\prec')$ and prove that $(C',\prec')$ is the completion of $(P',<')$. Finally the authors conclude that:
We intend to show that $(C',\prec')$ is a completion of $(P',<')$. Since $(P,<)$ and $(P',<')$ are isomorphic, it follows that $(P,<)$ has a completion.
I'm unable to understand why the authors have that conclusion.
How do we know that a completion of $(P,<)$ does exist?
How do we define a completion of $(P,<)$ explicitly from $(C',\prec')$?
Thank you for your clarification!
Consider the isomorphism $f\colon (P,<)\to (P',<')$. Now define $$ C=(C'\setminus P')\cup P $$ and a map $g\colon C\to C'$ by $$ g(x)=\begin{cases} f(x) & x\in P \\ x & x\notin P \end{cases} $$ Finally define, for $x,y\in C$, $x\prec y$ if and only if $g(x)\prec'g(y)$.
Now prove that $g\colon (C,\prec)\to(C,\prec')$ is an isomorphism and that $(C,\prec)$ has the required properties.