Let $(C, \prec)$ be the completion of $(P, <)$. Then, if $c, d \in C$, there is some $p \in P$ such that $c \prec p \prec d$.

Question

I'm studying the book Introduction to Set Theory by Hrbacek and Jech and came across this theorem:

5.3 Theorem Let $(P, \lt )$ 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 $p \lt q$ if and only if $p \prec q$ ($\prec$ coincides with $\lt$ On $P$).
(c) $P$ is dense in $C$, i.e., for any $p,q \in P$ such that $p\lt q$, there is $c\in C$ with $p\prec c\prec q$.
(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 $(C^*, \prec ^* )$ is a complete linearly ordered set which satisfies (a)-(d), then there is an isomorphism $h$ between $(C, \prec )$ and $(C^*, \prec ^*)$ such that $h(x)=x$ for each $x \in P$. The linearly ordered set $(C, \prec )$ is called the completion of $(P, \lt )$.

Specifically I'm interested in property (c); I would like to reverse the roles of $P$ and $C$ and prove the following:

Proposition: For any $c, d \in C$ such that $c \prec d$, there is $p \in P$ with $c \prec p \prec d$.

It seems reasonable, since in practice $P$ is a model for the rationals and $C$ is a model for the real numbers, so this proposition would simply say "between any pair of real numbers there is always a rational number". In fact, the authors use this result without ever proving it a couple of lines below.

However, I am unable to prove the result. I tried the contrapositive method but got confused with the quantifiers; then I tried contradiction, but couldn't quite finish the argument.

I would like to ask for any hint that could help me prove this result, since I'm pretty much stuck here. Thank you.

Answer 1

As the comment stated, there must be a typo in the definition of a dense subset given in condition (c). In fact, the correct definition is the following:

$P$ is dense in $C$, i.e., for any $c,d \in C$ such that $c \prec d$ there is $p \in P$ with $c \prec p \prec d$. $\tag{*}\label{*}$

Even if you accept the definition mentioned in the book, the above property ($\ref{*}$) and the uniqueness part of the theorem do not necessarily hold. For example, consider the following sets:$$A= \{ x \in \mathbb{Q} \mid x \lt 1 \} \cup \{ x \in \mathbb{Q} \mid x \gt 4 \}, \\ B= (-\infty , 1 ] \cup [2, 3] \cup [4, \infty).$$ $(A, \lt )$ is a dense linearly ordered set, $(B, \lt )$ is a complete linearly ordered set, and neither $(A, \lt )$ nor $(B, \lt )$ has endpoints. It can be easily seen that the other conditions of the theorem are satisfied. However, we can see that our required property, $\ref{*}$, does not hold; for example, for $c=2, d=3 \in B$, there is no $p \in A$ such that$$2 \lt p \lt 3.$$The uniqueness part of the theorem does not hold as well. For example, $C=(\mathbb{R}, \lt )$, as a completion of $(A, \lt )$, also satisfies the conditions of the theorem, but the isomorphism map constructed in the proof of the uniqueness cannot map $(B, \lt)$ onto $(\mathbb{R}, \lt )$ (in fact, it is an embedding).

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Addendum

Please note that in topology a dense set is defined as follows.

A subset $A$ of a topological space $X$ is dense in $X$ if for every non-empty open set $U$ of $X$ we have $A \cap U \neq \varnothing$.

Now, if we apply the above definition to the order topology (without endpoints), whose open sets are arbitrary unions of the sets $\{ x \in X \mid a \lt x \lt b \}$ for any $a,b \in X$, we can conclude that

$A$ is dense in X if for every $a \lt b$ in $X$ there is some $c\in A$ such that $a \lt c \lt b$.

It is worth noting that Thomas Jech in his book Set Theory states the definition of a dense subset as follows:

A set $D \subset P$ is a dense subset if for all $a \lt b$ in $P$ there exists a $d \in D$ such that $a \lt d \lt b$.

I hope with the above reasons and evidence you conclude that such a statement in condition (c) is just a typo.