Classifying the $1$-types over $(\mathbb{Q}, <)$

Question

Consider the $1$-types over $\mathbb{Q}$ as a dense linear order without endpoints. They are similar to cuts: for any $1$-type $p$ over $\mathbb{Q}$, we can define $L_p = \{a \in \mathbb{Q} \; | \; a < x \in p\}$ and $R_p = \{b \in \mathbb{Q} \; | \; x < b \in p\}$. It seems then possible to classify the $1$-types as follows:

(i) for some $a \in \mathbb{Q}$, $v = a \in p$. Then $p$ is isolated by this formula and is thus realized in $\mathbb{Q}$.

(ii) $p$ is omitted in $\mathbb{Q}$. Then there are at least three kinds of types: $L_p = \varnothing$ and $R_p = \mathbb{Q}$, $L_p = \mathbb{Q}$ and $R_p = \varnothing$, and, for $r \in \mathbb{R} \setminus \mathbb{Q}$, the type which corresponds to the cut corresponding to $r$, i.e. $L_p = \{a \in \mathbb{Q} \; | \; a<r\}$ and $R_p = \{a \in \mathbb{Q} \; | \; r<a\}$.

Now, I had thought that this exhausted the classification of $1$-types, but Marker (p. 122 of his book) claims that there are two other kinds of types as well: if $c \in \mathbb{Q}$, we also have the case $L_p = \{a \in \mathbb{Q} ; | \; a \leq c\}$ and $R_p = \{b \in \mathbb{Q} \; | \; c < b\}$ and the case $L_p = \{a \in \mathbb{Q} \; | \; a < c\}$ and $R_p = \{b \in \mathbb{Q} \; | \; c \leq b\}$. But I don't understand why these two other types are omitted in $\mathbb{Q}$. In fact, they seem to be isolated by the formula $v = c$. What am I missing?

Answer 1

Let $a\notin \mathbb{Q}$ and consider the structure $B=\mathbb{Q}\cup \{a\}$ where $a$ is "an infinitesimal", that is $0<a$ and for $\epsilon \in \mathbb{Q}, \epsilon >0$, we also have $\epsilon > a$.

This clearly uniquely defines an order on $B$, and we may look at the following type, which is realized by $a$ in $B$, $p(x) = \{(c< x), c\in \mathbb{Q}, c\leq 0\}\cup\{(x<c), c\in \mathbb{Q}, c>0\}$.

It's easy to check that this is a type over $\mathbb{Q}$ (a finite subset $q\subset p(x)$ only involves finitely many rationals, and so you can realize $q$ by taking a positive rational strictly smaller than all the involved positive rationals)

Now for $c\leq 0$, $(c<x)\in p(x)$ and for $c>0$, $(x<c)\in p(x)$, and $(0=x)\notin p(x)$, so this corresponds to $L_p = \{c\in \mathbb{Q}\mid c\leq 0\}, R_p =\{c\in \mathbb{Q}\mid c>0\}$, it is yet another type that is omitted in $\mathbb{Q}$. Of course replacing $0$ by any rational $b$ gives the same result and the other types you mentioned.

Now if you take any $1$-type $p(x)$, it is realized in some elementary extension $B$ of $\mathbb{Q}$, by, say $a\in B$. $B$ is then a total order, so you can compare $a$ to every rational, and (using least upper bounds and so on in $\mathbb{R}$) you can make a rather long disjunction of cases and see that you now have all the types (I mean yours and Marker's additional ones) : if you're a $1$-type, you're either a rational (isolated), an irrational real, an infinitely large quantity, an immediate successor, or an immediate predecessor (all of these being omitted of course)