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

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. $$L_p = \{a \in \mathbb{Q} \; | \; a and $$R_p = \{a \in \mathbb{Q} \; | \; r.

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?

• Well $v=c$ doesn't belong to the type in question. It's been a while since I last dealt with this stuff, but if you consider an extension $B$ of $\mathbb{Q}$ with a point $a$ that's the successor to the rational $c$, then in that extension the type is realized but not by a point of $\mathbb{Q}$ Commented Feb 6, 2019 at 17:02
• @Max why $v=c$ doesn't belong to the type? Commented Feb 6, 2019 at 17:25
• Well in the given extension, $a\neq c$, so $v=c$ can't be in the type. Look at it the other way : fix your extension $B$ with $a$ an immediate successor to $0$ (to be specific). Then define $p(x)$ to be the type of $a$ over $\mathbb{Q}$ (in $A$). Clearly $x=0$ is not in $p$ ! Commented Feb 6, 2019 at 17:49
• Yeah, kind of. But the point is that its type over $\mathbb{Q}$ (in $B$) is none of the types you mentioned, but it corresponds to the ones the Marker adds Commented Feb 6, 2019 at 18:26
• @Max You should add that as an answer. Commented Feb 6, 2019 at 18:26

Let $$a\notin \mathbb{Q}$$ and consider the structure $$B=\mathbb{Q}\cup \{a\}$$ where $$a$$ is "an infinitesimal", that is $$0 and for $$\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 $$B$$, $$p(x) = \{(c< x), c\in \mathbb{Q}, c\leq 0\}\cup\{(x0\}$$.

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 $$c\leq 0$$, $$(c and for $$c>0$$, $$(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)