Lecture in O-minimality

By Santiago Camacho for MA595, U Illinois, Fall 2015

We will use this format to publish lectures in o-minimality online. This page may be readable in various web browsers but it will look its best in firefox. We follow very closely the Book "Tame topology and o-minimal structures" by Lou van den Dries.

1.1. Chapter 1

Let

X, Y, Z

be sets and

x, y, z

variables raging over

X, Y, Z

respectively and

φ (x, y, z)

and

ψ (x, y, z)

be conditions on a point

(x, y, z) \in X \times Y \times Z

. We define

Φ, Ψ \subseteq X \times Y \times Z

Φ = {(x, y, z) \in X \times Y \times Z : φ (x, y, z) holds}

Ψ = {(x, y, z) \in X \times Y \times Z : ψ (x, y, z) holds}

Throughout the letures, for given conditions

φ

and

ψ

we use the notations

$φ \lor ψ,$
$φ \land ψ,$
$\neg φ,$
$\exists z φ (x, y, z),$
$\forall z φ (x, y, z) .$

The associated sets would correspond to

The union,
the intersection,
the complement,
the projection on the first two variables,
the complement of the projection of the complement.

Usually all variables will range over powers of the same set. for example

X = ℝ^{m}, Y = ℝ^{n}, Z = ℝ^{p}

Definition: A boolean algebra on a set

X

is a collection of subsets of

X

that is closed under finite unions and complements.

It is easy to see that Boolean algebras are closed under finite intersections and they always contain both

X

and

\emptyset

Definition We define a natural partial order

\leq

in Boolean algebras by

A \leq B \overset{}{\Leftrightarrow} A \cap B = A

An atom of a boolean algebra is a minimal element under the previously mentioned order that lies above

\emptyset

Definition: Let

A_{1}, \dots, A_{n}

be a collection of subsets of

X

. We define the boolean algebra on

X

generated by

A_{1}, \dots, A_{n}

to be

ℬ (A_{1}, \dots, A_{n}; X) = {⋃_{Δ \in Γ} (⋂_{i \in Δ} A_{i} \cap ⋂_{i \notin Δ} X \ A_{i}) : Γ \subset ℙ ({1, \dots, n})}

Where

ℙ (Y)

denotes the power set of

Y

. That is the collection of subsets of

Y

. Thus a boolean algebra generated by

n

elements has at most

2^{n}

atoms and has at most

2^{2^{n}}

elements.

1.11.1. Facts on Structures

Definition: A structure on a nonempty set

R

is a family

S = (S_{m})_{m \in ℕ}

such that for each

m \geq 0

;

$S_{m}$ is a boolean algebra of subsets of $R^{m}$ ;
if $A \in S_{m}$ , then $R \times A$ and $A \times R$ are both in $S_{m + 1}$ ;
${(x_{1}, \dots, x_{m}) \in R^{m} : x_{1} = x_{m}} \in S_{m}$ ;
If $A \in S_{m + 1}$ , then $π (A) \in S_{m}$ , were $π : R^{m + 1} \to R^{m}$ is the projection on the first $m$ coordinates.

We use the notation

(R, S)

to denote a structure, and we call the subsets of

R^{m}

that are in

S_{m}

definable sets.

Examples

$ℝ$ equipped with semialgebraic sets is an example of a structure.
Let $Ω$ be an algebrailcally closed field and $K$ a subfield. Take $S_{m}$ to be the $K$ -constructible subsets of $Ω^{m}$ . That is, finite unions of sets of the form ${x \in Ω^{m} : f_{1} (x) = f_{2} (x) = \dots = f_{n} (x) = 0 g (x) \neq 0}$ where $f_{1}, \dots, f_{n}$ , and $g$ are all polynomials in $m$ variables.

In order to check that these are in fact structures the main challenge is to prove closure under projections. Usually technics that help with this kind of issue lay in the realm of quantifier elimination. Nevertheless there are also times in which the problem of checking wheter a collection of sets form a structure lies in checking that said collection of sets is closed under taking complements. Sometimes this type of hurdles can be overcommend by technics such as Gabrielov's theorem.

Notation: Let

A \subseteq R^{m}

and

B \subseteq R^{n}

for

m, n \in ℕ

. We say that a function

f : A \to B

d e f i n a b l e

if its graph is a definable set.

Lemma 1.1.

If $A \in S_{m}$ and $B \in S_{n}$ , then $a \times B \in S_{m + n}$ .
For $1 \leq 1 \leq j \leq m$ we have that $δ_{i, j} := {(x_{1}, \dots, x_{m}) \in R^{m} : x_{i} = x_{j}}$ belong to $S_{m}$ .
For $B \in S_{m}$ and $i (1), \dots, i (n) \in {1, \dots, m}$ we have that $A \subseteq R^{m}$ given by
$(x_{1}, \dots, x_{m}) \in A \overset{}{\Leftrightarrow} (x_{i (1)}, \dots, x_{i (n)}) \in B$
is definable.

$A \times R^{n}$ and $R^{m} \times B$ are in $S_{m + n}$ so their intersection belongs to S
$A = {(x_{1}, \dots, x_{j - i + 1}) : x_{1} = x_{j - i + 1}} \in S_{j - i + 1}$ so take the cross product with $R^{i - 1}$ on the left and $R^{m - j}$ .
$(x_{1}, \dots, x_{m}) \in A$ if and only if $\exists y_{1} \dots \exists y_{n} (x_{i (1)} = y_{1}$ and $\dots$ and $x_{i (n)} = y_{n}$ and $(y_{1}, \dots, y_{n}) \in B .$

Lemma 1.2. Let

X \subseteq R^{m}

and

f : X \to R^{n}

be a map belonging to

S

$X \in S_{m}$ ;
if $A \subset X$ , $A \in S_{m}$ , then $f (A) \in s_{n}$ and $f ∣_{A}$ belongs to $S$ ;
if $B \in S_{n}$ , then $f^{- 1} (B) \in S_{m}$ ;
if $f$ is injective, then $f^{- 1}$ belongs to $S$
if $f (X) \subset T \subseteq R^{n}$ and $g : T \to R^{p}$ is a second map belonging to $S$ , then $g \circ f : X \to R^{p}$ belongs to $S$ .

2.2. O-minimal structures

Given ordered sets

R_{1}

and

R_{2}

and a map

f : R_{1} \to R_{2}

we say that:
f is strictly increasing if

x < y \to f (x < f (y)

f is increasing if

x < y \to f (x) \leq f (y)

f is strictly decreasing if

x < y \to f (x) > f (y)

f is decreasing if

x < y \to f (x) \geq f (y)

f is strictly monotone if it is strictly increasing or strictly decreasing, and
f is monotone if it is increasing or decreasing.

A linearly ordered set

R

is called dense if for all

a, b \in R

with

a < b

there is a

c \in R

such that

a < c < b

A subset

X

of a linearly ordered set is called

\ c o n v e x

if for all

a, b \in X

c \in R

such that

a < c < b

then

c \in X

Let

(R, <)

be a dense linearly ordered nonempty set without endpoints. We add two endpoints

- \infty, \infty

such that

R_{\infty} = R \cup {- \infty, \infty}

Notation: An interval is an open set of the form

(a, b) := {x \in R : a < x < b}

for

- \infty \leq a < b \leq \infty

The notation for intervals is the same as the notation for ordered pairs, but hopefully the readers of this notes will be able to identify when are we reffering to each of them in context.

We equip

R

with the interval topology (i.e. the topology in which the intervals form a base). We further equip

R^{n}

for

n \in ℕ

with the product topology ( The interval in which the ``boxes''

(a_{1}, b_{1}) \times \dots \times (a_{n}, b_{n})

form a base). It is easy to check that

R^{n}

i a hausdorf space with this topology.

Given a set

A

subset of a topological space we denote its topological closure by

cl (A)

and its topological interior as

int (A)

Given functions

f, g : X \to R_{\infty}

on a set

X \subseteq R^{m}

we put

(f, g) := {(x, r) \in X \times R : f (x) < r < g (x)}

[f, g] := {(x, r) \in X \times R : f (x) \leq r \leq g (x)}

We write

f < g

to indicate that

f (x) < g (x)

for all

x \in X

Let

(R, <)

be a dense linearly ordered nonempty set without endpoint. An o-minimal structure on

(R, <)

is a structure

S

R

such that it satisfies the following two conditions.

${(x, y) \in R^{2} : x < y} \in S_{2}$
The sets in $S_{1}$ are exactly the finite unions of points and intervals.

If the conditions above are satisfied we say that

(R; <, S)

is an o-minimal structure .

Unles stated otherwise we will assume from now on that

R

is always equipped with an o-minim al structure and 'definable' stands for 'definable with parameters'.

Lemma 2.1. Let

A \subset R

be definable. Then :

$inf (A)$ and $sup (A)$ exist in $R_{\infty}$ . (This property is called Dedekind Completeness).
the boundary $m b o x b d (A) := {x \in R : each interval containing x intersects both A and R - A} = Cl (A) - int (A) .$ is finite. Further more, if $a_{1} < \dots < a_{l}$ are the points of the boundary of $A$ , then each of the intervals $(a_{i}, a_{i} + 1)$ is completely contained, or disjoint from $A$ , for $i = 0, \dots, k$ and $a_{o} = - \infty$ and c.

Note that the first numeral in the previous lemma does not imply that every subset of

R

has a supremum (or infimum).

Lemma 2.2.

If $A \subset R^{m}$ is definable, then so are its closure and interior.
If $A \subseteq B \subseteq R^{m}$ are definable and $A$ is open in $B$ , then there is a definable open $U \subseteq R^{m}$ with $U \cap B = A$ .

$(x_{1}, \dots, x_{m}) \in A$ iff
$\forall z_{1} \dots \forall z_{m} \forall y_{1} \dots \forall y_{m} (z_{i} < x_{i} < y_{i} \to \exists w_{1} \dots \exists w_{m} (z_{i} < w_{i} < y_{i} and (w_{1}, \dots, w_{m}) \in A))$
The interior gets taken care of in a similar way.
Let $U$ be the union of boxes such that the intersection of the box with $B$ is contained in $A$ . Note that a point $(x_{1}, \dots, x_{m})$ is in $U$ iff there are $a_{1}, \dots, a_{m}$ and $b_{1}, \dots, b_{m}$ such that $a_{i} < x_{i} < b_{i}$ and for all $w_{1}, \dots, w_{m}$ such that $a_{i} < w_{i} < b_{i}$ we have that $(w_{1}, \dots, w_{m})$ is a p[oint in $B$ then it is a point in $A$ .

Definition: we say that

X \subseteq R

is definably connected if

X

is not the union of two nonempty disjoint open subsets of

X

Example:

ℚ

is not connected with the order topology, but it is definably connected in the structure

〈 ℚ; <, {q}_{q \in ℚ} 〉

Lemma 2.3.

The definable connected subsets of $R$ are the following; the empty set, the intervals, the sets $[a, b), (a, b], [a, b]$ .
The image of a definably connected set under a definably connected set under a definably continuous map $f : X \to R^{n}$ is definably connected.
If $X$ and $Y$ are definable subsets of $R^{m}$ X l(X) $a n d$ X $i s d e f i n a b l y c o n n e c t e d, t h e n$ Y $i s d e f i n a b l y c o n n e c t e d .$
If $X a n d$ Y $a r e d e f i n a b l y c o n n e c t e d s u b s e t s o f$ R^m $a n d$ X emptyset, then $X \cup Y$ is definably connected.

Item 1 is an easy consequence of o-minimality. The contrapositive of item 2 can be proven almost immediately.
For 3: Let

Y = A \cup B

definably open sets. Then either

A \cap X = \emptyset

B \cap X = \emptyset

. Without loss of generality let

A \cap X = \emptyset

A \cap cl (X) = \emptyset

A \cap Y = \emptyset

.
For 4: Let

X \cup Y = A \cup B

definably connected sets. Thus either
i)

A \cap X = \emptyset

and

A \cap Y = \emptyset

, or
ii)

B \cap X = \emptyset

and

B \cap Y = \emptyset

, or
iii)

A \cap X = \emptyset

and

B \cap Y = \emptyset

, or

B \cap X = \emptyset

and

A \cap Y = \emptyset

Corollary 2.1. For a definably continuous

f : [a, b] \to R

the Intermediate Value Theorem holds. That is; If

f (a) < c < f (b)

, then there is

d

with

a < d < b

such that

f (d) = c

Exercise: Let

S

be a definable subset of

R^{m}

show the following;

${x \in R^{m} : S_{x} is open}$ is definable,
${(x, y) \in R^{m + n}; y \in int (S_{x})}$ is definable. (Note that this is not the same as the interior of $S$ .)

3.3. O-minimal ordered groups and rings

Definition: An ordered group is a group

〈 G; \cdot 〉

equipped with a linear order

<

such that

x < y \to (x \cdot z < y \cdot z \land z \cdot x < z \cdot y)

Examples:

$〈 ℤ; <, + 〉$
$〈 ℝ \ {0}; <, \times 〉$

Proposition: Suppose that

〈 R; <, S 〉

is an o-minimal structure and

S

contains a binary operation

\cdot

R

such that

〈 R, <, \dot{〉}

is an ordered group. Then

〈 R; \dot{〉}

is abelian, divisible and torsion-free.

To prove this proposition we begin by stating a lemma.

Lemma 3.1. The only definable subsets of

R

that are also subgroups are

{1}

and

R

First show thatany definable subgroup of

R

must be convex. Otherwise you may find an ncreasing sequence

(a_{n})

where the odd number indexed elements lie outside of

G

and the even indexed elements lie inside

G

. Assume then that

G

is different from

{1}

and that there is

s \in R \ G

. Take

g > 1

with

g \in G

, then

s \in v g

lies in the convex hull of any group containing

g

but it is nopt in

G

, contradicting the first part.

The proof of the proposition goes as follows:

Abelian: Fix

r \in R

and consider

C_{r} := {x \in R : x r = r x}

. this is clearly a definable subgroup containing

r

. Thus by the previous lemma

C_{r} = R

Torsion free: This is true for all ordered groups.

Divisible: Fix

n \in ℤ^{>}

. The set

{x^{n} : x \in R}

is a definable subgroup of

R

and it is not trivial since

〈 R; \cdot 〉

is torsion free.

Remark: Let

〈 R; <, + 〉

be a non-tricial ordered abelian group. Then

+ : R^{2} \to R

and

- : R \to R

are both continuous operations in the order topology.

Definition: An ordered ring is a ring (associative with multiplicative identity 1) equipped with a liner order

<

such that

$0 < 1$
$<$ is a translation invariant. That is, $x < y$ gives $x + z < x + y$ for all $x, y, z \in R$
$<$ is invariant under multiplication of positive elements. That is $x < y$ and $z > 0$ gives $x z < y z$ .

Observations: The following are useful facts about ordered rings.

$〈 R;, <, + 〉$ is an ordered group,
The ring has no zero diisors,
$x^{2} \geq 0$ for all $x \in R$ .
$k \mapsto k \cdot 1 : ℤ \to R$ is a strictly increasing ring embeddng.

This follows directly from the definition of ordered ring.
Suppose $a$ and $b$ are n on-zero elements such that $a b = 0$ . We may assume that $a > 0$ . Then we have two cases. $b > 0$ implies $a b > 0$ and $b < 0$ implies $a b < 0$ , both clear contradictions with $a b = 0$ .
Case; 1 $x = 0$ , so $x^{2} = 0$ . Case 2; $x > 0$ so $x^{2} = x \cdot x > 0$ . Case 3; $X < 0$ , so $- x^{2} = - x \cdot x < 0$ so $x^{2} > 0$ .
Properties 1 and 2 of ordered rings give you the ``Strictly increasing'' part. It remains to show that $0 \in ℤ$ maps to $0 \in R$ . This follows from $〈 R; + 〉$ being torsion free.

This concludes the proof.

If moreover the ring is a division ring , that is there is a multiplicative right inverse for all non-zero elements, then this inverse is unique, and it is also a left inverse. Moreover the map

x \mapsto x^{- 1}

preserves the sign of

x

. In the order topology on

R

(and product topology on powers of

R

) we get that

\cdot

and

^- 1

are both continuous.

Definition: An ordered field is an ordered division ring with commutative multiplication.

Examples:

(ℚ; +, \times, <), (ℝ; +, \times, <), (ℝ ((t)); +, \times, <)

Non-examples

(ℂ; +, \times)

since there is no order preserving the field operations;

(ℤ; +, \times, <)

since it is not a field.

Definition: A real closed field is an ordered field such that if

f (X) \in R [X]

, and

a < b

elements in

R

are such that

f (a) < 0 < f (b)

, then there is

c \in (a, b)

such that

f (c) = 0

The reader should note that this is in fact the Intermediate Value Property. There are many other characterizations of "real closed fields", and they all relate to properties that the field of Real numbers satisfy, hence the name.

Proposition: Let

(R; <, S)

be an o-minimal structure and let

S

contain binary operations

+ : R^{2} \to R

and

\times : R^{2} \to R

so that

(R; <, +, \times)

is an ordered ring. Then

(R; <, +, \times)

is in fact a real closed field.

[Division Ring] Let $r \in R$ then $r R = {r x : x \in R}$ forms an additive subgroup of $R$ . Thus $r R = {0}$ if $r = 0$ and $R$ otherwise. In particular $r x = 1$ for some $x \in R$ .
[Commutativity of $\times$ ] The positive cone of $R$ , $Pos (R) = {x \in R; x > 0}$ is a definable subset of $R$ , and $(Pos (R); \times)$ is an ordered group. In fact it is an o-minimal group, since everything definable in $(Pos (R); \times)$ is definable in $(R; <, +, \times)$ . Thus as a group it is commutative, and the extention of the operation to $R$ preserves this property.
[I.V.P.] Let $P$ be a polynomial with coefficients in $R$ in one indeterminate. By definable completeness, the definable subset ${x \in (a, b) : P (x) < 0}$ of $(a, b)$ has a supremum. Since $+$ and $\times$ are continuous with respect to the order topology, this supremum $c$ satisfies that $P (c) = 0$

This concludes our proof

Note that this directly shows that

(ℚ; <, +, \times)

is not o-minimal, since

X^{2} - 2

has no root in

ℚ

4.4. Model Theoretic Structures

Definition: A language

ℒ

is a disjoint union of three (possibly empty) sets

ℒ = L_{c} \cup L_{F} \cup L_{P}

L_{c}

will be a called set of constant symbols ,

L_{F}

will be called a set of function symbols , each with an associated arity. That is for each

f \in L_{F}

we have

n_{f} \in ℕ

and we call

f

n_{f}

-ary function symbol.

L_{P}

will be called a set of relations or predicates or propositional symbols (all interchangeable terms). Each element in

L_{P}

has an associated arity.

Examples:

$ℒ_{a b} {0, +, -}$ The language of Abelian groups,
$ℒ_{a b} (<) = ℒ \cup {<}$ The language of ordered abelian groups,
$ℒ_{g r a p h} = {E}$ the language of graphs,
$ℒ_{r i n g} = {0, 1, +, \times, -}$ the language of rings.

We say that a language

ℒ ʹ

expands

ℒ

ℒ \subseteq ℒ ʹ

. For example

ℒ_{r i n g}

expands

ℒ_{a b}

Given a language we come up with a set of symbols we call variables and denote by

V a r

that is disjoint from

ℒ

We construct

ℒ

- terms in the following way

The elements of $V a r$ are terms,
the elements of $L_{c}$ are $ℒ$ -terms
the element $f t_{1} \dots t_{n}$ for $f \in L_{F}$ of arity $n$ and $t_{1}, \dots, t_{n}$ terms, is considered an $ℒ$ -term.

Examples:

$0$ in $ℒ_{a b}$ ,
$+ 0 x$ for $x \in V a r$ in $ℒ_{a b}$ (usually written as $0 + x$ ),
$(1 + 1) \times x$ in $ℒ_{r i n g}$ ,
$0 + 1$ in $ℒ_{r i n g}$

We construct

ℒ

- formulas as follows

If $t_{1}, t_{2}$ are $ℒ$ -terms, then $t_{1} = t_{2}$ is an $ℒ$ -formula,
if $t_{1}, \dots, t_{n}$ are $ℒ$ -terms and $P$ is an $n$ -ary predicate, then $P t_{1} \dots t_{n}$ is an $ℒ$ -formula,
if $φ$ is an $ℒ$ -formula, then $\neg φ$ is an $ℒ$ -formula,
if $φ_{1}$ and $φ_{2}$ are $ℒ - f o r m u l a s, t h e n$ wedge $i s a n$ L $- f o r m u l a,$
if $φ$ is an $ℒ$ -formula and $x$ is a variable, then $\exists x φ$ is an $ℒ$ -formula.

Examples:

$x_{1} = x_{2}$ in all languages
$< 1 + 0 x$ ( $1 < 0 + x$ ) for $x \in V a r$ in $ℒ_{r i n g}$ ,
$\exists x x \times x - 1 = 0$ in $ℒ_{r i n g}$ ,
$\neg \exists x \neg (0 + x = x)$ in $ℒ_{a b}$ .

Definition: We say that a variable

x

occuring in a formula

φ

is bounded in

φ

if any of the following occurs

$φ$ is $\exists x ψ$ for some formula $ψ$ ,
$φ$ is $\neg ψ$ for some formula $ψ$ in which $x$ is bounded,
$φ$ is $φ_{1} \land φ_{2}$ for formulas $φ_{1}$ and $φ_{2}$ in which $x$ appears in at least one, and is bounded in both.

Definition: Let

φ

be an

ℒ

-formula, we call

φ

a sentence if for every variable

x

appearing in it is bounded.

Examples: (3) and (4) in the previous set of examples are sentences in their respective languages.

Definition: Given a language

ℒ = L_{c} \cup L_{F} \cup L_{P}

, an

ℒ

- structure is a tuple

ℛ = (R; (c^{ℛ})_{c \in L_{c}}, (f^{ℛ})_{f \in L_{F}}, (P^{ℛ})_{P \in L_{P})}

such that

$R$ is a set (With care this can be taken to be a set theoretic proper class, but this is not necessary for this set of notes),
$c^{ℛ}$ is an element of $R$ for each $c \in L_{c}$ ,
$f^{ℛ}$ is a function f^ R :R^{n_f} $,$
$P^{ℛ}$ is a subset of $R^{n_{P}}$ .