\end{document}

## Lecture in O-minimality

## By Santiago Camacho for MA595, U Illinois, Fall 2015

\mbox

\begin{document} \maketitle 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. \section{Chapter 1} Let $X,Y,Z$ be sets and $x,y,z$ variables raging over $X,Y,Z$ respectively and $\phi(x,y,z)$ and $\psi(x,y,z)$ be conditions on a point $(x,y,z) \in X\times Y\times Z$. We define $\Phi, \Psi \subseteq X\times Y \times Z$ as \[\Phi = \{(x,y,z)\in X\times Y \times Z: \phi(x,y,z) \mbox{ holds } \}\] \[\Psi = \{(x,y,z)\in X\times Y \times Z: \psi(x,y,z) \mbox{ holds } \}\] Throughout the letures, for given conditions $\phi$ and $\psi$ we use the notations \begin{enumerate} \item $\phi \vee \psi,$ \item $\phi \wedge \psi,$ \item $\neg \phi,$ \item $\exists z \phi(x,y,z),$ \item $\forall z \phi(x,y,z).$ \end{enumerate} The associated sets would correspond to \begin{enumerate} \item The union, \item the intersection, \item the complement, \item the projection on the first two variables, \item the complement of the projection of the complement. \end{enumerate} Usually all variables will range over powers of the same set. for example \[ X= \mathbb{R}^m,\ Y= \mathbb{R}^n,\ Z=\mathbb{R}^p\]A \textbf{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 \[A\leq B \iff A\cap B = A\] An \textbf{atom} of a boolean algebra is a minimal element under the previously mentioned order that lies above $\emptyset$.DefinitionLet $A_1,\ldots,A_n$ be a collection of subsets of $X$. We define the boolean algebra on $X$ generated by $A_1,\ldots,A_n$ to be \[\mathcal{B}(A_1,\ldots,A_n;X) = \left \{ \bigcup_{\Delta \in \Gamma} \left( \bigcap_{i\in \Delta} A_i \cap \bigcap_{i\notin\Delta}X\setminus A_i \right) :\Gamma\subset \mathbb{P}(\{1,\ldots,n\})\right\} \] Where $\mathbb{P}(Y)$ denotes the \textbf{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. \subsection{Facts on Structures}Definition:A \textbf{structure} on a nonempty set $R$ is a family $\mathbf{S} = (S_m)_{m\in \mathbb{N}}$ such that for each $m\geq 0$; \begin{enumerate} \item $S_m$ is a boolean algebra of subsets of $R^m$; \item if $A\in S_m$, then $R\times A$ and $A\times R$ are both in $S_{m+1}$; \item $\{(x_1,\ldots,x_m)\in R^m: x_1=x_m\} \in S_m$; \item If $A\in S_{m+1}$, then $\pi(A)\in S_{m}$, were $\pi:R^{m+1}\rightarrow R^m$ is the projection on the first $m$ coordinates. \end{enumerate} We use the notation $(R,\mathbf{S})$ to denote a structure, and we call the subsets of $R^m$ that are in $S_m$ \textbf{definable} sets.Definition:Examples\begin{itemize} \item $\mathbb{R}$ equipped with semialgebraic sets is an example of a structure. \item Let $\Omega$ be an algebrailcally closed field and $K$ a subfield. Take $S_m$ to be the $K$-constructible subsets of $\Omega^m$. That is, finite unions of sets of the form $\{x\in \Omega^m: f_1(x)=f_2(x)= \cdots =f_n(x)=0 \ g(x)\neq 0\}$ where $f_1,\ldots,f_n$, and $g$ are all polynomials in $m$ variables. \end{itemize} 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 $m,n \in \mathbb{N}$. We say that a function $f:A\rightarrow B$ is $\textbf{definable}$ if its graph is a definable set. \begin{lemma} \begin{enumerate} \item If $A\in S_m$ and $B\in S_n$, then $a\times B \in S_{m+n}$. \item For $1\leq 1 \leq j \leq m$ we have that $\delta_{i,j}:= \{(x_1,\ldots,x_m)\in R^m:x_i=x_j\}$ belong to $S_m$. \item For $B\in S_m$ and $i(1),\ldots, i(n)\in \{1,\ldots,m\}$ we have that $A\subseteq R^m$ given by \[(x_1,\ldots,x_m)\in A \iff (x_{i(1)},\ldots, x_{i(n)})\in B\] is definable. \end{enumerate} \end{lemma} \begin{proof} \begin{enumerate} \item $A\timesR^n$ and $R^m\times B$ are in $S_{m+n}$ so their intersection belongs to $\mathbf{S} \item $A = \{(x_1,\ldots, x_{j-i+1}): x_1 = x_{j-i+1}\} \in S_{j-i+1}$ so take the cross product with $R^{i-1}$ on the left and $R^{m-j}$. \item $(x_1,\ldots, x_m)\in A$ if and only if $\exists y_1\ldots\exists y_n(x_{i(1)} = y_1$ and $\ldots$ and $ x_{i(n)} = y_n$ and $(y_1,\ldots,y_n)\in B.$ \end{enumerate} \end{proof} \begin{lemma} Let $X\subseteq R^m$ and $f:X\rightarrow R^n$ be a map belonging to $\mathbf{S}$. \begin{enumerate} \item $X\in S_m$; \item if $A\subset X$, $A\in S_m$, then $f(A)\in s_n$ and $f|_A$ belongs to $\mathbf{S}$; \item if $B\in S_n$, then $f^{-1}(B) \in S_m$; \item if $f$ is injective, then $f^{-1}$ belongs to $\mathbf{S}$ \item if $f(X)\subset T\subseteq R^n$ and $g:T\rightarrow R^p$ is a second map belonging to $\mathbf{S}$, then $g\circ f:X\rightarrow R^p$ belongs to $\mathbf{S}$. \end{enumerate} \end{lemma} %%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%% \section{O-minimal structures} Given ordered sets $R_1$ and $R_2$ and a map $f:R_1\rightarrow R_2$ we say that:\\ f is \textbf{strictly increasing} if $x\lt y \rightarrow f(x\lt f(y)$\\ f is \textbf{increasing} if $x\lt y \rightarrow f(x)\leq f(y)$\\ f is \textbf{strictly decreasing} if $x\lt y \rightarrow f(x)\gt f(y)$\\ f is \textbf{decreasing} if $x\lt y \rightarrow f(x)\geq f(y)$\\ f is \textbf{strictly monotone} if it is strictly increasing or strictly decreasing, and\\ f is \textbf{monotone} if it is increasing or decreasing. A linearly ordered set $R$ is called \textbf{dense} if for all $a,b\in R$ with $a\lt b$ there is a $c\in R$ such that $a\lt c\lt b$. A subset $X$ of a linearly ordered set is called $\convex$ if for all $a,b \in X$, $c\in R$ such that $a\lt c\lt b$ then $c\in X$. Let $(R,\lt)$ 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 \textbf{interval} is an open set of the form $(a,b):=\{x\in R: a\lt x\lt b\}$ for $-\infty \leq a \lt b \leq \infty$ 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 $n\in \mathbb{N}$ with the product topology ( The interval in which the ``boxes'' $(a_1,b_1)\times \cdots \times (a_n,b_n)$ 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 \textbf{topological closure} by $\mbox{cl}(A)$ and its \textbf{topological interior} as $\mbox{int}(A)$. Given functions $f,g:X\rightarrow R_\infty$ on a set $X\subseteq R^m$ we put \[(f,g) := \{(x,r)\in X\times R: f(x) \lt r \lt g(x)\}\] \[[f,g] := \{(x,r)\in X\times R: f(x)\leq r\leq g(x)\}\] We write $f\lt g$ to indicate that $f(x)\lt g(x)$ for all $x\in X$. Let $(R,\lt)$ be a dense linearly ordered nonempty set without endpoint. An \textbf{o-minimal structure on} $(R,\lt)$ is a structure $\mathbf{S}$ on $R$ such that it satisfies the following two conditions. \begin{enumerate} \item $\{(x,y)\in R^2: x\lt y\}\in S_2$ \item The sets in $S_1$ are exactly the finite unions of points and intervals. \end{enumerate} If the conditions above are satisfied we say that $(R;\lt,\mathbf{S})$ is an \textbf{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'. \begin{lemma} Let $A\subset R$ be definable. Then : \begin{enumerate} \item $\inf(A)$ and $\sup(A)$ exist in $R_\infty$. (This property is called \textit{Dedekind Completeness). \item the boundary $\mbox\mbox{bd}(A) := \{x\in R: \mbox{ each interval containing }x \mbox{ intersects both } A \mbox{ and } R-A\} = \mbox{Cl}(A) - \mbox{int}(A).$ is finite. Further more, if $a_1\lt\ldots\lt a_l$ 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 $i=0,\ldots, k$ and $a_o=-\infty$ and c. \end{enumerate} \end{lemma} Note that the first numeral in the previous lemma does not imply that every subset of $R$ has a supremum (or infimum). \begin{lemma} \begin{enumerate} \item If $A\subset R^m$ is definable, then so are its closure and interior. \item 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 $U\cap B = A$. \end{enumerate} \end{lemma} \begin{proof} \begin{enumerate} \item $(x_1,\ldots,x_m)\in A$ iff \[\forall z_1\ldots \forall z_m \forall y_1\ldots \forall y_m (z_i\lt x_i \lt y_i \rightarrow \exists w_1\ldots \exists w_m (z_i\lt w_i\lt y_i \mbox{ and } (w_1,\ldots,w_m)\in A))\] The interior gets taken care of in a similar way. \item Let $U$ be the union of boxes such that the intersection of the box with $B$ is contained in $A$. Note that a point $(x_1,\ldots, x_m)$ is in $U$ iff there are $a_1,\ldots, a_m$ and $b_1,\ldots,b_m$ such that $a_i\lt x_i \lt b_i$ and for all $w_1,\ldots, w_m$ such that $a_i\lt w_i\lt b_i$ we have that $(w_1,\ldots,w_m)$ is a p[oint in $B$ then it is a point in $A$. \end{enumerate} \end{proof}we say that $X\subseteq R$ is \textbf{definably connected} if $X$ is not the union of two nonempty disjoint open subsets of $X$.Definition:Example:$\mathbb{Q}$ is not connected with the order topology, but it is definably connected in the structure $\langle \mathbb{Q}; \lt, \{q\}_{q\in \mathbb{Q}}\rangle$ \begin{lemma} \begin{enumerate} \item The definable connected subsets of $R$ are the following; the empty set, the intervals, the sets $[a,b), (a,b], [a,b]$. \item The image of a definably connected set under a definably connected set under a definably continuous map $f:X\rightarrow R^n$ is definably connected. \item If $X$ and $Y$ are definable subsets of $R^m$ X\subseteq Y\subseteq cl(X)$ and $X$ is definably connected, then $Y$ is definably connected. \item If $X and $Y$ are definably connected subsets of $R^m$ and $X\cap Y\neq \emptyset, then $X\cup Y$ is definably connected. \end{enumerate} \end{lemma} \begin{proof} 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 $A\cap X = \emptyset$ or $B\cap X= \emptyset$. Without loss of generality let $A\cap X= \emptyset$ so $A\cap \mbox{cl}(X) = \emptyset$ so $A\cap Y = \emptyset$.\\ 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 \\iv) $B\cap X = \emptyset$ and $A\cap Y = \emptyset$. \end{proof} \begin{corollary} For a definably continuous $f:[a,b]\rightarrow R$ the Intermediate Value Theorem holds. That is; If $f(a)\lt c\lt f(b)$, then there is $d$ with $a\lt d \lt b$ such that $f(d)=c$. \end{corollary}Exercise:Let $S$ be a definable subset of $R^m$ show the following; \begin{enumerate} \item $\{x\in R^m:S_x \mbox{ is open }\}$ is definable, \item $\{(x,y)\in R^{m+n}; y\in \mbox{int}(S_x)\}$ is definable. (Note that this is not the same as the interior of $S$.) \end{enumerate} \section{O-minimal ordered groups and rings}An \textbf{ordered group} is a group $\langle G;\cdot\rangle$ equipped with a linear order $\lt$ such that $x\lt y \rightarrow (x\cdot z\lt y\cdot z \wedge z\cdot x \lt z\cdot y)$Definition:Examples:\begin{enumerate} \item $\langle \mathbb{Z}; \lt, +\rangle$ \item $\langle \mathbb{R}\setminus \{0\}; \lt, \times\rangle$ \end{enumerate}Proposition:Suppose that $\langle R; \lt, \mathbf{S}\rangle$ is an o-minimal structure and $\mathbf{S}$ contains a binary operation $\cdot$ on $R$ such that $\langle R,\lt,\dot\rangle$ is an ordered group. Then $\langle R; \dot\rangle$ is abelian, divisible and torsion-free. To prove this proposition we begin by stating a lemma. \begin{lemma} The only definable subsets of $R$ that are also subgroups are $\{1\}$ and $R$. \end{lemma} \begin{proof} 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 $s\in R\setminus G$. Take $g\gt 1$ with $g\in G$, then $s\inv g$ lies in the convex hull of any group containing $g$ but it is nopt in $G$, contradicting the first part. \end{proof} The proof of the proposition goes as follows: \begin{proof}Abelian:Fix $r\in R$ and consider $C_r:= \{x\in R: xr=rx\}$. 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 $n\in \mathbb{Z}^{\gt}$. The set $\{x^n:x\in R\}$ is a definable subgroup of $R$ and it is not trivial since $\langle R;\cdot\rangle$ is torsion free. \end{proof}Let $\langle R;\lt, +\rangle$ be a non-tricial ordered abelian group. Then $+:R^2\rightarrow R$ and $-:R\rightarrow R$ are both continuous operations in the order topology.Remark:An \textbf{ordered ring} is a ring (associative with multiplicative identity 1) equipped with a liner order $\lt$ such that \begin{enumerate} \item $0\lt 1$ \item $\lt$ is a translation invariant. That is, $x\lt y$ gives $x+z \lt x+y$ for all $x,y,z \in R$ \item $\lt$ is invariant under multiplication of positive elements. That is $x\lt y$ and $z\gt 0$ gives $xz\lt yz$. \end{enumerate}Definition:Observations:The following are useful facts about ordered rings. \begin{enumerate} \item $\langle R;,\lt,+\rangle$ is an ordered group, \item The ring has no zero diisors, \item $x^2 \geq 0$ for all $x\in R$. \item $k\mapsto k\cdot 1: \mathbb{Z}\rightarrow R$ is a strictly increasing ring embeddng. \end{enumerate} \begin{proof} \begin{enumerate} \item This follows directly from the definition of ordered ring. \item Suppose $a$ and $b$ are n on-zero elements such that $ab=0$. We may assume that $a>0$. Then we have two cases. $b\gt 0$ implies $ab\gt 0$ and $b\lt 0$ implies $ab\lt 0$, both clear contradictions with $ab=0$. \item Case; 1 $x= 0$, so $x^2 = 0$. Case 2; $x>0$ so $x^2=x\cdot x \gt 0$. Case 3; $X\lt 0$, so $-x^2=-x \cdot x \lt 0$ so $x^2 \gt 0$. \item Properties 1 and 2 of ordered rings give you the ``Strictly increasing'' part. It remains to show that $0\in \mathbb{Z}$ maps to $0\in R$. This follows from $\langle R; +\rangle$ being torsion free. \end{enumerate} This concludes the proof.\end{proof} If moreover the ring is a \textbf{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.An \textit{ordered field} is an ordered division ring with commutative multiplication.Definition:Examples:$(\mathbb{Q};+,\times,\lt),\ (\mathbb{R};+,\times,\lt),\ (\mathbb{R}((t));+,\times,\lt)$Non-examples$(\mathbb{C};+,\times)$ since there is no order preserving the field operations; $(\mathbb{Z};+,\times,\lt)$ since it is not a field.A \textit{real closed field} is an ordered field such that if $f(X)\in R[X]$, and $a\lt b$ elements in $R$ are such that $f(a) \lt 0\lt f(b)$, 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.Definition:Let $(R;\lt, \mathbf{S})$ be an o-minimal structure and let $\mathbf{S}$ contain binary operations $+:R^2\rightarrow R$ and $\times:R^2\rightarrow R$ so that $(R;<,+,\times)$ is an ordered ring. Then $(R;\lt,+,\times)$ is in fact a real closed field. \begin{proof} \begin{enumerate} \item[Division Ring] Let $r\in R$ then $rR=\{rx:x\in R\}$ forms an additive subgroup of $R$. Thus $rR = \{0\}$ if $r=0$ and $R$ otherwise. In particular $rx=1$ for some $x\in R$. \item[Commutativity of $\times$] The positive cone of $R$, $\mbox{Pos}(R) = \{x\in R;x\gt 0\}$ is a definable subset of $R$, and $(\mbox{Pos}(R);\times)$ is an ordered group. In fact it is an o-minimal group, since everything definable in $(\mbox{Pos}(R);\times)$ is definable in $(R;\lt,+,\times)$. Thus as a group it is commutative, and the extention of the operation to $R$ preserves this property. \item[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 $P(c)=0$ \end{enumerate} This concludes our proof \end{proof} Note that this directly shows that $(\mathbb{Q};\lt,+,\times)$ is not o-minimal, since $X^2-2$ has no root in $\mathbb{Q}$. \section{Model Theoretic Structures}Proposition:A \textit{language} $\mathcal{L}$ is a disjoint union of three (possibly empty) sets \[\mathcal{L}= L_c\cup L_{F} \cup L_P\] $L_c$ will be a called set of \textit{constant symbols}, $L_{F}$ will be called a set of \textit{function symbols}, each with an associated arity. That is for each $f\in L_F$ we have $n_f\in \mathbb{N}$ and we call $f$ an $n_f$-ary function symbol. $L_P$ will be called a set of \textit{relations} or \textit{predicates} or \textit{propositional symbols} (all interchangeable terms). Each element in $L_P$ has an associated arity.Definition:Examples:\begin{itemize} \item $\mathcal{L}_{ab}\{0,+,-\}$ The language of Abelian groups, \item $\mathcal{L}_{ab}(\lt) = \mathcal{L}\cup \{\lt\}$ The language of ordered abelian groups, \item $\mathcal{L}_{graph} = \{E\}$ the language of graphs, \item $\mathcal{L}_{ring} = \{0,1,+,\times,-\}$ the language of rings. \end{itemize} We say that a language $\mathcal{L}'$ expands $\mathcal{L}$ if $\mathcal{L}\subseteq \mathcal{L}'$. For example $\mathcal{L}_{ring}$ expands $\mathcal{L}_{ab}$. Given a language we come up with a set of symbols we call variables and denote by $Var$ that is disjoint from $\mathcal{L}$ We construct $\mathcal{L}$-\textit{terms} in the following way \begin{itemize} \item The elements of $Var$ are terms, \item the elements of $L_c$ are $\mathcal{L}$-terms \item the element $ft_1\ldots t_n$ for $f\in L_F$ of arity $n$ and $t_1,\ldots,t_n$ terms, is considered an $\mathcal{L}$-term. \end{itemize}Examples:\begin{enumerate} \item $0$ in $\mathcal{L}_{ab}$, \item $+0x$ for $x\in Var$ in $\mathcal{L}_{ab}$ (usually written as $0+x$), \item $(1+1)\times x$ in $\mathcal{L}_{ring}$, \item $0+1$ in $\mathcal{L}_{ring}$ \end{enumerate} We construct $\mathcal{L}$-\textit{formulas} as follows \begin{itemize} \item If $t_1, t_2$ are $\mathcal{L}$-terms, then $t_1=t_2$ is an $\mathcal{L}$-formula, \item if $t_1, \ldots, t_n$ are $\mathcal{L}$-terms and $P$ is an $n$-ary predicate, then $Pt_1\ldots t_n$ is an $\mathcal{L}$-formula, \item if $\phi$ is an $\mathcal{L}$-formula, then $\neg \phi$ is an $\mathcal{L}$-formula, \item if $\phi_1$ and $\phi_2$ are $\mathcal{L}-formulas, then $\phi_1\wedge \phi_2$ is an $\mathcal{L}$-formula, \item if $\phi$ is an $\mathcal{L}$-formula and $x$ is a variable, then $\exists x \phi$ is an $\mathcal{L}$-formula. \end{itemize}Examples:\begin{enumerate} \item $x_1=x_2$ in all languages \item $\lt1+0x$ ($1\lt0+x$) for $x\in Var$ in $\mathcal{L}_{ring}$, \item $\exists x x\times x -1 =0$ in $\mathcal{L}_{ring}$, \item $\neg\exists x \neg (0+x= x)$ in $\mathcal{L}_{ab}$. \end{enumerate}We say that a variable $x$ occuring in a formula $\phi$ is \textit{ bounded in } $\phi$ if any of the following occurs \begin{enumerate} \item $\phi$ is $\exists x \psi$ for some formula $\psi$, \item $\phi$ is $\neg \psi$ for some formula $\psi$ in which $x$ is bounded, \item $\phi$ is $\phi_1\wedge \phi_2$ for formulas $\phi_1$ and $\phi_2$ in which $x$ appears in at least one, and is bounded in both. \end{enumerate}Definition:Let $\phi$ be an $\mathcal{L}$-formula, we call $\phi$ a \textit{sentence} if for every variable $x$ appearing in it is bounded.Definition:Examples:(3) and (4) in the previous set of examples are sentences in their respective languages.Given a language $\mathcal{L} = L_c \cup L_F\cup L_P$, an $\mathcal{L}$-\textit{structure} is a tuple $\mathcal{R}=(R;(c^\mathcal{R})_{c\in L_c}, (f^\mathcal{R})_{f\in L_F}, (P^\mathcal{R})_{P\in L_P)$ such that \begin{itemize} \item $R$ 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), \item $c^\mathcal{R}$ is an element of $R$ for each $c\in L_c$, \item $f^\mathcal{R}$ is a function f^\mathcal{R}:R^{n_f}\rightarrow R$, \item $P^\mathcal{R}$ is a subset of $R^{n_P}$. \end{itemize}Definition:\end{document}