[proofplan] We prove the equivalence by separating the cardinal arithmetic from the model-theoretic content. First, stability in every cardinal satisfying $\kappa^{|T|}=\kappa$ immediately gives stability in some infinite cardinal. Second, a formula with the order property produces, over a model of any large prescribed size, more complete finite-tuple types than stability permits. Finally, if no formula has the order property, the local counting lemma for stable formulas bounds each local type space by $\kappa^{|T|}$, and the product over all formulas still has size $\kappa$. [/proofplan] [step:Choose a cardinal satisfying the fixed point equation] Let $\mathbb{N}:=\{1,2,3,\dots\}$ denote the set of positive integers, and let $\aleph_0:=|\mathbb{N}|$ denote its cardinality. Let $\lambda:=|T|$, where $|T|$ is understood as in the statement to be the cardinality of the set of formulas of $T$ up to logical equivalence. For a finite tuple $x$ of object variables and a model $M\models T$, write $S_x(M)$ for the set of complete $x$-types over $M$, meaning maximal consistent collections of formulas in the variables $x$ with parameters from $M$. The phrase "$T$ is stable in $\kappa$" means that for every model $M\models T$ with $|M|=\kappa$ and every finite tuple $x$ of object variables, one has $|S_x(M)|\le \kappa$. If $\lambda$ is finite, set $\kappa_0:=\aleph_0$; then $\kappa_0^{\lambda}=\kappa_0$. If $\lambda$ is infinite, set $\kappa_0:=2^{\lambda}$; then cardinal arithmetic gives \begin{align*} \kappa_0^{\lambda}=(2^{\lambda})^{\lambda}=2^{\lambda\cdot \lambda}=2^{\lambda}=\kappa_0. \end{align*} Thus condition 2 implies condition 1 by applying condition 2 to this infinite cardinal $\kappa_0$. [/step] [step:Use an order pattern to manufacture many complete types] We prove the contrapositive of condition 1 implying condition 3, using the convention stated in condition 1 that the stability cardinal is an infinite cardinal $\kappa\ge |T|$. Suppose that a formula $\varphi(x;y)$ has the [order property](/page/Order%20Property) with respect to $T$, where $x$ and $y$ are finite tuples of variables. By definition, for each $n\in\mathbb{N}$ there are tuples $a_1,\dots,a_n$ and $b_1,\dots,b_n$ in a model of $T$ such that $\varphi(a_i;b_j)$ holds exactly when $i<j$. Let $\kappa$ be any infinite cardinal with $\kappa\ge |T|$. Choose a dense linear order $I$ without endpoints, of cardinality $\kappa$, with $2^{\kappa}$ Dedekind cuts. We use the [Hausdorff Cut Theorem for Linear Orders](/page/Hausdorff%20Cut%20Theorem%20for%20Linear%20Orders): for every infinite cardinal $\kappa$, there is a dense linear order without endpoints of size $\kappa$ whose set of Dedekind cuts has cardinality $2^\kappa$. By the [Compactness Theorem](/page/Compactness%20Theorem), applied to constants $(c_i)_{i\in I}$ and $(d_i)_{i\in I}$ indexed by $I$, there is a model $N\models T$ containing tuples $(c_i,d_i)_{i\in I}$ such that \begin{align*} N\models \varphi(c_i;d_j) \quad\Longleftrightarrow\quad i<j. \end{align*} Choose an elementary submodel $M\preceq N$ of cardinality $\kappa$ containing all parameters $d_i$; this exists by the [Downward Lowenheim-Skolem Theorem](/page/Downward%20Lowenheim-Skolem%20Theorem) because the parameter set has cardinality $\kappa$ and $|T|\le \kappa$. For each Dedekind cut $C\subset I$, meaning an initial segment whose complement is also nonempty and such that every element of $C$ is below every element of $I\setminus C$, define a partial [type](/page/Type) over $M$ in the finite tuple of variables $x$ by \begin{align*} p_C(x):=\{\neg\varphi(x;d_i): i\in C\}\cup\{\varphi(x;d_i): i\in I\setminus C\}. \end{align*} Let $\Delta\subset p_C(x)$ be finite. Let $P\subset I\setminus C$ be the finite set of indices $i$ for which $\varphi(x;d_i)$ occurs in $\Delta$, and let $R\subset C$ be the finite set of indices $i$ for which $\neg\varphi(x;d_i)$ occurs in $\Delta$. Since $C$ is a cut, every element of $R$ is below every element of $P$. Since $I$ is dense and has no endpoints, there is $j\in I$ such that every element of $R$ is below $j$ and $j$ is below every element of $P$; when $P$ or $R$ is empty, the no-endpoints property supplies the required one-sided choice. Then, for each $i\in P$, the inequality $j<i$ gives $N\models\varphi(c_j;d_i)$, and for each $i\in R$, the inequality $i<j$ gives $N\models\neg\varphi(c_j;d_i)$. Thus $c_j$ realizes $\Delta$. Hence every finite subset of $p_C(x)$ is satisfiable, and the [Compactness Theorem](/page/Compactness%20Theorem) makes $p_C(x)$ consistent with $T$. Extend $p_C$ to a complete type $q_C\in S_x(M)$. If $C\ne C'$, then some $i\in I$ lies in exactly one of the two cuts, and the formula $\varphi(x;d_i)$ separates $q_C$ from $q_{C'}$. Hence the map $C\mapsto q_C$ is injective, so \begin{align*} |S_x(M)|\ge 2^{\kappa}>\kappa. \end{align*} Therefore $T$ is not [stable in $\kappa$](/page/Stability), because stability for a complete theory bounds complete type spaces in every finite tuple of object variables. Since this holds for every infinite $\kappa\ge |T|$, $T$ is not stable in any cardinal allowed by condition 1. This proves that condition 1 implies condition 3. [guided] The goal is to show that an order pattern is incompatible with stability. Stability in $\kappa$ for a complete theory means that for every model $M\models T$ of cardinality $\kappa$ and every finite tuple of object variables $x$, the space $S_x(M)$ of complete $x$-[types](/page/Type) over $M$ has cardinality at most $\kappa$. In this theorem condition 1 explicitly restricts the relevant stability cardinals to infinite cardinals satisfying $\kappa\ge |T|$. Thus it is enough to build, for an arbitrary infinite $\kappa\ge |T|$, one model $M$ of size $\kappa$ over which there are more than $\kappa$ complete types in the same finite tuple of variables $x$ used by the order-property formula. Assume $\varphi(x;y)$ has the [order property](/page/Order%20Property). This means that arbitrarily long finite linear orders can be encoded by instances of $\varphi$: for every $n\in\mathbb{N}$ there are tuples $a_1,\dots,a_n$ and $b_1,\dots,b_n$ such that $\varphi(a_i;b_j)$ holds exactly when $i<j$. Compactness upgrades these finite patterns to an indexed pattern over any chosen linear order $I$. We choose $I$ to be a dense linear order without endpoints, of cardinality $\kappa$, with $2^{\kappa}$ Dedekind cuts. This is supplied by the [Hausdorff Cut Theorem for Linear Orders](/page/Hausdorff%20Cut%20Theorem%20for%20Linear%20Orders), which states that every infinite cardinal $\kappa$ admits such a dense linear order. Adding constants $(c_i)_{i\in I}$ and $(d_i)_{i\in I}$, every finite part of the desired diagram is realised by the finite order-property pattern, so the [Compactness Theorem](/page/Compactness%20Theorem) gives a model $N\models T$ satisfying \begin{align*} N\models \varphi(c_i;d_j) \quad\Longleftrightarrow\quad i<j. \end{align*} Now we put the parameters into a model of the right size. Let $A:=\{d_i:i\in I\}$. Since $|A|=\kappa$ and $\kappa\ge |T|$, the [Downward Lowenheim-Skolem Theorem](/page/Downward%20Lowenheim-Skolem%20Theorem) gives an elementary submodel $M\preceq N$ with $A\subseteq M$ and $|M|=\kappa$. Each Dedekind cut $C$ of $I$ determines a partial [type](/page/Type) \begin{align*} p_C(x):=\{\neg\varphi(x;d_i): i\in C\}\cup\{\varphi(x;d_i): i\in I\setminus C\}. \end{align*} Why is this consistent? Take a finite subset $\Delta\subset p_C(x)$. It mentions only finitely many indices from $I$. Let $P\subset I\setminus C$ be the finite set of indices appearing positively, and let $R\subset C$ be the finite set of indices appearing negatively. Since $C$ is an initial segment, every element of $R$ is below every element of $P$. Because $I$ is dense and has no endpoints, we can choose $j\in I$ after all elements of $R$ and before all elements of $P$; if $P=\varnothing$, choose $j$ after the finitely many elements of $R$, and if $R=\varnothing$, choose $j$ before the finitely many elements of $P$. The corresponding tuple $c_j$ satisfies exactly those finitely many requirements: for $i\in P$ we have $j<i$, so $N\models\varphi(c_j;d_i)$, while for $i\in R$ we have $i<j$, so $N\models\neg\varphi(c_j;d_i)$. Hence every finite subset is satisfiable, and the [Compactness Theorem](/page/Compactness%20Theorem) makes $p_C$ consistent. Extend $p_C$ to a complete type $q_C\in S_x(M)$. Distinct cuts give distinct complete types: if $C\ne C'$, choose $i\in I$ belonging to exactly one of them. Then one type contains $\varphi(x;d_i)$ and the other contains $\neg\varphi(x;d_i)$. Since there are $2^{\kappa}$ cuts, we obtain at least $2^{\kappa}$ complete $x$-types over $M$: \begin{align*} |S_x(M)|\ge 2^{\kappa}>\kappa. \end{align*} Thus $T$ is not [stable in $\kappa$](/page/Stability), because stability bounds complete type spaces in every finite tuple of variables. Since $\kappa\ge |T|$ was arbitrary, this contradicts stability in any cardinal allowed by condition 1, so a theory stable in some such infinite cardinal cannot have a formula with the order property. [/guided] [/step] [step:Bound complete types when no formula has the order property] Assume no formula has the order property with respect to $T$. Let $\kappa$ be an infinite cardinal satisfying $\kappa^{|T|}=\kappa$, let $M\models T$ be a model with $|M|=\kappa$, and let $x$ be any finite tuple of object variables. For a formula $\varphi(x;y)$, define $S_\varphi(M)$ to be the set of maximal consistent assignments of truth values to the instances $\varphi(x;b)$ with $b\in M^{|y|}$, equivalently the set of complete local $\varphi$-types over $M$. [claim:Local stable formulas have only $\kappa$ many local types] For every formula $\varphi(x;y)$, the local type space $S_{\varphi}(M)$ has cardinality at most $\kappa$. [/claim] [proof] Fix $\varphi(x;y)$. Since the global assumption says that no formula has the order property with respect to $T$, this particular formula $\varphi(x;y)$, with its displayed object tuple $x$ and parameter tuple $y$, has no order property. We use the theorem-level [Local Stability Theorem for Formulas](/page/Local%20Stability%20Theorem%20for%20Formulas) in the following precise counting form: if $T$ is complete, $M\models T$, and $\varphi(x;y)$ has no order property with respect to $T$, then every $p\in S_\varphi(M)$ is represented by a definition scheme chosen from a set of at most $|T|$ formulas, and each such scheme uses a tuple of parameters from $M$ of length at most $|T|$. Thus, for each $p\in S_\varphi(M)$, there are a formula $d_\varphi(y;z)$ from this family and a tuple $e\in M^{|z|}$ with $|z|\le |T|$ such that, for each $b\in M^{|y|}$, \begin{align*} \varphi(x;b)\in p \quad\Longleftrightarrow\quad M\models d_{\varphi}(b;e). \end{align*} The hypotheses of this local theorem are satisfied: $T$ is complete by the theorem statement, $M\models T$ by construction in this step, and $\varphi$ has no order property by the current global assumption. The theorem also supplies the cardinal bound on schemes relative to the convention that $|T|$ counts formulas of the language up to $T$-equivalence, so there are at most $|T|$ choices for $d_\varphi$. For each chosen scheme, the number of possible parameter tuples $e$ is at most $|M|^{|T|}=\kappa^{|T|}=\kappa$. Therefore \begin{align*} |S_{\varphi}(M)|\le |T|\cdot \kappa\le \kappa\cdot \kappa=\kappa. \end{align*} [/proof] A complete type over $M$ in the fixed finite tuple of variables $x$ is determined by its restrictions to the local type spaces $S_{\varphi}(M)$ as $\varphi(x;y)$ ranges over formulas with that object tuple $x$. There are $|T|$ many such formulas up to logical equivalence and variable renaming. Hence \begin{align*} |S_x(M)|\le \prod_{\varphi}|S_{\varphi}(M)|\le \kappa^{|T|}=\kappa. \end{align*} Since $M\models T$ of size $\kappa$ and the finite tuple $x$ were arbitrary, $T$ is stable in $\kappa$ in the complete-theory sense. This proves condition 3 implies condition 2. [/step] [step:Combine the implications to obtain the equivalence] We have shown that condition 2 implies condition 1, that condition 1 implies condition 3, and that condition 3 implies condition 2. These three implications form a cycle through all three conditions, so the conditions are equivalent. This completes the proof. [/step]