[proofplan] We pass from the parameter set $A$ to its $F$-linear span $W=\operatorname{span}_F(A)$, whose cardinality is bounded by $|A|+|F|$. The key model-theoretic point is that over $A$, an element has only two possible kinds of behaviour: either it is equal to a vector in $W$, or it is linearly independent over $W$. Linear automorphisms fixing $W$ show that all elements of the second kind have the same complete type over $A$. Thus the $1$-types over $A$ consist of at most one realised type for each $w\in W$, together with one generic independent type. [/proofplan] [step:Bound the size of the linear span of the parameter set] Let $W \subset V$ denote the $F$-linear span of $A$: \begin{align*} W := \operatorname{span}_F(A) = \left\{\sum_{i=1}^{n} \lambda_i a_i : n \in \mathbb{N},\ \lambda_i \in F,\ a_i \in A\right\}. \end{align*} For each $n \in \mathbb{N}$, the set of sums of length $n$ has cardinality at most $|F|^n |A|^n$. Since $A$ is infinite, \begin{align*} |W| \leq \sum_{n=1}^{\infty} |F|^n |A|^n \leq \aleph_0 \cdot \max\{|F|,|A|\} = |F|+|A|. \end{align*} Here the final equality uses the standard arithmetic of infinite cardinals. [/step] [step:Separate realised types from independent types over the span] Let $\mathfrak{C} \models T$ be a sufficiently saturated elementary extension of $V$, and regard $A \subset V \subset \mathfrak{C}$. Define \begin{align*} W_{\mathfrak{C}} := \operatorname{span}_F(A) \subset \mathfrak{C}. \end{align*} Since every element of $W_{\mathfrak{C}}$ is a finite $F$-linear combination of elements of $A$, we have $W_{\mathfrak{C}}=W$ as sets. If $b \in W$, then $b$ realizes the type containing the formula $x=b$, where $b$ is named by an $\mathcal{L}_F(A)$-term of the form $\sum_{i=1}^{n}\lambda_i a_i$. Hence there is at most one complete $1$-type over $A$ for each $b\in W$ realized inside $W$. It remains to show that all elements of $\mathfrak{C}\setminus W$ have the same complete type over $A$. [guided] The span $W=\operatorname{span}_F(A)$ is the part of the [vector space](/page/Vector%20Space) already algebraically forced by the parameters. If an element $b$ lies in $W$, then $b$ is literally named by a finite linear expression in parameters from $A$: \begin{align*} b=\sum_{i=1}^{n}\lambda_i a_i \end{align*} for some $n\in\mathbb{N}$, coefficients $\lambda_i\in F$, and parameters $a_i\in A$. Therefore the formula \begin{align*} x=\sum_{i=1}^{n}\lambda_i a_i \end{align*} isolates the equality information of that realised element over $A$. Now suppose $b\in\mathfrak{C}\setminus W$. This means precisely that $b$ is not a finite $F$-linear combination of parameters from $A$. Equivalently, the set $W\cup\{b\}$ is obtained from $W$ by adding one new linearly independent vector. Thus $b$ satisfies all inequations \begin{align*} x\neq w \end{align*} with $w\in W$, and more generally no nonzero affine linear equation over $W$ can force $b$ back into $W$. The proof below turns this intuition into equality of complete types by using linear automorphisms fixing $W$. [/guided] [/step] [step:Use linear automorphisms to identify all independent elements] Let $b,c\in\mathfrak{C}\setminus W$. Since $b\notin W$ and $c\notin W$, both singleton sets $\{b\}$ and $\{c\}$ are linearly independent over $W$. Choose an $F$-basis $B_W$ of $W$. Extend $B_W\cup\{b\}$ to an $F$-basis $B_b$ of $\mathfrak{C}$, and extend $B_W\cup\{c\}$ to an $F$-basis $B_c$ of $\mathfrak{C}$. Since $\mathfrak{C}\models T$ is infinite-dimensional and both bases extend the same basis $B_W$ by one additional vector and then by enough remaining basis elements, there is a bijection \begin{align*} \theta:B_b\to B_c \end{align*} such that $\theta(w)=w$ for every $w\in B_W$ and $\theta(b)=c$. Extend $\theta$ uniquely to an $F$-[linear map](/page/Linear%20Map) \begin{align*} \sigma:\mathfrak{C}\to\mathfrak{C}. \end{align*} Because $\theta$ is a bijection between bases, $\sigma$ is an $F$-linear automorphism. Since the language $\mathcal{L}_F$ consists exactly of the abelian group operations and scalar multiplication maps by elements of $F$, every $F$-linear automorphism is an $\mathcal{L}_F$-automorphism. Moreover, $\sigma$ fixes every element of $W$, and hence fixes every element of $A$. Therefore $\sigma$ is an automorphism of $\mathfrak{C}$ fixing $A$ pointwise and sending $b$ to $c$. It follows that \begin{align*} \operatorname{tp}(b/A)=\operatorname{tp}(c/A). \end{align*} Thus all elements outside $W$ realize a single complete $1$-type over $A$. [/step] [step:Count the possible complete $1$-types over $A$] Every complete $1$-type over $A$ is realized in the sufficiently saturated elementary extension $\mathfrak{C}$. Let $p\in S_1^T(A)$, and choose $b\in\mathfrak{C}$ realizing $p$. If $b\in W$, then $p$ is one of the realised types determined by equality with an element of $W$. There are at most $|W|$ such types. If $b\notin W$, then by the previous step $p$ is the unique generic type of an element linearly independent over $W$. Hence \begin{align*} |S_1^T(A)| \leq |W|+1. \end{align*} Using the cardinal bound from the first step, \begin{align*} |S_1^T(A)| \leq |W|+1 \leq |A|+|F|. \end{align*} This proves the desired estimate. [/step] [step:Derive $\kappa$-stability from the type-counting estimate] Let $\kappa$ be an infinite cardinal with $\kappa\geq |F|+\aleph_0$. Let $M\models T$ satisfy $|M|=\kappa$, and let $A\subset M$ satisfy $|A|\leq\kappa$. If $A$ is finite, then $\operatorname{span}_F(A)$ has cardinality at most $|F|+\aleph_0\leq\kappa$, and the same argument above gives $|S_1^T(A)|\leq\kappa$. If $A$ is infinite, the estimate already proved gives \begin{align*} |S_1^T(A)|\leq |A|+|F|\leq \kappa. \end{align*} Therefore, for every parameter set $A$ of size at most $\kappa$, there are at most $\kappa$ complete $1$-types over $A$. Hence $T$ is $\kappa$-stable for every infinite $\kappa\geq |F|+\aleph_0$. [/step]