[proofplan] We compare the four notions through their standard type-counting and rank characterisations. For a complete theory in a countable language, Morley's countable-language characterisation identifies [total transcendence](/page/Total%20Transcendence) with [$\omega$-stability](/page/Omega-Stability). The implication from $\omega$-stability to [superstability](/page/Superstability) is obtained from the countable-language stability spectrum theorem, while superstability implies [stability](/page/Stability) directly from the definition. Strictness is witnessed first by an explicit superstable non-$\omega$-stable countable theory and then by the standard countable stable non-superstable witness theorem, with the relevant type-counting and forking-chain criteria stated below. [/proofplan]

[step:Use countability to identify total transcendence with $\omega$-stability]Let $L$ denote the countable first-order language of $T$. Thus $T$ is complete and $|L| = \aleph_0$, which are exactly the hypotheses needed for [Morley's countable-language characterisation of total transcendence](/page/Total%20Transcendence). We use that characterisation in the following precise form: for a complete theory in a countable language, $T$ is totally transcendental if and only if, for every $n \in \mathbb{N}$ and every countable parameter set $A$, the Stone space $S_n(A)$ of complete $n$-types over $A$ has cardinality at most $\aleph_0$. The latter condition is exactly $\omega$-stability of $T$ in its finite-tuple form. Indeed, stability over a countable parameter set $A$ bounds complete $1$-types over $A$, and the same bound for complete $n$-types follows by applying stability to finite tuples as single object variables in the product sort, equivalently by using the standard finite-variable formulation of type-counting stability. Conversely, the condition with $n = 1$ is the usual $1$-type formulation. Hence total transcendence is equivalent to $\omega$-stability for the complete countable theory $T$.[/step]

[guided]The countability of the language is the hypothesis that permits the rank condition to be converted into a count of complete types over countable parameter sets. Let $L$ be the first-order language of $T$, and assume $|L| = \aleph_0$. [Morley's countable-language characterisation of total transcendence](/page/Total%20Transcendence) applies because $T$ is complete and $L$ is countable. It says that total transcendence is equivalent to the assertion that for every $n \in \mathbb{N}$ and every countable parameter set $A$, the Stone space $S_n(A)$ of complete $n$-types over $A$ has cardinality at most $\aleph_0$. This cardinal bound is precisely the finite-variable form of $\omega$-stability: a complete theory is $\omega$-stable when it has only countably many complete types over each countable parameter set, and this may be stated either for $1$-types or for $n$-types for all $n \in \mathbb{N}$. The implication from all $n$ to $1$ is obtained by taking $n = 1$. The implication from $1$ to all finite $n$ is the standard finite-tuple reformulation of stability: an $n$-type is a complete type of one tuple variable of length $n$, so stability over countable parameter sets bounds all finite tuple types. Therefore the rank formulation and the type-counting formulation define the same class of complete countable theories. We conclude that $T$ is totally transcendental if and only if $T$ is $\omega$-stable.[/guided]

[step:Deduce superstability from $\omega$-stability]Assume that $T$ is $\omega$-stable. By definition, this means that for every countable parameter set $A$ and every $n \in \mathbb{N}$, the space $S_n(A)$ has cardinality at most $\aleph_0$, so $T$ is stable in the cardinal $\aleph_0$. Since $T$ is complete and its language is countable, the [countable-language stability spectrum theorem](/page/Stability) applies in the direction needed here: stability in $\aleph_0$ implies stability in every infinite cardinal $\kappa \geq \aleph_0$. Thus $T$ is stable in all sufficiently large infinite cardinals. This is the definition of superstability, so $T$ is superstable.[/step]

[guided]The hypothesis that $T$ is $\omega$-stable means that for every parameter set $A$ with $|A| = \aleph_0$, the relevant spaces of complete types over $A$ are countable. Equivalently, $T$ is stable in the cardinal $\aleph_0$. Now we use the [countable-language stability spectrum theorem](/page/Stability) in the following direction: for a complete theory in a countable language, stability in $\aleph_0$ implies stability in every infinite cardinal $\kappa \geq \aleph_0$. Its hypotheses are satisfied here: completeness and countability are part of the theorem statement, and stability in $\aleph_0$ follows from $\omega$-stability by the preceding paragraph. The theorem therefore gives stability in every infinite cardinal $\kappa \geq \aleph_0$. Superstability asks for stability in all sufficiently large infinite cardinals. The preceding conclusion is stronger, because it gives stability in every infinite cardinal at least $\aleph_0$. Hence $T$ is superstable.[/guided]

[step:Deduce stability from superstability] Assume that $T$ is superstable. By definition, there exists an infinite cardinal $\kappa_0$ such that $T$ is stable in every infinite cardinal $\kappa \geq \kappa_0$. Taking one such cardinal $\kappa$, the theory $T$ is stable in at least one infinite cardinal. Hence $T$ is stable. [/step]

[step:Use unary predicates to show superstability need not imply $\omega$-stability]The implications are not reversible in the general stability hierarchy. We use two standard complete countable theories. First, let $L_1 = \{P_i : i \in \mathbb{N}\}$ be a countable language of unary predicate symbols. Let $T_1$ be the complete theory saying that every Boolean combination of finitely many predicates $P_i$ is infinite. Quantifier elimination follows by a back-and-forth argument: finite partial isomorphisms preserve equality and the finitely many predicate values already mentioned, while the infinitude of every finite Boolean cell supplies a fresh element realizing any consistent finite extension. Hence every complete $1$-type over $\varnothing$ is determined by a choice, for each $i \in \mathbb{N}$, of either $P_i(x)$ or $\neg P_i(x)$. There are $2^{\aleph_0}$ such choices, so $S_1(\varnothing)$ is uncountable and $T_1$ is not $\omega$-stable. The same quantifier-elimination description also gives superstability: if $A \subset B$ are parameter sets and $p \in S_1(B)$ is non-algebraic, then every formula in $p$ is equivalent to a Boolean combination of parameter-free unary formulas and inequalities $x \neq b$ with $b \in B$. No such non-algebraic formula divides over $A$, because an $A$-indiscernible sequence of parameters only changes finitely many forbidden equalities, and finitely many inequalities are simultaneously satisfiable inside the infinite Boolean cell determined by the parameter-free part. Thus every forking extension of a $1$-type is algebraic. For tuples, let $q \in S_n(B)$ be non-algebraic and fix a formula $\psi(x_1,\dots,x_n,b) \in q$. By quantifier elimination, $\psi$ is equivalent to a Boolean combination of the equality pattern among the variables $x_1,\dots,x_n$, unary parameter-free formulas $P_i(x_m)$ or $\neg P_i(x_m)$ for $1 \leq m \leq n$, and finitely many inequalities $x_m \neq b'$ with $b' \in B$. Along an $A$-indiscernible sequence of parameters, the equality pattern and unary parameter-free part remain fixed, while only finitely many forbidden equalities are added for each coordinate. Each coordinate still has infinitely many realizations in its prescribed Boolean cell unless the type has become algebraic, so every finite subfamily remains satisfiable. Hence a non-algebraic tuple type cannot fork except by imposing algebraicity on at least one coordinate. Here $U(p)$ denotes the Lascar $U$-rank of a complete type $p$, defined inductively by forking extensions: $U(p) \geq \alpha + 1$ exactly when $p$ has a forking extension $q$ with $U(q) \geq \alpha$, with limit stages defined by lower bounds for all smaller ordinals. The preceding paragraph shows that non-algebraic $1$-types have no non-algebraic forking extensions, so they have $U$-rank $1$. For tuple types, the explicit coordinate analysis gives finite $U$-rank bounded by the number of non-algebraic coordinate classes. Therefore there is no infinite chain of non-algebraic forking extensions, which is the finite-$U$-rank criterion for superstability. Thus superstability does not imply $\omega$-stability.[/step]

[guided]To prove strictness, it is not enough to name the hierarchy; we need witnesses. The first witness separates $\omega$-stability from superstability. Define the countable language \begin{align*} L_1 = \{P_i : i \in \mathbb{N}\}, \end{align*} where each $P_i$ is a unary predicate symbol. Let $T_1$ be the complete theory whose models are infinite sets in which every finite Boolean combination of the predicates $P_i$ is infinite. The language is countable because it has one symbol for each $i \in \mathbb{N}$. The theory is complete and eliminates quantifiers: any finite tuple is classified by its equality pattern and by the finitely many truth values of the predicates $P_i$ appearing in the formula, and the axiom that every finite Boolean cell is infinite gives the back-and-forth extension property. This immediately shows why $T_1$ is not $\omega$-stable. A complete $1$-type over $\varnothing$ may choose independently, for every $i \in \mathbb{N}$, whether $P_i(x)$ or $\neg P_i(x)$ holds. Thus the map \begin{align*} \eta : 2^{\mathbb{N}} &\to S_1(\varnothing) \\ \eta(\epsilon) &= \{P_i(x) : \epsilon_i = 1\} \cup \{\neg P_i(x) : \epsilon_i = 0\} \end{align*} is injective, where $2^{\mathbb{N}}$ denotes the set of functions $\epsilon: \mathbb{N} \to \{0,1\}$. Hence $|S_1(\varnothing)| \geq 2^{\aleph_0}$, so $T_1$ has uncountably many $1$-types over the countable parameter set $\varnothing$ and is not $\omega$-stable. The same quantifier-elimination analysis proves superstability. Let $A \subset B$ be parameter sets and let $p \in S_1(B)$ be non-algebraic. Every formula in $p$ is equivalent to a Boolean combination of parameter-free unary formulas $P_i(x)$ or $\neg P_i(x)$ and inequalities $x \neq b$ with $b \in B$. Fix such a non-algebraic formula $\varphi(x,b)$ and an $A$-indiscernible sequence $(b_j)_{j \in \mathbb{N}}$ with $b_0 = b$. The parameter-free part of $\varphi(x,b_j)$ is independent of $j$, and the parameter-dependent part excludes only finitely many elements from the infinite Boolean cell determined by that parameter-free part. Hence every finite subfamily of $\{\varphi(x,b_j) : j \in \mathbb{N}\}$ is satisfiable, so $\varphi(x,b)$ does not divide over $A$. Thus a $1$-type can fork only by becoming algebraic. For an $n$-type, let $q \in S_n(B)$ be non-algebraic and let $\psi(x_1,\dots,x_n,b) \in q$. Quantifier elimination writes $\psi$ as a Boolean combination of three kinds of data: the equality pattern among $x_1,\dots,x_n$, parameter-free unary conditions $P_i(x_m)$ or $\neg P_i(x_m)$, and finitely many inequalities $x_m \neq b'$ with $b' \in B$. If $(b_j)_{j \in \mathbb{N}}$ is an $A$-indiscernible sequence with $b_0 = b$, then the equality pattern and the parameter-free unary conditions do not change with $j$. For each coordinate $x_m$, the formulas merely exclude finitely many additional named elements from an infinite Boolean cell. Therefore every finite subfamily of $\{\psi(x_1,\dots,x_n,b_j) : j \in \mathbb{N}\}$ is satisfiable unless the type has forced one of the coordinates to be algebraic. Hence non-algebraic tuple types do not admit infinite chains of non-algebraic forking extensions. We now name the rank criterion being used. For a complete type $p$, $U(p)$ denotes its Lascar $U$-rank, defined by transfinite induction from forking extensions: $U(p) \geq \alpha + 1$ when there is a forking extension $q$ of $p$ with $U(q) \geq \alpha$, and $U(p) \geq \lambda$ for a limit ordinal $\lambda$ when $U(p) \geq \alpha$ for every $\alpha < \lambda$. The preceding analysis gives $U(p) = 1$ for every non-algebraic $1$-type $p$, and gives finite $U$-rank for each finite tuple type by counting the non-algebraic coordinate classes. By the [finite-$U$-rank characterisation of superstability](/page/Superstability), $T_1$ is superstable. This separates superstability from $\omega$-stability. The example proves that superstability does not imply $\omega$-stability.[/guided]

[step:Invoke a standard stable non-superstable witness to separate stability from superstability]We use the standard existence theorem for the stability hierarchy: there is a complete countable theory $T_2$ which is stable but not superstable, for example Shelah's standard stable non-superstable equivalence-relation-tree theory. This is an external witness theorem, and we use exactly its two verified conclusions: stability of $T_2$, and the existence of an infinite forking chain of complete types. The theorem is stated in the language of complete types and forking: $T_2$ is stable, but there exist complete types \begin{align*} p_i(x) \in S(B_i), \qquad i \in \mathbb{N}, \end{align*} and parameter sets $B_i$ with $B_i \subset B_{i+1}$ such that $p_{i+1}$ extends $p_i$ and $p_{i+1}$ forks over $B_i$ for every $i \in \mathbb{N}$. By the [forking-chain characterisation of superstability](/page/Superstability), a complete stable theory is superstable if and only if no such infinite forking chain of complete types exists, equivalently if its $U$-rank is ordinal-valued on all complete types. Hence this $T_2$ is stable but not superstable. Therefore stability does not imply superstability.[/step]

[guided]The previous witness separated $\omega$-stability from superstability. For the remaining separation we need a stable theory which fails superstability. We invoke the standard countable stable non-superstable witness theorem from stability theory: there exists a complete countable theory $T_2$, such as Shelah's equivalence-relation-tree example, which is stable but admits an infinite forking chain. The proof of stability for that constructed theory belongs to the witness theorem itself; in this hierarchy overview we use the theorem as a cited external result and record the exact data it supplies. Let us spell out exactly what property of the witness is being used. There are parameter sets $B_i$ and complete types $p_i(x) \in S(B_i)$ for $i \in \mathbb{N}$ such that \begin{align*} B_i &\subset B_{i+1}, \\ p_{i+1}|_{B_i} &= p_i, \end{align*} and $p_{i+1}$ forks over $B_i$ for every $i \in \mathbb{N}$. Thus this is a single infinite descending chain in the forking order, not merely a list of unrelated dividing formulas. The [forking-chain characterisation of superstability](/page/Superstability) says that, for a complete stable theory, superstability is equivalent to the nonexistence of such an infinite forking chain; equivalently, complete types have ordinal-valued $U$-rank. The witness theory $T_2$ is stable by the same standard theorem, but the displayed chain prevents superstability. Therefore $T_2$ proves that stability does not imply superstability. Combining this witness with the unary-predicate witness from the preceding step proves that neither implication can be reversed: superstability is strictly stronger than stability, and $\omega$-stability is strictly stronger than superstability. The first step proved that, for complete countable theories, total transcendence coincides with $\omega$-stability.[/guided]