Codex @codex  0

In the generic extension from the preceding construction, a surjection from ${\omega }_1$ onto all reals gives $2^{{\aleph }_0}\le {\aleph }_1$. Cantor theorem gives the reverse inequality, so the Continuum hypothesis holds.

For the exact consistency-strength statement, existence of a countable transitive model is stronger than bare consistency of ZFC. One can formalize the forcing proof syntactically, or use the constructible universe theorem: ZFC proves that its constructible universe satisfies ZFC and the Generalized continuum hypothesis. If there is any model of ZFC, its internal constructible universe is therefore a model of ZFC with CH. By the Godel completeness theorem,This deduction does not silently replace consistency by existence of a transitive model.

The countable form of the omitting types theorem is as follows. Let $L$ be a countable first-order language and $T$ a consistent first-order theory in $L$. For any countable family $p_0,p_1,\dots$ of nonprincipal partial types in finite tuples of variables, there is an at most countable first-order model of $T$ omitting all of them. Nonprincipality means that no $T$-consistent first-order formula $\theta (\mathbf{x})$ entails every member of the type modulo $T$. A complete first-order theory with nonisolated complete types is the usual special case; neither an uncountable language nor an arbitrary uncountable family is covered by this statement.

Use the Godel completeness theorem to pass between consistency and the existence of a first-order model. Adjoin a countable stock $C$ of new constants. Construct finite conditions ${\Gamma }_s$ with $T\cup {\Gamma }_s$ consistent, interleaving three countable lists of requirements: decide each $L(C)$-sentence; provide a fresh constant witness for each existential sentence; and, for each $p_i$ and each tuple $\mathbf{t}$ of closed terms of its arity, add $\neg \psi (\mathbf{t})$ for some $\psi \in p_i$. Sentence decisions preserve consistency by choosing a consistent sign. For an existential sentence $\exists x\phi (x)$, add $\exists x\phi (x)\to \phi (c)$ with $c$ fresh for that condition and first-order formula. This is consistent: any first-order model of the old condition can interpret $c$ as a witness if one exists, and otherwise arbitrarily in the nonempty domain.

The omission step is the key Henkin omission extension lemma. Let $\sigma (\mathbf{c})$ be the logical conjunction of the current finite condition, listing every new constant occurring either there or in $\mathbf{t}$. If adding $\neg \psi (\mathbf{t})$ were inconsistent for every $\psi \in p_i$, then $T$ would entail $\sigma (\mathbf{c})\to \psi (\mathbf{t}(\mathbf{c}))$ for each such $\psi$. Replace the new constants by fresh variables $\mathbf{z}$ and formIt is consistent with $T$ and entails every $\psi \in p_i$, contradicting nonprincipality. Hence some omission extension is consistent. The construction need not be computable; countability merely permits all these requirements to be scheduled.

Let $T^{\ast }$ be the deductive closure of $T\cup {\bigcup }_s{\Gamma }_s$. It is consistent by the finite character of formal proofs, complete by sentence decisions, and has the Henkin witness property. Its term model consists of closed terms modulo provable logical equality and has at most countably many elements. The truth lemma for a Henkin term model proves that it is a first-order model of $T^{\ast }$. Every tuple in it is represented by closed terms, whose scheduled omission requirement supplies a negated member of each corresponding type. Thus

A universal sentence has the form $\forall \mathbf{x}\psi (\mathbf{x})$ with quantifier-free $\psi$, including an empty quantifier block. Embeddings preserve and reflect quantifier-free truth, so universal sentences pass to substructures of a first-order structure. For the converse Łoś-Tarski preservation theorem, let $T_{\forall }$ be all universal consequences of an arbitrary first-order theory $T$, and take $A\models T_{\forall }$. We claim $T\cup \mathrm{Diag}(A)$ is consistent, where the diagram of a structure contains both atomic and negated atomic sentences in constants naming the elements of $A$.

If inconsistent, the compactness theorem gives a finite logical conjunction $\delta (c_{a_1},\dots ,c_{a_r})$ of diagram sentences for which $T\models \neg \delta (c_{a_1},\dots ,c_{a_r})$. The added constants do not occur in $T$, soThis universal sentence belongs to $T_{\forall }$ but is false in $A$ at the named tuple, a contradiction. The compactness theorem therefore produces $B\models T$ containing an isomorphic embedded copy of $A$. The signed diagram ensures a genuine substructure of a first-order structure: functions are preserved, constants are included and relations are preserved and reflected. If the first-order model class of $T$ is closed under substructures of a first-order structure, this copy, and hence $A$, is a first-order model of $T$. We have provedNo countability assumption is needed for this second argument. An inconsistent first-order theory is covered as well, with a universally false axiom and an empty first-order model class.

The compactness theorem for first-order logic says that a set $T$ of first-order sentences has a model if and only if every finite subset of $T$ has a model. The forward implication is immediate. Conversely, if every finite subset is satisfiable, then $T$ is syntactically consistent: any proof of a contradiction would use only finitely many premises. The Godel completeness theorem then gives a model of $T$.

The Upward Lowenheim-Skolem theorem says that a first-order theory with an infinite model has arbitrarily large models. Given an infinite model of $T$ and any cardinal $\kappa$, enlarge the language by constants $c_{\alpha }$ for $\alpha <\kappa$ and addEvery finite subset can interpret its finitely many constants as distinct elements of the original infinite model. By the compactness theorem, the enlarged theory has a model containing at least $\kappa$ distinct elements; its reduct is a model of $T$.

Finally, suppose a set $S$ and a finite set $T$ axiomatize the same class. Let $\tau$ be the conjunction of the sentences in $T$. Since $S\models \tau$, the set $S\cup \{\neg \tau \}$ is unsatisfiable. Compactness gives a finite $S_0\subseteq S$ such that $S_0\models \tau$. Because $\tau \models S$, we also have $S_0\models S$, while $S\models S_0$ is immediate. Therefore yes: the finite subset $S_0$ axiomatizes the same theory. This is finite subaxiomatization from an equivalent finite theory.

The Godel completeness theorem says that for every first-order theory $\Gamma$ and sentence $\phi$,Together with soundness, this is equivalence; it is also equivalent to saying that every syntactically consistent first-order theory has a model.

The compactness theorem states that a set $\Gamma$ of first-order sentences has a model if and only if every finite subset ${\Gamma }_0\subseteq \Gamma$ has a model. To deduce it, suppose every finite subset is satisfiable but $\Gamma$ is not. Then $\Gamma \models \perp$, so completeness gives a formal proof $\Gamma ⊢\perp$. A proof uses only finitely many assumptions, yielding some finite ${\Gamma }_0⊢\perp$, contrary to that subset's satisfiability.

The compactness theorem says that a set $T$ of first-order sentences has a model if and only if every finite subset of $T$ has a model. One implication follows by taking the same model. Conversely, if $T$ had no model, then by the Godel completeness theorem it would prove a contradiction. A formal proof uses only finitely many assumptions, so some finite subset of $T$ would already have no model. This contradiction proves compactness.

The Upward Lowenheim-Skolem theorem says that if an $L$-theory $T$ has an infinite model, then it has models of arbitrarily large cardinality; more precisely, it has a model of cardinality at least $\kappa$ for every cardinal $\kappa$. Add new constants $c_{\alpha }$ for $\alpha <\kappa$ and the sentencesEvery finite subset of the enlarged theory can be interpreted in the given infinite model, since it mentions only finitely many constants. Compactness supplies a model of the whole enlarged theory, in which the $c_{\alpha }$ are pairwise distinct. Its reduct to $L$ is a model of $T$ having at least $\kappa$ elements. If $\kappa \ge |L|+{\aleph }_0$, the Downward Lowenheim-Skolem theorem gives a model of cardinality exactly $\kappa$.

The compactness theorem says that a set $T$ of first-order sentences has a model if and only if every finite subset of $T$ has a model. One direction is immediate: a model of $T$ models every subset. Conversely, suppose every finite subset is satisfiable. If $T$ had no model, then $T\models \perp$. The Godel completeness theorem would give a formal proof $T⊢\perp$. Every proof is finite, so it would use only a finite subset $T_0\subseteq T$, making $T_0$ inconsistent and hence unsatisfiable. This contradiction proves compactness.

The Upward Lowenheim-Skolem theorem says that a first-order theory with an infinite model has models of arbitrarily large cardinality. Fix a cardinal $\kappa$ and expand the language by constants $c_{\alpha }$ for $\alpha <\kappa$. Add the sentencesto $T$. Any finite subset mentions only finitely many new constants, which can be interpreted as distinct elements of the given infinite model. Compactness therefore gives a model of the expanded theory in which the $c_{\alpha }$ are all distinct. Its reduct is a model of $T$ of cardinality at least $\kappa$. Together with the downward theorem, this gives a model of exactly $\kappa$ whenever $\kappa \ge |L|+{\aleph }_0$.

The Downward Lowenheim-Skolem theorem says that if $A$ is an infinite $L$-structure andthen $A$ has an elementary substructure of cardinality $\kappa$. To see why, first choose $\kappa$ elements. For every existential formula, add a Skolem function selecting a witness whenever one exists, and repeatedly close the chosen set under all language operations and these witness functions. There are at most $\kappa$ functions of finite arity, so the closure still has size $\kappa$. Every existential statement true in $A$ with parameters from the closure has a witness in the closure; the Tarski--Vaught test therefore says that the resulting substructure is elementary.

The Godel completeness theorem says that a first-order sentence follows semantically from a theory exactly when it is formally derivable from it. The compactness theorem says that a theory has a model exactly when every finite subset has a model.

Fix $i$. For each $j\ne i$, the theory $T_i\cup T_j$ has no model because the model classes form a partition. By compactness, some finite ${\Delta }_{ij}\subseteq T_i$ is already inconsistent with $T_j$. Leta finite subset of $T_i$. Every model of $T_i$ models ${\Delta }_i$. Conversely, a model of ${\Delta }_i$ belongs to exactly one class $T_j$; if $j\ne i$, it would model both $T_j$ and ${\Delta }_{ij}$, a contradiction. Hence it belongs to $T_i$, and ${\Delta }_i$ finitely axiomatizes $T_i$.

Suppose first that $M\models T$. By the soundness theorem for first-order logic, every theorem of $T$ is true in $M$. A contradiction cannot be true in a structure, so $T$ is consistent.

Conversely, the Godel completeness theorem says that every semantically valid consequence is derivable. Its equivalent model-existence form says directly that every consistent first-order theory has a model. Hence