Exploration 1: The canonical model of a theory

November 2, 2023.

The goal of this exploration is to show what it takes for a theory $T$ to have a model.

Each section will prove that $T$ has a model for increasingly wider classes of $T$, starting with $T$ only containing atomic sentences, and ending with $T$ containing first-order sentences.

In this section, we construct the canonical model for a set of atomic sentences $T$.

For now, we will only discuss theories $T$ where $T$ is a set of atomic sentences (atomic formulas with no variables $\bar{x}$). We’ll prove that there is always a model for $T$, the canonical model of $T$.

Outline:

• First, we simplify the problem to finding a model of $T$ when $T$ is “$=$-closed”.
• Later, we define how to make $T$ $=$-closed by extending $T$ to its “$=$-closure”.
• The canonical model of $T$ is the model of the $=$-closure of $T$, which also models the subset $T$.

Let’s start. We’d like to extend our theory $T$ to give it some properties that make it easier to find a model for it. Here are two such properties:

• (reflexivity) $t=t$ is in $T$, for all closed terms $t$ of $L$, and
• (congruence) $s=t$ in $T$ implies $\phi(s)\in T$ iff $\phi(t)\in T$, for all closed terms $s,t$ of $L$ and atomic formulas $\phi$(x) of $L$.

Intuitively, (congruence) says if two terms are equal, then you should be able to swap them in any sentence in $T$ and the result will still be in $T$.

If a set of atomic sentences $T$ has both (reflexivity) and (congruence), then it is $=$-closed (in $L$). So we’ll go ahead and extend our set of atomic sentences to its $=$-closure, i.e. its minimal $=$-closed extension.

Now let’s use these properties to find a model for $T$.

Building the canonical model

First of all, what is the underlying set? By definition of $\models$, we know that to model equalities in $T$, we must interpret both sides of the equality to be the same. An easy way to do this is to have the elements of $A$ be equivalence classes where $t_1=t_2\in T$ puts $t_1$ and $t_2$ in the same equivalence class.

We prove $t_1=t_2\in T$ is an equivalence relation on $t_i$. By (reflexivity) we already have reflexivity. By (congruence) if $t_1=t_2$ is in $T$ then we can swap the terms to get $t_2=t_1$ also in $T$, so we have symmetry, and by a similar argument we have transitivity. So $t_1=t_2\in T$ is a valid equivalence relation.

Therefore, the underlying set for our model shall be equivalence classes of closed terms, where $t_1$ and $t_2$ are equivalent if $t_1=t_2\in T$.

Since terms are now equivalence classes, our constant symbols must also be equivalence classes, and our function symbols and relation symbols all must operate on equivalence classes. We do this by lifting the relations $R$ in $L$ to take equivalence classes instead of terms.

Let’s show that this definition preserves the truth of relations in $T$. This requires showing that in $L$, whenever $t_1$ and $t_2$ are terms in the same equivalence class (of equality in T), $R(t_1)$ is the same as $R(t_2)$. But this is true because of (congruence). More generally, we must show $R(\bar{s})=R(\bar{t})$ for all length-$n$ sequences of closed terms $\bar{s}$ and $\bar{t}$ where corresponding $s_i$ and $t_i$ are in the same equivalence class. But this is true by applying (congruence) $n$ times. Thus, lifting the relation symbols from $L$ to apply to equivalence classes is well-defined, and gives us a set of relation symbols for $A$.

What about the constants and function symbols of $A$? The constant elements $c$ and function symbols $F$ of $L$ can be lifted from $L$ into $A$ by having them return equivalence classes of terms rather than single terms. We again apply (congruence) recursively to show $F(\bar{t}_1)$ and $F(\bar{t}_2)$ are in the same equivalence class, and so this is well-defined.

Summary

In summary, by defining this $L$-structure $A$ on the $=$-closure of $T$, we’ve built the canonical model of $T$ that, by construction, models our given set of atomic sentences $T$.

In this section, we prove that every model of $T$ is isomorphic to the canonical model, assuming $T$ is composed of atomic sentences.

First we define homomorphisms between $L$-structures. The idea of a homomorphism is that it set every symbol in an $L$-structure $A$ to a corresponding symbol in another $L$-structure $B$ (symbols may overlap).

An isomorphism is an invertible homomorphism. This necessarily means each unique symbol in $A$ maps to a corresponding unique symbol in $B$, i.e. it’s a renaming.

We’d like to prove that the canonical model is unique – that any two canonical models $M$, $N$ are equal. But models generally aren’t unique since you can always swap around the names of its elements to get an equivalent model. So we prove instead that the canonical model is unique up to isomorphism – any two canonical models are renamings of each other.

Suppose that $M$ is the canonical model for $T$ and $N$ is some other arbitrary model of $T$. Define the homomorphism $f:M\to N$ that maps equivalence classes $[t]\in M$ to the element $t^N\in N$.

• We show that $f$ is a homomorphism:
  • (congruence) means all constants and functions are preserved under $f$.
  • $M$ and $N$ both being models of $T$ means all relations are preserved under $f$.
  • Therefore the interpretation of each symbol in $L$ is preserved under $f$.
• To show that $f$ is well-defined, we need to show that whenever $t_1,t_2$ are in the same equivalence class, $N$ must interpret $t_1$ and $t_2$ as the same element ($t_1^N=t_2^N$). But this is true because being in the same equivalence class means $t_1=t_2\in T$, so $N\models T$ implies $N\models t_1=t_2$ implies $t_1^N=t_2^N$ by definition of $\models$.
• The reverse shows that $f$ is injective: whenever $t_1^N=t_2^N$, $t_1=t_2\in T$ must be true by (reflexivity), and therefore $t_1,t_2$ are in the same equivalence class.
• $f$ is also surjective since every element $n\in N$ is an interpretation of some term $t_1$ (since $N\models T$), and therefore is mapped to by $[t_1]\in M$.

Therefore $f$ is an isomorphism. Since we’ve constructed an isomorphism from the canonical model $M$ to an arbitrary model $N$ of $T$, this means every model of $T$ is isomorphic to the canonical model, i.e. the canonical model is unique up to isomorphism.

In this section, we show certain first-order theories are also modelled by the canonical model.

The language $L$ we work with is often a subset of $L_{\infty\omega}$ (“infinitary logic”), an extension of first-order logic. The first symbol $\infty$ means that the maximum size of conjunctions $\land$ and disjunctions $\lor$ in formulas is infinite. The second symbol $\omega$ means that the maximum number of quantifiers can be large but finite.

By taking all sentences in $L_{\infty\omega}$ true for a structure $A$, we obtain its first-order theory $Th(A)$. So every structure has a first-order theory. Conversely, does every first-order theory have a model, like we showed before for atomic sentence theories? The answer turns out to be no, not every first-order theory has a model.

But by extending some atomic sentence theory $T$ with certain first-order sentences, it’s possible to still have the same canonical model be a model of the extended $T$, as long as the resulting first-order theory maintains the following consistency properties:

1. Consistency: $\phi\in T$ implies $\lnot\phi\notin T$ for atomic sentences $\phi$
2. Congruence: $\phi(s)\in T$ iff $\phi(t)\in T$ for atomic formulas $\phi$ and closed terms $s,t$ of $L_{\infty\omega}$, whenever $s=t\in T$
3. Reflexivity: $(t=t)\in T$ for every closed term $t$ of $L_{\infty\omega}$
4. Double negation: $\lnot\lnot\phi\in T$ implies $\phi\in T$
5. Closure under $\land$: $\bigwedge\Phi\in T$ implies $\Phi\subseteq T$
6. Its inverse: $\lnot\bigwedge\Phi\in T$ implies $\exists\psi\in\Phi\ldotp\lnot\psi\in T$
7. Closure under $\lor$: $\bigvee\Phi\in T$ implies $\exists\psi\in\Phi\ldotp\psi\in T$
8. Its inverse: $\lnot\bigvee\Phi\in T$ implies $\forall\psi\in\Phi\ldotp\lnot\psi\in T$
9. Closure under $\forall$: $\forall x\ldotp\phi(x)\in T$ implies $\phi(t)\in T$ for every closed term $t$ of $L_{\infty\omega}$
10. Its inverse: $\lnot\forall x\ldotp\phi(x)\in T$ implies $\lnot\phi(t)\in T$ for some closed term $t$ of $L_{\infty\omega}$
11. Closure under $\exists$: $\exists x\ldotp\phi(x)\in T$ implies $\phi(t)\in T$ for some closed term $t$ of $L_{\infty\omega}$
12. Its inverse: $\lnot\exists x\ldotp\phi(x)\in T$ implies $\lnot\phi(t)\in T$ for every closed term $t$ of $L_{\infty\omega}$

The motivation for these properties: #1 and #4 encode negation, #2 and #3 encode $=$-closure, #5/#6 and #7/#8 encode $\lor$ and $\land$, #9/#10 and #11/#12 encode $\forall$ and $\exists$. These properties are known as consistency properties of $T$, because they ensure that $T$ is consistent even under the various properties of our logical connectives. Such theories $T$ satisfying consistency properties are called Hintikka sets (for $L_{\infty\omega}$).

The 8th property ($\lnot\bigvee\Phi\in T$ implies $\forall\psi\in\Phi\ldotp\lnot\psi\in T$) is more commonly known as the Henkin property, which is special for reasons we’ll get to soon.

But why do we care about this specific assortment of properties? And why are these properties enough to encode everything we need about their respective operations?

• To answer the first question: recall that we want the canonical model to fit $T$, and so we need to have properties on $T$ that make sure that the canonical model remains a model of $T$ after adding non-atomic sentences. Non-atomic sentences in the language $L_{\infty\omega}$ are precisely atomic sentences that are bound together by the logical connectives $\lnot,\lor,\land,\exists,\forall$. So we need to separately ensure that the properties of $\lnot,\lor,\land,\exists,\forall$ are held in the $=$-closure of $T$, recalling that the $=$-closure of $T$ is the actual theory that we construct the canonical model on; see the previous section on the canonical model for atomic sentences.
• To answer the second question: recall that for the canonical model $A$ to model $T$, we need to ensure that every sentence in $T$ is in $R^A$. When $T$ had only atomic sentences (i.e. single relations), it was easy to map each to the corresponding relation in $R^A$. But now that $T$ has non-atomic sentences, we need to ensure that what we know in first-order logic holds in $T$ as well. For example, if $A$ models $T$ and $T$ contains $\phi$ and $\psi$, we know logically that $A\models\phi\land\psi$ by definition of $\models$, but that’s not necessarily true in $T$ unless we make it true by adding the corresponding property. By encoding the axioms of our logical connectives $\lnot,\lor,\land,\exists,\forall$ in this manner, we arrive at the consistency properties above. (The two properties for $=$-closedness, #2 and #3, correspond to the fact that $T$ must be a $=$-closure, as mentioned in the previous bulletpoint.)

Hopefully that clears up the motivation for Hintikka sets; now we can say:

(Theorem 2.3.3) If $T$ is a Hintikka set for $L$, the canonical model for (atomic sentences in $T$) models $T$.

Theorem 2.3.3 implies that any theory $T$ that can be extended to a Hintikka set has a model (the canonical model). “Just extend it to a Hintikka set” is a lot to ask since there’s so many properties to fulfill. Luckily, all the properties above are true if the following smaller set of properties are true:

(Theorem 2.3.4) If

• (finite satisfiability) every finite subset of $T$ has a model,
• (completeness) $\phi$ or $\lnot\phi$ is in $T$ but not both, for every sentence $\phi$ in $L$
• (Henkin property): every $\exists x\ldotp\phi(x)$ in $T$ implies $\phi(t)$ in $T$ for some closed term $t$ of $L$

…then $T$ is a Hintikka set for $L$.

If a theory $T$ can be extended to get all three properties, then the extension is a Hintikka set, and therefore $T$ has a model. Proof that these three properties imply a Hintikka set:

Two of the consistency properties follow immediately:

• (Completeness) implies (#1) consistency: $\phi\in T$ implies $\lnot\phi\notin T$ for atomic sentences $\phi$.
• The (Henkin property) is exactly (#8) closure under $\exists$: $\exists x\ldotp\phi(x)\in T$ implies $\phi(t)\in T$ for some closed term $t$.

To derive the remaining consistency properties quickly, we first prove three short lemmas:

• (finite consistency) Every finite subset of $T$ is consistent (no contradictions).
  • Proof: Every finite subset of $T$ has a model (finite satisfiability), and inconsistent theories have no model, so every finite subset of $T$ cannot be inconsistent.
• (finite strong completeness) Every sentence logically derivable ($\tt$) from a finite subset of $T$ is in $T$.
  • Proof: Let $U$ be a finite subset of $T$. If $U\tt\phi$, then $U\cup\{\lnot\phi\}$ is finite but inconsistent, and therefore fails to be a subset of $T$ (finite consistency). This means $\lnot\phi\notin T$, which by (completeness) implies $\phi\in T$.
• (finite strong completeness corollary): if $\{\phi\}\tt\psi$, then $\phi\in T\to\psi\in T$.
  • That is, if $\psi$ follows from $\phi$ via deductions on logical operators, then we know that the property $\phi\in T\to\psi\in T$ holds on $T$.
  • Proof: if $\phi\in T$ then $\{\phi\}$ is a finite subset of $T$, so $\psi\in T$ immediately follows from (finite strong completeness).
  • This trivially extends for multiple hypotheses, $\{\phi_1,\phi_2,\ldots\}\tt\psi$, as well as none, $\{\}\tt\psi$.

Since the majority of our consistency properties are in the form $\phi\in T\to\psi\in T$ where $\{\phi\}\tt\psi$, we can use the (finite strong completeness corollary) to immediately prove them.

• $\{\}\tt(t=t)$ proves reflexivity: $(t=t)\in T$ for every closed term $t$.
  • (Reflexivity is an axiom in our language, so we can always derive it from nothing.)
• $\{\lnot\lnot\phi\}\tt\phi$ (for every sentence $\phi$) proves double negation: $\lnot\lnot\phi\in T$ implies $\phi\in T$.
• $\{\land\Phi\}\tt\phi$ (for every $\phi$ in $\Phi$) proves closure under $\land$: $\bigwedge\Phi\in T$ implies $\Phi\subseteq T$.
• $\{\lnot\lor\Phi\}\tt\lnot\psi$ (for every $\psi$ in $\Phi$) proves inverse of closure under $\lor$: $\lnot\bigvee\Phi\in T$ implies $\forall\psi\in\Phi\ldotp\lnot\psi\in T$.
• $\{\forall x\ldotp\phi(x)]\}\tt\phi(t)$ (for every closed term $t$) proves closure under $\forall$: $\forall x\ldotp\phi(x)\in T$ implies $\phi(t)\in T$ for every closed term $t$.
• $\{\lnot\exists x\ldotp\phi(x)\}\tt\lnot\phi(t)$ (for every closed term $t$) proves inverse of closure under $\exists$: $\lnot\exists x\ldotp\phi(x)\in T$ implies $\lnot\phi(t)\in T$ for every closed term $t$.
• $\{\lnot\forall x\ldotp\phi(x)\}\tt\exists x\ldotp\lnot\phi(x)$ proves $\exists x\ldotp\lnot\phi(x)\in T$. Then, using the (Henkin property) on $\exists x\ldotp\lnot\phi(x)$, we get inverse of closure under $\forall$: $\lnot\forall x\ldotp\phi(x)\in T$ implies $\lnot\phi(t)\in T$ for some closed term $t$.
• $\{s=t,\phi(s)\}\tt\phi(t)$ and $\{s=t,\phi(t)\}\tt\phi(s)$ (for every atomic formula $\phi(x)$) prove both sides of congruence: $\phi(s)\in T$ iff $\phi(t)\in T$ for atomic formulas $\phi(x)$ and closed terms $s,t$, whenever $(s=t)\in T$.

We’ve proved 10 out of 12 properties; the remaining ones are closure under $\lor$ and the inverse of closure under $\land$. They both require showing some $\psi\in\Phi$ exists in $T$ without specifying which $\psi$. Since we can’t derive ($\tt$) this unknown $\psi$, we can’t use (finite strong completeness) here. Instead, we’ll prove by contradiction – not having the property lets us construct an inconsistent finite subset of $T$, violating (finite consistency).

• Closure under $\lor$: $\bigvee\Phi\in T$ implies $\exists\psi\in\Phi\ldotp\psi\in T$.
  • Towards contradiction, assume $\bigvee\Phi\in T$ but $\lnot\exists\psi\in\Phi\ldotp\psi\in T$. Using (completeness), this implies $\forall\psi\in\Phi\ldotp\lnot\psi\in T$.
  • $\Phi$ is finite since we only allow finite conjunction/disjunction in first-order logic.
  • But then $T$ has an inconsistent finite subset $\{\bigvee\Phi,\lnot\psi_0,\lnot\psi_1,\ldots\}$, which contradicts (finite consistency). This finishes the proof.
• Inverse of closure under $\land$: $\lnot\bigwedge\Phi\in T$ implies $\exists\psi\in\Phi\ldotp\lnot\psi\in T$.
  • Let $\lnot\Phi$ denote “$\Phi$ but with $\lnot$ prepended to each $\psi\in\Phi$”.
  • $\{\lnot\bigwedge\Phi\}\tt\bigvee\lnot\Phi$ proves $\lnot\bigwedge\Phi\in T$ implies $\bigvee\lnot\Phi\in T$, by way of (finite strong completeness).
  • Then, closure under $\lor$ gives $\exists\psi\in\Phi\ldotp\lnot\psi\in T$, as required.

Theorem 2.3.4 tells you a set of minimum requirements to make an extension of $T$ into a Hintikka set: every finite subset must have a model, it must be a complete theory, and it must be closed under $\exists$ (Henkin property). If you can extend $T$ with these properties, you have a model for $T$. We will use this in the next section.

In this section, we show the absolute minimum requirement for a first-order theory to have a canonical model. (Compactness theorem)

Previously, we found that not all first-order theories have a model, but certain extensions of atomic sentences as first-order theories do have a model (the canonical model) if they can be extended to get a Hintikka set: specifically, if (1) every finite subset of $T$ has a model, and $T$ is (2) complete and (3) closed under $\exists$, then it has a model.

The following theorem shows that (2) and (3) are unnecessary; (1) alone is enough to find a model for first-order $T$, again by taking the canonical model of a larger (cleverly-crafted) theory. This fundamental result of model theory is known as the compactness theorem for first-order logic, and is the more general way to find when a given first-order theory has a model.

The compactness theorem: If every finite subset of $T$ has a model, then $T$ has a model. (where $T$ is a theory in a first-order language $L$)

Proof.

• Given that every finite subset of $T$ has a non-empty model, WTS $T$ has a model.
• The first step is to extend $T$ to a Hintikka set $T^+$. We will do so by extending the language $L$ to a larger first-order language $L^+$, and then later taking the $L$-reduct of the model of $T^+$ to get a model of the original theory $T$ in $L$.
• To extend $T$ into a Hintikka set $T^+$, we should satisfy the three requirements from the previous section:
  1. every finite subset of $T^+$ has a model
  2. every sentence $\phi$ of $L^+$ appears in $T^+$ as either $\phi$ or $\lnot\phi$
  3. every sentence $\exists x\ldotp\psi(x)$ in $T^+$ implies $\psi(t)$ in $T^+$, for some closed term of $t$
• The plan for satisfying these requirements is twofold. The first step is extending $L$, as discussed. We add an infinite set of $\kappa$ new unique constants $c_i$ to $L$, to be used as the closed terms $t$ for the third condition above. To ensure there are enough constants for each $\exists$ in $T^+$, $\kappa$ must be at least the cardinality of $L$ (so there’s at least one per term $x\in L$) plus the cardinality of $T$ (so there’s at least one per sentence $\psi\in T$). The resulting language $L^+$ has cardinality $\kappa$. Let $\phi_i$ represent the sentences of $L^+$.
• The second step towards our Hintikka set is to define a chain of theories $T_i$ in $L^+$, where, starting with $T_0=T$, each theory $T_{i+1}$ attempts to add one more sentence $\phi_i$ to the previous theory $T_i$ in a way that continues to satisfy the three conditions above and is therefore a Hintikka set. Then the union of the chain $T_\kappa = \bigcup T_i$ (essentially the “final” theory after all sentences $\phi_i$ have been added) is a Hintikka set. The precise mechanism of adding each $\phi_i$ will be described soon.

Then $T_\kappa$ (the theory obtained after considering every sentence $\phi_i$) will achieve the above three requirements of being a Hintikka set by ensuring

1. every finite subset of each $T_i$ has a model. Then $T_\kappa$ satisfies the first requirement. We can ensure this if, when we try to add each $\phi_i$, we only actually add $\phi_i$ if every finite subset of the resulting $T_{i+1}$ has a model. Otherwise, $T_{i+1}=T_i$.
2. either $\phi_i$ or $\phi_j$ is in $T_\kappa$ (where $\phi_j=\lnot\phi_i$).

• We can ensure this by tweaking the way we add $\phi_i$. When we find that $\phi_i$ cannot be added since some finite subset $U_i$ of $T_i\cup\{\phi_i\}$ has no model (i.e. $U_i\setminus\{\phi_i\}$ can derive $\lnot\phi_i$), we’ll add $\lnot\phi_i$ instead.
• In this second case we know that all finite subsets $U_j\subseteq T_i\cup\{\lnot\phi_i\}$ have a model, because if some finite subset $U_j$ doesn’t have a model, it means $U_j\setminus\{\lnot\phi_i\}$ can derive $\phi_i$. And since $U_i,U_j$ are subsets of $T_i$, their union $U_i\cup U_j$ must have a model, which is impossible if $U_i$ derives $\phi_i$ and $U_j$ derives $\{\lnot\phi_i\}$.
• Therefore we’re still preserving (1), and since we’re either adding $\phi_i$ or $\lnot\phi_i$ for every $\phi_i$ in the language, we ensure completeness by construction.
• Note: if we’ve already added $\phi_i$ in the form $\lnot\phi_j$ for some $\phi_j$, we’ll just skip adding $\phi_i$ (so $T_{i+1}=T_i$).

3. every $\exists x\ldotp\psi(x)$ in $T_\kappa$ implies $\psi(t)$ in $T_\kappa$ for some $t$ in $L^+$

• This is the reason we extended $L$ to $L^+$ – the constants $c_i$ we add will be used for $t$.
• We’ll again modify the mechanism of adding each $\phi_i$: whenever we add some $\phi_i$ that happens to be in the form $\exists x\ldotp\psi(x)$, we additionally add $\psi(c_i)$ where $c_i$ is the constant added for $\phi_i$ and therefore has not been used up to now. Note that it’s not possible that $\phi_i$ = some $\lnot\phi_j$ we added earlier since we know $\phi_i$ is in the form $\exists x\ldotp\psi(x)$. This trivially completes the proof of (3) since we explicitly made it so $\psi(t)$ exists for every $\exists x\ldotp\psi(x)$.
• However, by modifying the mechanism of adding each $\phi_i$, we will have to prove (1) and (2) again. We will only prove (1), since (1) implies (2) using the same argument before.
• (1 again) every finite subset of our new $T_i$ (= old $T_i$ with $\psi(c_i)$ added) has a model.
  • In the case where $\phi_i$ is not in the form $\exists x\ldotp\psi(x)$, we don’t add $\psi(c_i)$, so by the old argument (1) still holds.
  • In the case where $\phi_i$ is in the form $\exists x\ldotp\psi(x)$, we add $\psi(c_i)$ as well. This changes things – we must show that the new $T_{i+1}=T_i\cup\{\exists x\ldotp\psi(x),\psi(c_i)\}$ has a model.
  • From the old argument, we know that every arbitrary subset of the resulting $T_{i+1}=T_i\cup\{\exists x\ldotp\psi(x)\}$ has a model $A$.
  • Let $U$ be one of those subsets that contains $\exists x\ldotp\psi(x)$, which has a model $A\models U$.
  • The plan is to define a new $L^+$-structure $B$ modelling the same theory $U$ such that $B\models\psi(a)$ means the same thing as $B\models\psi(c_i)$, thus showing that every subset containing $\psi(c_i)$ of our new $T_{i+1}=T_i\cup\{\exists x\ldotp\psi(x),\psi(c_i)\}$ has a model $B$.
  • Since $A\models\exists x\ldotp\psi(x)$, there is some element $a\in A$ such that $A\models\psi(a)$. We can define $B$ as identical to $A$, except it interprets the constant symbol $c_i$ as $a$. In other words, $B\models\psi(a)$ means the same thing as $B\models\psi(c_i)$. We ensured that $c_i$ is a fresh variable since this is the first time we’re considering adding the corresponding $\phi_i$. So any previous interpretation of $c_i$ is irrelevant, making $B$ still a model of $U$
  • Since $A\models\psi(a)$ by definition of $A$, we have $B\models\psi(a)$, which means the same thing as $B\models\psi(c_i)$, thus showing that every subset containing $\psi(c_i)$ of our new $T_{i+1}=T_i\cup\{\exists x\ldotp\psi(x),\psi(c_i)\}$ has a model $B$.

Finally, we’ve constructed some $T_\kappa$ that is a Hintikka set of $L^+$ by virtue of satsifying the three conditions from the previous section. Therefore it has a model (the canonical model of its atomic sentences). Then $T_\kappa$ includes $T$ by its construction (a chain on top of $T$), so we can take the $L$-reduct of the model of $T_\kappa$ to get a model of the original theory $T$ in $L$.

There are many many proofs of the compactness theorem, and this is just one of them, called the Henkin construction. For more proofs of the compactness theorem, see: https://mathoverflow.net/a/45501

The converse of the compactness theorem is quite trivial. If a theory $T$ has a model $M$, then $M$ models every finite subset of $T$ since they are subsets of $T$.

“$T$ is satisfiable” is another way to say “$T$ has a model”. So the compactness theorem can also be stated “$T$ is satisfiable if and only if $T$ is finitely satisfiable.”

In the end we’ve found the minimum requirement for $T$ to have a canonical model – we just need to prove that every finite subset of $T$ has a model, which is often done by construction.

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