Dense total orders without endpoints

David Pierce

2004.11.01

These notes repeat and supplement the 2004.10.26 lecture of Math 406 (Introduction to mathematical logic and model-theory).

If L is a signature (of first-order logic), 𝔄 is an L-structure, and σ is a sentence of L, then we have defined what it means if σ is true in 𝔄. In this case, we write

𝔄 ⊧ σ.

Having defined truth, we can define logical consequence. Let Sn_{L} be the set of sentences of L. The L-structure 𝔄 is a model of a subset Σ of Sn_{L} if each sentence in Σ is true in 𝔄; then we can write

𝔄 ⊧ Σ.

If a sentence σ is true in every model of Σ, then σ is a (logical) consequence of Σ, and we can write

Σ ⊧ σ.

If ∅ ⊧ σ, then we can write just

⊧ σ;

in this case, σ is a validity.

Two sentences are (logically) equivalent if each is a logical consequence of the other.

1 Lemma. Let σ and τ be sentences of L.

(∗) {σ} ⊧ τ if and only if ⊧ (σ → τ), for all σ and τ in Sn_{L}.

(†) σ and τ are equivalent if and only if ⊧ (σ → τ) ∧ (τ → σ).

(‡) Logical equivalence is an equivalence-relation on Sn_{L}.

Proof. Exercise.

Instead of the formula (ϕ → χ) ∧ (χ → ϕ), let us write

ϕ ↔ χ.

By the lemma, σ and τ are logically equivalent if and only if (σ ↔ τ) is a validity. We may blur the distinction between logically equivalent sentences, identifying σ with ¬¬σ for example.

Instead of ¬∃v ¬ϕ, we may write

∀v ϕ.

Then ¬∀v ϕ is (equivalent to) ∃v ¬ϕ.

If fv(ϕ) = {u_{0}, …, u_{n−1}}, and 𝔄 ⊧ ∀u_{0} ⋯ ∀u_{n−1} ϕ, we may write just

𝔄 ⊧ ϕ.

Here, the sentence ∀u_{0} ⋯ ∀u_{n−1} ϕ is the (universal) generalization of ϕ. Now we can define Σ ⊧ ϕ for arbitrary formulas ϕ (although Σ should still be a set of sentences); we can also say that arbitrary formulas ϕ and χ are (logically) equivalent if

⊧ (ϕ ↔ χ).

For the formula ϕ with free variables x_{0}, …, x_{n−1}, if we have

𝔄 ⊧ ∃u_{0} ⋯ ∃u_{n−1} ϕ,

then we can say that ϕ is satisfied in 𝔄.

It can happen then that 𝔄 |̸= ϕ and 𝔄 |̸= ¬ϕ. However, if σ is a sentence, then either σ or ¬σ is true in 𝔄.

2 Example. Each of the following formulas is true in every group:

x · (y · z) = (x · y) · z,
\begin{aligned}x · 1 & = x, \\ 1 · x & = x, \end{aligned}  \begin{matrix}x · x^{−1} = 1, \\ x^{−1} · x = 1. \end{matrix}

If Σ ⊆ Sn_{L}, let

Con_{L}(Σ) = {σ ∈ Sn_{L} : Σ ⊧ σ}.

3 Lemma. Con_{L}(Con_{L}(Σ)) = Con_{L}(Σ).

Proof. Since Σ ⊆ Con_{L}(Σ), we have Con_{L}(Σ) ⊆ Con_{L}(Con_{L}(Σ)). Suppose σ ∈ Con_{L}(Con_{L}(Σ)). Then Con_{L}(Σ) ⊧ σ. But if 𝔄 ⊧ Σ, then 𝔄 ⊧ Con_{L}(Σ), so in this case 𝔄 ⊧ σ. Thus σ ∈ Con_{L}(Σ).

A subset T of Sn_{L} is a theory of L if Con_{L}(T) = T. A subset Σ of a theory T is a set of axioms for T if

T = Con_{L}(Σ);

we may also say then that Σ axiomatizes T.

4 Example. The theory of groups is axiomatized by

∀x ∀y ∀z x · (y · z) = (x · y) · z,
\begin{aligned}∀x x · 1 & = x, \\ ∀x 1 · x & = x, \end{aligned}  \begin{matrix}∀x x · x^{−1} = 1, \\ ∀x x^{−1} · x = 1. \end{matrix}

If 𝔄 is an L-structure, let

Th(𝔄) = {σ ∈ Sn_{L} : 𝔄 ⊧ σ}.

5 Lemma. Th(𝔄) is a theory.

Proof. Say Th(𝔄) ⊧ σ. Since 𝔄 ⊧ Th(𝔄), we have 𝔄 ⊧ σ, so σ ∈ Th(𝔄).

We can now call Th(𝔄) the theory of 𝔄. Note that, if T is Th(𝔄), then

T ⊧ σ ⟺ T |̸= ¬σ

for all sentences σ. An arbitrary theory T need not have this property; if it does, then T is complete. So, the theory of a structure is always complete. The set Sn_{L} is a theory, but it is not complete by this definition. Complete theories are ‘maximal’ in the following sense:

6 Lemma. Let T be a theory of L.

(∗) If T has no model, then T is Sn_{L} itself.

(†) If T has a model, namely 𝔄, then T is included in a complete theory, namely Th(𝔄).

(‡) If T has a model, then

T ⊧ σ ⟹ T |̸= ¬σ

for all σ in Sn_{L}.

(§) Hence, to prove that T is complete, it is enough to show that T has models and

T |̸= σ ⟹ T ⊧ ¬σ

for all σ in Sn_{L}.

Proof. If T is a theory with no models, and σ is a sentence, then σ is true in every model of T, so T ⊧ σ, whence σ ∈ T. The second statement is obvious. The third statement follows since {σ, ¬σ} has no models. The last statement is now obvious.

We can also speak of the theory of a class of L-structures. If K is such a class, then Th(K) is the set of sentences of L that are true in every structure in K.

In particular, if Σ ⊆ Sn_{L}, then we can define

Mod(Σ)

to be the class of all models of Σ. Then

Th(Mod(Σ)) = Con_{L}(Σ).

7 Example. By definition, a group is just a model of the theory of groups, as axiomatized in 4. Hence this theory is Th(K), where K is the class of all groups.

In general, if we have some sentences, how might we show that the theory that they axiomatize is complete? If the theory is not complete, this is easy to show:

8 Example. The theory of groups is not complete, since the sentence

∀x ∀y xy = yx

is true (by definition) only in abelian groups, but there are non-abelian groups (such as the group of permutations of three objects). The theory of abelian groups is not complete either, since (in the signature {+, −, 0}) the sentence

∀x (x + x = 0 → x = 0)

is true in (ℤ, +, −, 0), but false in (ℤ/2ℤ, +, −, 0).

Let TO be the theory of strict total orders; this is axiomatized by the universal generalizations of:

¬(x < x),
x < y → ¬(y < x),
x < y ∧ y < z → x < z,
x < y ∨ y < x ∨ x = y.

This theory is not complete, since (ω, <) and (ℤ, <) are models of TO with different complete theories (exercise).

Let TO^{∗} be the theory of dense total orders without endpoints, namely, TO^{∗} has the axioms of TO, along with the universal generalizations of:

∃z (x < z ∧ z < y),
∃y y < x,
∃y x < y.

The theory TO^{∗} has a model, namely (ℚ, <). We shall show that TO^{∗} is complete. In order to do this, we shall first show that the theory admits (full) elimination of quantifiers.

An arbitrary theory T admits (full) elimination of quantifiers if, for every formula ϕ of L, there is an open formula χ of L such that

T ⊧ (ϕ ↔ χ)

—in words, ϕ is equivalent to χ modulo T.

9 Lemma. An L-theory T admits quantifier-elimination, provided that, if ϕ is an open formula, and v is a variable, then ∃v ϕ is equivalent modulo T to an open formula.

Proof. Use induction on formulas. Specifically:

Every atomic formula is equivalent modulo T to an open formula, namely itself.

Suppose ϕ is equivalent modulo T to an open formula α. Then T ⊧ (¬ϕ ↔ ¬α); but ¬α is open.

Suppose also χ is equivalent modulo T to an open formula β. Then

T ⊧ ((ϕ → χ) ↔ (α → β));

but (α → β) is open.

Finally, T ⊧ (∃v ϕ ↔ ∃v α) (exercise); but by assumption, ∃v α is equivalent to an open formula γ; so T ⊧ (∃v ϕ ↔ γ) (exercise). This completes the induction.

The lemma can be improved slightly. Every open formula is logically equivalent to a formula in disjunctive normal form:

⋁_{i<m} ⋀_{j<n} α_{i}^{(j)},

where each α_{i}^{(j)} is either an atomic or a negated atomic formula. (See § 2.6 of this year’s notes for Math 111.) This formula in disjunctive normal form can also be written

⋁_{i<m} ⋀ Σ_{i}

where Σ_{i} = {a_{i}^{(j)} : j < n}. Note that

⊧ (∃v ⋁_{i<m} ⋀ Σ_{i} ↔ ⋁_{i<m} ∃v ⋀ Σ_{i})  (1)

(exercise). The formulas ∃v ⋀ Σ_{i} are said to be primitive. In general, a primitive formula is a formula

∃u_{0} ⋯ ∃u_{n−1} ⋀ Σ,

where Σ is a finite non-empty set of atomic and negated atomic formulas. (Remember that ⋀ Σ is just an abbreviation for ϕ_{0} ∧ … ∧ ϕ_{n−1}, where the formulas ϕ_{i} compose Σ; so Σ must be finite since formulas must have finite length. Also, formulas have positive length, so Σ must be non-empty. However, the notation ⋀ ∅ could be understood to stand for a validity.)

Using (1), we can adjust the induction above to show that T admits quantifierelimination, provided that every primitive formula with one (existential) quantifier is equivalent modulo T to an open formula.

Henceforth suppose L is {<}, and TO ⊆ T; so T is a theory of total orders. Then we can improve 9 even more. Indeed, the atomic formulas of L now are x = y and x < y, where x and y are variables. Moreover,

TO ⊧ (¬(x < y) ↔ (x = y ∨ y < x)),
TO ⊧ (¬(x = y) ↔ (x < y ∨ y < x)).

Hence, in L, any formula is equivalent, modulo TO, to the result of replacing each negated atomic sub-formula with the appropriate disjunction of atomic formulas. If this replacement is done to a formula in disjunctive normal form, then the new formula will have a disjunctive normal form that involves no negations. So T admits quantifier-elimination, provided that every formula

∃v ⋀ Σ

is equivalent, modulo T, to an open formula, where now Σ is a set of atomic formulas.

Using this criterion, we shall show that TO^{∗} admits quantifier-elimination:

10 Theorem. TO^{∗} admits full elimination of quantifiers.

Proof. Let Σ be a finite, non-empty set of atomic formulas (in the signature {<}). Let X be the set of variables appearing in formulas in Σ; that is,

X = ⋃_{α∈Σ} fv(α).

Then X is a finite non-empty set; say

X = {x_{0}, …, x_{n}}.

Suppose 𝔄 is an L-structure, and a⃗ ∈ A^{n+1}. If α is an atomic formula of L with variables from X, we can let α(a⃗) be the result of replacing each x_{i} in α with a_{i}. Then we can let

Σ(a⃗) = {α(a⃗) : α ∈ Σ}.

Suppose in fact

𝔄 ⊧ TO ∪ {⋀ Σ(a⃗)}.

Let us define Σ_{(𝔄,a⃗)} as the set of atomic formulas α such that fv(α) ⊆ X and 𝔄 ⊧ α(a⃗). Then

Σ ⊆ Σ_{(𝔄,a⃗)}.

Moreover, once Σ has been chosen, there are only finitely many possibilities for the set Σ_{(𝔄,a⃗)}. Let us list these possibilities as

Σ_{0}, …, Σ_{m−1}.

Now, possibly m = 0 here. In this case,

TO ⊧ (∃v ⋀ Σ ↔ v ≠ v),

so we are done. Henceforth we may assume m > 0. If 𝔅 ⊧ TO ∪ {⋀ Σ(b⃗)}, then

𝔅 ⊧ ⋀ Σ_{i}(b⃗)

for some i in m. Therefore

TO ⊧ (⋀ Σ ↔ ⋁_{i<m} ⋀ Σ_{i}),

and hence

TO ⊧ (∃v ⋀ Σ ↔ ⋁_{i<m} ∃v ⋀ Σ_{i}).

Therefore, for our proof of quantifier-elimination, we may assume that Σ is one of the sets Σ_{(𝔄,a⃗)} (so that, in particular, m = 1).

Now partition Σ as Γ ∪ Δ, where no formula in Γ, but every formula in Δ, contains v. There are two extreme possibilities:

(∗) Suppose Γ = ∅. Then X = {v} (since if x ∈ X∖{v}, then (x = x) ∈ Γ). Also, Σ = Δ = {v = v}, so

⊧ (∃v ⋀ Σ ↔ v = v),

and we are done in this case.

(†) Suppose Δ = ∅. Then v ∉ X, and

⊧ (∃v ⋀ Σ ↔ ⋀ Σ),

so we are done in this case.

Henceforth, suppose neither Γ nor Δ is empty. Then

⊧ (∃v ⋀ Σ ↔ ⋀ Γ ∧ ∃v ⋀ Δ).

We shall show that

TO^{∗} ⊧ (∃v ⋀ Σ ↔ ⋀ Γ),  (2)

which will complete the proof. To show (2), it is enough to show

TO^{∗} ⊧ (⋀ Γ → ∃v ⋀ Δ).

But this follows from the definition of TO^{∗}:

Indeed, remember that Σ is Σ_{(𝔄,a⃗)}. Hence, for all i and j in n + 1, we have

a_{i} < a_{j} ⟺ (x_{i} < x_{j}) ∈ Σ;
a_{i} = a_{j} ⟺ (x_{i} = x_{j}) ∈ Σ.

We have v ∈ X. We can relabel the elements of X as necessary so that v is x_{n} and

a_{0} ⩽ … ⩽ a_{n−1}.

(Here, a_{i} ⩽ a_{i+1} means a_{i} < a_{i+1} or a_{i} = a_{i+1} as usual.) Suppose 𝔅 ⊧ TO^{∗}, and B^{n} contains b⃗ such that 𝔅 ⊧ ⋀ Γ(b⃗). We have to show that there is c in B such that 𝔅 ⊧ ⋀ Δ(b⃗ , c). Now, for all i and j in n, we have

b_{i} < b_{j} ⟺ a_{i} < a_{j};
b_{i} = b_{j} ⟺ a_{i} = a_{j}.

Because 𝔅 is a model of TO^{∗} (and not just TO), we can find c as needed according to the relation of a_{n} with the other a_{i}:

(∗) If a_{n} = a_{i} for some i in n, then let c = b_{i}.
(†) If a_{n−1} < a_{n}, then let c be greater than b_{n−1}.
(‡) If a_{n} < a_{0}, then let c be less than b_{0}.
(§) If a_{k} < a_{n} < a_{k+1}, then we can let c be such that b_{k} < c < b_{k+1}.

This completes the proof that TO^{∗} admits quantifier-elimination.

We have proved more than quantifier-elimination: we have shown that, modulo TO^{∗}, the formula ∃v ⋀ Σ is equivalent to v ≠ v or v = v or an open formula with the same free variables as ∃v ⋀ Σ. In the proof, we introduced v ≠ v simply as a formula ϕ such that 𝔄 |̸= ϕ for every structure 𝔄. Such a formula corresponds to a nullary Boolean connective, namely an absurdity (the negation of a validity). We used 0 as such a connective; but let us now use ⊥.

Likewise, instead of v = v, we can use, as a validity, the nullary Boolean connective ⊤. From the last proof, therefore, we have:

11 Porism. In the signature {<}, with the nullary connectives ⊥ and ⊤ allowed, every formula is equivalent modulo TO^{∗} to an open formula with the same free variables.

In a signature of first-order logic without constants, an open sentence consists entirely of Boolean connectives, with no propositional variables; so it is either an absurdity or a validity. As a consequence, we have:

12 Theorem. TO^{∗} is a complete theory.

Proof. By the porism, every sentence is equivalent to an open sentence; as just noted, such a sentence is an absurdity or a validity. Suppose TO^{∗} ⊧ (σ ↔ ⊥). But ⊧ (σ ↔ ⊥) ↔ ¬σ; so TO^{∗} ⊧ ¬σ. Similarly, if TO^{∗} ⊧ (σ ↔ ⊤), then TO^{∗} ⊧ σ. Hence, for all sentences σ, if TO^{∗} |̸= σ, then TO^{∗} ⊧ ¬σ. Therefore TO^{∗} is complete by 6.