证明实数扩张模型Th(ℛ)无ℚ上素模型的技术问询

Let's walk through this step by step, starting with the basics and building up to the key argument.

Background Setup #

First, let's formalize our structure and theory:

• We define ℛ as the expanded real ordered set: ℛ = ⟨ℝ; <, Q⟩, where Q is a unary predicate that picks out exactly the rational numbers ℚ.
• Let T = Th(ℛ), the complete first-order theory of this structure.
• Our goal is to show T has no prime model over ℚ. Recall that a theory has a prime model over a set A if and only if isolated types are dense in the type spaces Sₙ(A) for all n. We only need to focus on n=1 here (the space of 1-types over ℚ) to reach our conclusion.

The Key Formula φ(x) #

The candidate formula we'll use is straightforward:
φ(x) ≡ ¬Q(x)
In plain terms, this says "x is irrational".

Step 1: φ(x) is Consistent with T #

This is trivial: ℛ itself contains plenty of irrationals (like √2 or π), so T proves that there exists an x satisfying φ(x). So φ(x) is definitely consistent with our theory.

Step 2: No Isolated Type Contains φ(x) #

Now, we need to show that there's no isolated type in S₁^T(ℚ) that includes φ(x). Let's suppose the opposite for contradiction: suppose there is an isolated type p(x) that contains φ(x).

Since p is isolated, there must be some formula ψ(x) (with parameters from ℚ) such that:

1. ψ(x) is in p, and
2. Every formula in p is logically implied by ψ(x) (plus T).

Since φ(x) is in p, ψ(x) must imply ¬Q(x) (otherwise, we could have a rational satisfying ψ(x), which would contradict φ(x) being in p).

Now, any formula ψ(x) in our language (with rational parameters) is equivalent to a Boolean combination of:

• Inequalities like x < q or q < x (for q ∈ ℚ),
• The predicate Q(x) or its negation.

Since ψ(x) implies ¬Q(x), we can rewrite ψ(x) as ¬Q(x) ∧ χ(x), where χ(x) is a Boolean combination of those rational inequalities.

Here's the critical point: χ(x) defines a finite union of intervals in ℝ. Each of these intervals has rational endpoints (or extends to ±∞). Take any one of these intervals I that contains an irrational (which it must, since ψ(x) is consistent).

Now, pick two distinct irrationals a and b inside I. Both a and b satisfy ψ(x): they're irrational, and they're in I so they satisfy χ(x). But their types over ℚ are different! Because ℚ is dense in ℝ, there's some rational q where a < q < b. That means the type of a includes x < q, while the type of b includes q < x.

But wait—ψ(x) was supposed to isolate a unique type p(x). If both a and b satisfy ψ(x) but have different types, that's a contradiction. Therefore, our initial assumption (that such an isolated type p exists) is wrong.

Final Conclusion #

Since φ(x) is consistent with T, but no isolated type in S₁(ℚ) includes φ(x), the isolated types in S₁(ℚ) are not dense. By the prime model existence criterion, this means T has no prime model over ℚ.

内容的提问来源于stack exchange，提问作者R. Harlow