Questions: Vaught's Theorem on Number of Countable Models

Question 1

A logician claims to have constructed a complete first-order theory with exactly 2 non-isomorphic countable models. According to Vaught's theorem, this claim is:

• A. Impossible — Vaught's theorem rules out exactly 2 non-isomorphic countable models for any complete theory
• B. Possible — Vaught's theorem only constrains the number of models at uncountable cardinals
• C. Possible — if the theory is categorical at some uncountable cardinal, the countable count can be 2
• D. Impossible — complete theories always have exactly 1 countable model

Answer: (A)

Vaught's theorem states the number of non-isomorphic countable models of a complete theory is either 1 (ℵ₀-categorical) or ≥ ℵ₀. Exactly 2 is specifically ruled out. The theorem says nothing special about uncountable cardinals here — it is a constraint purely on the countable spectrum. Option D is also wrong: many complete theories have infinitely many non-isomorphic countable models.

Question 2

Which of the following counts of non-isomorphic countable models is consistent with Vaught's theorem for a complete theory?

• A. Exactly 3
• B. Exactly 7
• C. Exactly ℵ₀ (countably infinitely many)
• D. Exactly 2

Answer: (C)

Vaught's theorem says the count is either 1 or ≥ ℵ₀. Any finite number ≥ 2 is ruled out, including 2, 3, and 7. Exactly ℵ₀ satisfies ≥ ℵ₀, so it is allowed. The continuum 2^ℵ₀ is also allowed. Vaught's conjecture (still open) asks whether the count must be either ≤ ℵ₀ or exactly 2^ℵ₀ — but the theorem itself only rules out finite numbers ≥ 2.

Question 3

Vaught's theorem implies that if a complete theory has more than 1 non-isomorphic countable model, it should have uncountably many.

• T. True
• F. False

Answer: False

Vaught's theorem says the count is either 1 or ≥ ℵ₀ — not necessarily uncountable. A complete theory can have exactly ℵ₀ (countably infinitely many) non-isomorphic countable models. Vaught's conjecture conjectures a stronger dichotomy (either ≤ ℵ₀ or 2^ℵ₀), but that conjecture is unproved in general. The theorem only rules out finite counts between 2 and ℵ₀.

Question 4

The key combinatorial reason Vaught's theorem rules out exactly 2 non-isomorphic countable models is that non-isolated types generate a cascade of further non-isomorphic models that cannot stop at a finite count greater than 1.

• T. True
• F. False

Answer: True

This is the correct intuition. If two non-isomorphic countable models exist, their difference is witnessed by a non-isolated type — one realized in one model but not the other. The Omitting Types Theorem and the proliferation of partial types extending in incompatible directions guarantee infinitely many distinct realizations. The cascade cannot 'stop at 2' because each non-isolated type branches into more variants, each realizable in a new non-isomorphic model.

Question 5

Why does Vaught's theorem rule out exactly 2 non-isomorphic countable models? Explain the role of non-isolated types in the argument.

Model answer: If a theory has two non-isomorphic countable models, their difference is witnessed by a non-isolated type — a set of formulas describing element behavior that is not implied by any single formula. The Omitting Types Theorem guarantees that non-isolated types can be omitted or realized independently. Because the type is non-isolated, there are infinitely many incompatible extensions of it, each realizable in a distinct countable model. This cascade cannot halt at exactly 1 additional model — once you have one non-isolated type, you get infinitely many non-isomorphic models, making a count of exactly 2 impossible.

The proof connects the existence of exactly 2 models to the existence of non-isolated types, then shows non-isolated types force infinitely many non-isomorphic models. The interplay between the Omitting Types Theorem and type isolation is the technical core of why 'exactly 2' is forbidden — isolation either forces all types to be realized identically (giving 1 model) or allows infinite independent variation (giving ≥ ℵ₀).