Questions: Spectrum of a Theory and Vaught's Conjecture

Question 1

A complete countable theory T has infinitely many non-isomorphic countable models. What does Vaught's conjecture assert about I(ℵ₀, T)?

• A. I(ℵ₀, T) must equal ℵ₁, the first uncountable cardinal
• B. I(ℵ₀, T) cannot equal ℵ₁ — it must be either at most ℵ₀ or exactly 2^ℵ₀
• C. I(ℵ₀, T) can be any cardinal between ℵ₀ and 2^ℵ₀
• D. I(ℵ₀, T) must equal 2^ℵ₀ once it exceeds ℵ₀

Answer: (B)

Vaught's conjecture (1961, still open) asserts that for a complete countable theory, the number of countable models up to isomorphism cannot be exactly ℵ₁. It must be either countable (≤ ℵ₀) or maximal (2^ℵ₀). The conjecture is striking precisely because ℵ₁ is the 'obvious' intermediate value between ℵ₀ and 2^ℵ₀, and it is this natural candidate that is claimed to be forbidden.

Question 2

For uncountable cardinals κ, which values can I(κ, T) take for a complete first-order theory T?

• A. Any cardinal from 0 up to 2^κ, depending on the complexity of T
• B. Only 0, 1, or 2^κ — no intermediate values are possible
• C. Only ℵ₀ or 2^κ, since uncountable models are either few or many
• D. Any infinite cardinal ≤ 2^κ, but never 0

Answer: (B)

At uncountable cardinals, the spectrum is severely constrained: I(κ, T) can only be 0 (no model of that size), 1 (categoricity — exactly one model up to isomorphism), or 2^κ (the maximum). There is no middle ground. This 'all or nothing' structure at uncountable cardinalities has no analogue at ℵ₀, where the countable spectrum is far richer and more mysterious.

Question 3

Morley's theorem implies that if a complete countable theory is categorical in some uncountable cardinal, it is categorical in all uncountable cardinals.

• T. True
• F. False

Answer: True

Morley's theorem (1965) is one of the landmark results in model theory and states exactly this. Uncountable categoricity is a robust global property: it cannot hold at some uncountable cardinals and fail at others. This is the 'all or nothing' behavior at uncountable cardinals — a stark contrast to the countable case, where the spectrum can vary in complex ways.

Question 4

A stable theory can have spectrum I(κ, T) = 2^κ at many uncountable cardinals, just like an unstable theory.

• T. True
• F. False

Answer: False

Stable theories have tightly controlled spectra. At uncountable cardinals, their model counts are bounded by a polynomial in the cofinality of the cardinal — far below the maximum 2^κ. Unstable theories, by contrast, can achieve I(κ, T) = 2^κ at many cardinals. Stability, which reflects deep combinatorial properties (order properties, type definability), is precisely what prevents the spectrum from exploding to its maximum value.

Question 5

Why is ℵ₁ considered the 'obvious' intermediate value that Vaught's conjecture claims is forbidden, and why would its existence be surprising from the perspective of model theory?

Model answer: ℵ₁ is the first uncountable cardinal and sits immediately between ℵ₀ and 2^ℵ₀ (assuming the continuum hypothesis fails, 2^ℵ₀ > ℵ₁). A theory with exactly ℵ₁ countable models would be neither 'tame' (finitely many or countably many models) nor 'wild' (as many as possible). Vaught's conjecture asserts this intermediate complexity is forbidden — there is no 'medium' regime for countable model counts. Its existence would suggest a kind of combinatorial structure in the models that doesn't fit either the controlled behavior of ω-stable theories or the maximal complexity of unstable ones.

The conjecture has deep connections to the descriptive set-theoretic complexity of the isomorphism relation on countable models. If I(ℵ₀, T) = ℵ₁, the set of countable models (up to isomorphism) would have a Borel-but-non-analytic character that current model-theoretic tools cannot produce. This is why the conjecture remains open after 60+ years — it requires combining model theory with descriptive set theory in ways that are not yet fully understood.