A non-principal ultrafilter U is fixed on the index set ℕ. For each n ∈ ℕ, let Mₙ be the cyclic group ℤ/nℤ. Which first-order sentence is true in the ultraproduct ∏ᵤ Mₙ?
AEvery element has finite order, since each Mₙ is finite
BThe ultraproduct is the zero ring, because most Mₙ are small
CThe sentence 'there exists an element of order > k' holds for every standard k, because for each k the set {n : Mₙ has an element of order > k} is cofinite, hence in U
DNo first-order sentence is decidable in the ultraproduct without knowing U explicitly
By Łoś's Theorem, φ holds in ∏ᵤ Mₙ iff {n : Mₙ ⊨ φ} ∈ U. For any fixed k, all groups ℤ/nℤ with n > k contain an element of order > k, so the set {n > k} is cofinite. Because U is non-principal, every cofinite set is in U, so the ultraproduct contains elements of order exceeding every standard integer. Option A applies finite reasoning to an infinite construction — Łoś shows the ultraproduct can satisfy properties no individual component satisfies.
Question 2 Multiple Choice
You want to use ultraproducts to prove the Compactness Theorem: if every finite subset of a theory Σ is satisfiable, then Σ is satisfiable. You choose models Mₙ where Mₙ satisfies the first n sentences of Σ, then form ∏ᵤ Mₙ for a non-principal ultrafilter U. Why does each sentence φₖ of Σ hold in the ultraproduct?
ABecause every Mₙ satisfies φₖ, so the whole index set is in U
BBecause Mₙ ⊨ φₖ for all n ≥ k, making the set {n : Mₙ ⊨ φₖ} cofinite, hence in U
CBecause the ultraproduct takes a logical average of all models
DBecause φₖ is finitely satisfiable, and ultraproducts preserve finite properties
By construction, Mₙ ⊨ φₖ for every n ≥ k (since Mₙ satisfies the first n sentences). The set {n : n ≥ k} is cofinite. A non-principal ultrafilter contains every cofinite set, so {n : Mₙ ⊨ φₖ} ∈ U. By Łoś's Theorem, ∏ᵤ Mₙ ⊨ φₖ. Option A would be correct if every Mₙ satisfied φₖ, but Mₙ with n < k may not — only cofiniteness is needed, not universality.
Question 3 True / False
The ultraproduct ∏ᵤ Mᵢ satisfies a first-order sentence φ if and mainly if more than half the component structures satisfy φ.
TTrue
FFalse
Answer: False
The correct criterion is not majority vote but U-membership: φ holds in the ultraproduct iff {i : Mᵢ ⊨ φ} ∈ U. An ultrafilter is not a counting measure — it does not depend on proportions. A non-principal ultrafilter on ℕ can contain sets with density zero (if every cofinite superset of a set is in U) and can exclude sets with density one if those sets are not in the filter. The voting metaphor is useful but the rule is 'which sets the ultrafilter declares large,' not 'which option gets a numerical majority.'
Question 4 True / False
If Mᵢ ⊨ φ for every i ∈ I, then ∏ᵤ Mᵢ ⊨ φ for any ultrafilter U on I.
TTrue
FFalse
Answer: True
If φ holds in every component, then {i : Mᵢ ⊨ φ} = I. Every ultrafilter must contain the whole index set I (it is closed upward and contains the empty set's complement). So by Łoś's Theorem, ∏ᵤ Mᵢ ⊨ φ. This is the simplest case of Łoś and explains why ultraproducts of models of a complete theory are again models of that theory.
Question 5 Short Answer
Explain why the ultrafilter's maximality — the fact that for every set A ⊆ I, either A ∈ U or its complement Aᶜ ∈ U — is essential for Łoś's Theorem to work.
Think about your answer, then reveal below.
Model answer: Łoś's Theorem must assign a definite truth value to every first-order sentence in the ultraproduct. For an atomic formula, the ultraproduct satisfies it or it doesn't; there is no middle ground. Maximality ensures that for any index set S = {i : Mᵢ ⊨ φ}, exactly one of S or Sᶜ is in U. If U were merely a filter (not an ultrafilter), some sentences could have S ∉ U and Sᶜ ∉ U simultaneously, leaving the ultraproduct's truth value undefined. Maximality turns the ultrafilter into a complete decision procedure: it takes a definite stance on every subset, which translates into a definite truth value for every first-order sentence.
This is the deepest structural reason ultrafilters (not just filters) are needed. A filter guarantees certain sets are 'large' but may leave many sets undecided. Adding maximality collapses this to a total two-valued measure: every set is either large (in U) or small (complement in U). Łoś's proof uses this at each step of the induction on formula complexity, especially for negations: ¬φ holds in the ultraproduct iff φ does not, which requires the index set for φ and for ¬φ to be complementary with exactly one in U.