You have a free ultrafilter U on ℕ. Consider the set A = {2, 4, 6, 8, ...} of all even numbers. Which statement must be true?
AA ∈ U, because even numbers are just as numerous as odd numbers and any reasonable notion of largeness must include them
BA ∉ U, because A is not cofinite and free ultrafilters only contain cofinite sets
CExactly one of A or its complement (the odd numbers) belongs to U, but you cannot determine which without knowing more about U
DNeither A nor its complement belongs to U, since neither is cofinite
The defining property of an ultrafilter is maximality: for every subset A, exactly one of A or its complement belongs to the filter. So one of {evens, odds} is in U, but U's non-constructiveness means we cannot say which. Option B is wrong: free ultrafilters contain all cofinite sets but also many non-cofinite sets. Option D contradicts the maximality condition.
Question 2 Multiple Choice
Łoś's theorem guarantees that the ultraproduct ∏Mᵢ/U is always a well-formed model (satisfying every sentence or its negation). Why is this result specifically tied to U being an ultrafilter rather than merely a filter?
AFilters are not closed under finite intersections, so the construction would be incoherent
BAn ultrafilter decides every subset — exactly one of A or its complement is large — ensuring every first-order sentence is either true or false in the ultraproduct with no undecided cases
CUltrafilters are the only filters that contain cofinite sets, which are needed for the quotient to be well-defined
DFilters cannot be used to construct quotient structures; only ultrafilters generate equivalence relations
Łoś's theorem says φ holds in the ultraproduct iff φ holds on a U-large index set. For the ultraproduct to be a model, every sentence needs a definite truth value. A mere filter could have both {i : φ holds} and {i : ¬φ holds} be filter-small (not in the filter), leaving φ undecided. An ultrafilter's totality — exactly one of every set or its complement is large — eliminates this gap, guaranteeing every sentence is decided.
Question 3 True / False
A principal ultrafilter on ℕ is the collection of all subsets of ℕ that contain some fixed element n₀.
TTrue
FFalse
Answer: True
A principal ultrafilter is generated by a single point n₀: a set A is 'large' iff n₀ ∈ A. This is the simplest ultrafilter — it collapses to evaluation at n₀. Free (non-principal) ultrafilters, by contrast, are not generated by any single element; they contain all cofinite sets and exist only by the Axiom of Choice.
Question 4 True / False
A free ultrafilter on ℕ can in principle be explicitly constructed by a sufficiently detailed specification of which subsets it contains.
TTrue
FFalse
Answer: False
Free ultrafilters cannot be explicitly constructed. Their existence is proved via Zorn's lemma (a consequence of the Axiom of Choice) — you show that every filter on an infinite set can be extended to an ultrafilter, but the extension process makes non-canonical choices at every step. No explicit description of a free ultrafilter on ℕ can be written down; they are inherently non-constructive objects.
Question 5 Short Answer
Explain why the maximality condition — that for every subset A, exactly one of A or its complement belongs to the ultrafilter — is essential for making the ultraproduct construction work via Łoś's theorem.
Think about your answer, then reveal below.
Model answer: Łoś's theorem assigns truth values in the ultraproduct by 'voting': sentence φ is true iff φ holds on a U-large index set. For the ultraproduct to be a genuine model, every sentence must receive exactly one truth value. If U were merely a filter, the sets where φ holds and where ¬φ holds could both be U-small (absent from the filter), leaving φ genuinely undecided — the ultraproduct would not be a structure in the classical sense. Maximality guarantees that for every sentence, exactly one of the two index sets is U-large, so the ultraproduct always makes a complete decision, yielding a proper first-order model.
The key is the totality of the verdict: ultrafilters act as decisive voting systems. Every issue (every subset, every sentence) is resolved with no abstentions. This is what makes ultraproducts so powerful — they inherit a coherent first-order theory from the component structures, decided by the ultrafilter.