False statementConstruction: AI-adaptedVerification: AI-generatedverified 2026-09-26 (gpt-6-sol)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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FALSE, once the ultrafilter lemma is available: every ultrafilter is principal

Statement

FALSE. For every set $X$ , every ultrafilter $U$ on $X$ (Ultrafilter) is principal: there is an $x \in X$ with $U = {A \subseteq X : x \in A}$ .

What refutes the claim, and what that costs. The refutation below assumes the ultrafilter lemma, which this library proves from the Axiom of Choice (The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter).

Facts & Assumptions

Given: The natural numbers $N$ with $0$ , the successor $σ$ and the order $\leq$ (The natural numbers $N$ (von Neumann), Order on the natural numbers), and the Axiom of Choice in the form used by the ultrafilter lemma.

[A1]

A filter on $X$ contains $X$ , omits $\emptyset$ , and is closed under pairwise intersection and upward in $X$ ; a filter base is a nonempty, downward directed family of nonempty subsets (Filter on a set, Filter base and the filter it generates). The principal filter at $x$ is ${A \subseteq X : x \in A}$ , and it contains ${x}$ .

[L1]

The upward closure $⟨ B ⟩$ of a filter base $B$ on $X$ is a filter on $X$ , and $B \subseteq ⟨ B ⟩$ (The upward closure of a filter base is the smallest filter containing it).

[L2]

Every filter on a set is contained in an ultrafilter on that set, and an ultrafilter is in particular a filter (The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter, Ultrafilter).

[L3]

$\leq$ on $N$ is reflexive, transitive, antisymmetric and total ( $\leq$ is a linear order on $N$ ).

[L4]

$m < n ⟺ σ (m) \leq n$ , and $m < n$ means $m \leq n$ with $m \neq n$ (Discreteness: $σ (n)$ is the immediate successor, Order on the natural numbers).

Refutation

technique · direct

1.1construct

For $N \in N$ put $T_{N} = {n \in N : N \leq n}$ , the tail at $N$ , and let $B = {T_{N} : N \in N}$ .

2.1step 1.1L3

$B \neq \emptyset$ because $T_{0} \in B$ , and $\emptyset \notin B$ because $N \in T_{N}$ by reflexivity.

2.2step 1.1L3

$B$ is downward directed: given $M, N$ , totality gives say $M \leq N$ , and then transitivity gives $T_{N} \subseteq T_{M}$ , so the member $T_{N}$ of $B$ satisfies $T_{N} \subseteq T_{M} \cap T_{N}$ .

2.3step 1.1L4

For every $x \in N$ , ${x} \cap T_{σ (x)} = \emptyset$ : an element of the intersection equals $x$ and satisfies $σ (x) \leq x$ , which would give $x < x$ and hence $x \neq x$ .

3.1step 2.1step 2.2A1L1

$B$ is a filter base on $N$ , so $F = ⟨ B ⟩$ is a filter on $N$ and every tail $T_{N}$ belongs to it.

4.1step 3.1L2

By the ultrafilter lemma there is an ultrafilter $U$ on $N$ with $F \subseteq U$ ; in particular $T_{N} \in U$ for every $N \in N$ .

5.1step 4.1step 2.3A1L2

No singleton lies in $U$ : if ${x} \in U$ then ${x}$ and $T_{σ (x)}$ both lie in $U$ , hence so does their intersection $\emptyset$ , which a filter omits.

6.1step 5.1A1∎

The principal filter at $x$ contains ${x}$ , so $U$ is not the principal filter at any $x \in N$ ; thus $U$ is an ultrafilter on $N$ that is not principal, and the claim is refuted.

Remarks

What the refutation consumes. The ultrafilter $U$ is produced by The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter, which this library proves from the Axiom of Choice. The tail-filter construction above establishes the non-principal conclusion under exactly that theorem's hypotheses.
The filter used is the Fréchet filter in disguise. The standard witness is the filter of cofinite subsets of $N$ . A subset of $N$ is cofinite exactly when it contains a tail, so the filter generated by the tails and the filter of cofinite sets coincide; tails are used here because they need only the order on $N$ , whereas "cofinite" needs a theory of finite sets that this library has not yet developed. Nothing else changes.
On a finite $X$ the claim is true, which is why the intuition survives: a finite $X$ is a finite union of its singletons, primeness (Ultrafilters are prime: a union in $U$ has a member in $U$ ) puts one singleton ${x}$ into $U$ , and upward closure then makes $U$ the principal filter at $x$ . Stating that argument in full needs the notion of a finite set, so it is recorded here as motivation rather than as a proved item.

FALSE, once the ultrafilter lemma is available: every ultrafilter is principal

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources