ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-07-31

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For every nonempty $C \subseteq X$ , the supersets of $C$ form the filter generated by ${C}$ , and this filter is an ultrafilter exactly when $C$ is a singleton

Example

Let $X$ be a set and let $\emptyset \neq C \subseteq X$ . Define

$F_{C} : = {A \subseteq X : C \subseteq A} .$

Then $F_{C}$ is the filter generated by the one-member filter base ${C}$ . Moreover, $F_{C}$ is an ultrafilter exactly when $C$ is a singleton.

Facts & Assumptions

Given: A set $X$ , a nonempty subset $C \subseteq X$ , and the family $F_{C}$ displayed above.

[F1]

A filter on $X$ is a family $F \subseteq P (X)$ containing $X$ , omitting $\emptyset$ , closed under pairwise intersection, and closed upward in $X$ (Filter on a set).

[F2]

A filter base on $X$ is a nonempty family $B \subseteq P (X)$ that omits $\emptyset$ and is downward directed: for $B_{1}, B_{2} \in B$ there is $B_{3} \in B$ with $B_{3} \subseteq B_{1} \cap B_{2}$ (Filter base and the filter it generates).

[L1]

The upward closure $⟨ B ⟩ = {A \subseteq X : B \subseteq A for some B \in B}$ of a filter base is the smallest filter containing it (The upward closure of a filter base is the smallest filter containing it).

[L2]

A filter $U$ on $X$ is an ultrafilter exactly when, for every $A \subseteq X$ , exactly one of $A$ and $X ∖ A$ belongs to $U$ (Characterisation of ultrafilters: every set or its complement).

Verification

technique · direct

1.1

Every member of $F_{C}$ is a subset of $X$ , and $X \in F_{C}$ because $C \subseteq X$ .

given

1.2

The empty set is not in $F_{C}$ , since $C \subseteq \emptyset$ would contradict $C \neq \emptyset$ .

given

1.3

If $A, B \in F_{C}$ , then $C \subseteq A \cap B$ , so $A \cap B \in F_{C}$ .

given

1.4

If $A \in F_{C}$ and $A \subseteq B \subseteq X$ , then $C \subseteq B$ , so $B \in F_{C}$ .

given

1.5

The family ${C}$ is a filter base: it is nonempty, it omits $\emptyset$ because $C \neq \emptyset$ , and its only pair of members has $C \subseteq C \cap C$ .

givenF2

1.6

Its upward closure is $⟨ {C} ⟩ = {A \subseteq X : C \subseteq A} = F_{C}$ .

given

1.7

If $C = {x}$ and $A \subseteq X$ , then either $x \in A$ , giving $C \subseteq A$ , or $x \notin A$ , giving $C \subseteq X ∖ A$ .

given

1.8

If $C$ is not a singleton, choose $x \in C$ and $y \in C ∖ {x}$ . Then $C ⊈ {x}$ because $y \notin {x}$ , and $C ⊈ X ∖ {x}$ because $x \in C$ .

givenchoose

2.1

Steps 1.1 through 1.4 verify the four axioms in [F1], so $F_{C}$ is a filter on $X$ .

step 1.1step 1.2step 1.3step 1.4F1

2.2

By steps 1.5 and 1.6 and [L1], $F_{C}$ is the filter generated by ${C}$ .

step 1.5step 1.6L1

3.1

If $C = {x}$ , step 1.7 says that $F_{C}$ contains one member of every complementary pair, so [L2] makes it an ultrafilter.

step 1.7step 2.1L2

3.2

If $C$ is not a singleton, step 1.8 says that neither ${x}$ nor its complement belongs to $F_{C}$ , so [L2] says that $F_{C}$ is not an ultrafilter.

step 1.8step 2.1L2

4.1

Thus $F_{C}$ is the filter generated by ${C}$ , and it is an ultrafilter exactly when $C$ is a singleton.

step 2.2step 3.1step 3.2∎

Depends on

Used by

The subsets of X containing a fixed point x form the principal ultrafilter at x Example

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Filter (set theory) (Wikipedia) (standard reference, not scraped)
Ultrafilter (set theory) (Wikipedia) (standard reference, not scraped)
Ultrafilter (Wikipedia) (standard reference, not scraped)
B. Kaya, Ultrafilters and How to Use Them (standard reference, not scraped)

For every nonempty C⊆XC⊆X, the supersets of CC form the filter generated by {C}{C}, and this filter is an ultrafilter exactly when CC is a singleton

Example

Facts & Assumptions

Verification

Depends on

Used by

Dependency tree · two levels

Sources

For every nonempty $C \subseteq X$ , the supersets of $C$ form the filter generated by ${C}$ , and this filter is an ultrafilter exactly when $C$ is a singleton