ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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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.

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

$\mathcal F_C:=\{A\subseteq X:C\subseteq A\}.$

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

Facts & Assumptions

Given: A set $X$ , a nonempty subset $C\subseteq X$ , and the family $\mathcal F_C$ displayed above.

[F1]

A filter on $X$ is a family $\mathcal F\subseteq\mathcal P(X)$ 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 $\mathcal B\subseteq\mathcal P(X)$ that omits $\emptyset$ and is downward directed: for $B_1,B_2\in\mathcal B$ there is $B_3\in\mathcal B$ with $B_3\subseteq B_1\cap B_2$ (Filter base and the filter it generates).

[L1]

The upward closure $\langle\mathcal B\rangle=\{A\subseteq X:B\subseteq A\text{ for some }B\in\mathcal B\}$ 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 $\mathcal U$ on $X$ is an ultrafilter exactly when, for every $A\subseteq X$ , exactly one of $A$ and $X\setminus A$ belongs to $\mathcal U$ (Characterisation of ultrafilters: every set or its complement).

Verification

technique · direct

1.1

Every member of $\mathcal F_C$ is a subset of $X$ , and $X\in\mathcal F_C$ because $C\subseteq X$ .

given

1.2

The empty set is not in $\mathcal F_C$ , since $C\subseteq\emptyset$ would contradict $C\neq\emptyset$ .

given

1.3

If $A,B\in\mathcal F_C$ , then $C\subseteq A\cap B$ , so $A\cap B\in\mathcal F_C$ .

given

1.4

If $A\in\mathcal F_C$ and $A\subseteq B\subseteq X$ , then $C\subseteq B$ , so $B\in\mathcal F_C$ .

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 $\langle\{C\}\rangle=\{A\subseteq X:C\subseteq A\}=\mathcal F_C$ .

given

1.7

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

given

1.8

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

givenchoose

2.1

Steps 1.1 through 1.4 verify the four axioms in [F1], so $\mathcal F_C$ 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], $\mathcal F_C$ is the filter generated by $\{C\}$ .

step 1.5step 1.6L1

3.1

If $C=\{x\}$ , step 1.7 says that $\mathcal F_C$ 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 $\mathcal F_C$ , so [L2] says that $\mathcal F_C$ is not an ultrafilter.

step 1.8step 2.1L2

4.1

Thus $\mathcal F_C$ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 22 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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\subseteq XC⊆X, the supersets of CCC form the filter generated by {C}\{C\}{C}, and this filter is an ultrafilter exactly when CCC is a singleton

Example

Facts & Assumptions

Verification

Depends on

Used by

Dependency tree · next 3 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