CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck 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.

The intersection of the two principal ultrafilters on a two-point set is a filter but not an ultrafilter

Statement refuted

The intersection of two ultrafilters on the same set is again an ultrafilter.

On $X = {0, 1}$ , the intersection of the principal ultrafilters at $0$ and $1$ is the one-member filter ${X}$ , which is not an ultrafilter.

Facts & Assumptions

Given: The set $X = {0, 1}$ and the principal ultrafilters $U_{0} = {A \subseteq X : 0 \in A}$ and $U_{1} = {A \subseteq X : 1 \in A}$ .

[L1]

The subsets of $X$ containing $x$ form the principal ultrafilter at $x$ (The subsets of $X$ containing a fixed point $x$ form the principal ultrafilter at $x$ ).

[F1]

A filter contains $X$ , omits $\emptyset$ , and is closed under pairwise intersection and upward inclusion in $X$ (Filter on a set).

[L2]

A filter is an ultrafilter exactly when it contains one member of every complementary pair; the two alternatives are always exclusive (Characterisation of ultrafilters: every set or its complement).

Counterexample

technique · direct

1.1

By [L1], $U_{0}$ and $U_{1}$ are ultrafilters on $X$ .

givenL1

1.2

A subset $A \subseteq X$ lies in $U_{0} \cap U_{1}$ exactly when it contains both $0$ and $1$ , which on this two-point set holds exactly when $A = X$ . Thus $U_{0} \cap U_{1} = {X}$ .

given

1.3

The family ${X}$ is a filter: it contains $X$ , omits $\emptyset$ , its only pairwise intersection is $X$ , and its only superset inside $X$ is $X$ .

2.1

Neither ${0}$ nor its complement ${1}$ belongs to ${X}$ , so [L2] shows that this filter is not an ultrafilter.

step 1.3L2

3.1

Hence $U_{0} \cap U_{1}$ is a filter but not an ultrafilter, refuting the claim.

step 1.1step 1.2step 1.3step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

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

Ultrafilter (set theory) (Wikipedia) (standard reference, not scraped)
Ultrafilter (Wikipedia) (standard reference, not scraped)