CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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.

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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 $\mathcal U_0=\{A\subseteq X:0\in A\}$ and $\mathcal U_1=\{A\subseteq X:1\in A\}$ .

[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], $\mathcal U_0$ and $\mathcal U_1$ are ultrafilters on $X$ .

givenL1

1.2

A subset $A\subseteq X$ lies in $\mathcal U_0\cap\mathcal U_1$ exactly when it contains both $0$ and $1$ , which on this two-point set holds exactly when $A=X$ . Thus $\mathcal U_0\cap\mathcal U_1=\{X\}$ .

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 $\mathcal U_0\cap\mathcal U_1$ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 24 results over 12 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

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