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.

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

The union of the two principal ultrafilters on a two-point set is not a filter

Statement refuted

The union of any two filters on a set is again a filter.

On $X=\{0,1\}$ , let $\mathcal U_0$ and $\mathcal U_1$ be the principal ultrafilters at $0$ and $1$ . Their union is not a filter.

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]

For every $x\in X$ , the subsets of $X$ containing $x$ form the principal ultrafilter $\mathcal U_x$ (The subsets of $X$ containing a fixed point $x$ form the principal ultrafilter at $x$ ).

[F1]

A filter is closed under pairwise intersection and omits $\emptyset$ (Filter on a set).

[L2]

The union of a nonempty inclusion-chain of filters is a filter; comparability is used to place any two members in one filter before intersecting them (The union of a nonempty chain of filters is a filter).

Counterexample

technique · direct

1.1

By [L1], $\mathcal U_0$ and $\mathcal U_1$ are filters on $X$ .

givenL1

1.2

The singleton $\{0\}$ belongs to $\mathcal U_0$ and the singleton $\{1\}$ belongs to $\mathcal U_1$ , so both belong to $\mathcal U_0\cup\mathcal U_1$ .

given

1.3

The two filters are not comparable: $\{0\}\in\mathcal U_0\setminus\mathcal U_1$ and $\{1\}\in\mathcal U_1\setminus\mathcal U_0$ . Hence [L2] does not apply, and the example shows why its chain hypothesis is essential.

givenL2

2.1

Neither principal ultrafilter contains $\emptyset$ , so $\emptyset\notin\mathcal U_0\cup\mathcal U_1$ .

step 1.1F1

3.1

If $\mathcal U_0\cup\mathcal U_1$ were a filter, intersection closure applied to $\{0\}$ and $\{1\}$ would put $\emptyset=\{0\}\cap\{1\}$ in the union, contradicting step 2.1. Thus the union is not a filter.

step 1.2step 2.1F1

4.1

Therefore an arbitrary union of two filters, even two principal ultrafilters, need not be a filter.

step 1.1step 3.1∎

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: 23 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

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