LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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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.

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Under the ultrafilter lemma, an ultrafilter algebra maps each ultrafilter to its unique limit

Statement

Assume UL/BPI. For an ultrafilter algebra $ξ : β X \to X$ with its induced topology, every ultrafilter $U$ converges to exactly one point, namely $ξ (U)$ .

Facts & Assumptions

Given: UL/BPI, an ultrafilter algebra $ξ : β X \to X$ , and an ultrafilter $U$ on $X$ .

[L1]

In the induced topology, the closure of $A$ is $ξ [\hat{A}]$ (Under the ultrafilter lemma, closure in an ultrafilter-algebra topology is the image of ultrafilters containing the set).

[L2]

A filter converges to $p$ when every neighbourhood of $p$ belongs to the filter (Convergence and cluster points of a filter on a topological space).

[L3]

The ultrafilter extension principle says that every filter on a set is contained in an ultrafilter on that set (The ultrafilter extension principle (UL/BPI)).

Proof

technique · direct

1.1givenL2

If an induced-open neighbourhood $O$ contains $ξ (U)$ , the definition of induced-open gives $O \in U$ . Thus $U$ converges to $ξ (U)$ by [L2].

1.2L1L2

Let $x$ be any limit of $U$ . For each $A \in U$ , every neighbourhood of $x$ meets $A$ , so $x \in \overline{A} = ξ [\hat{A}]$ by [L1].

2.1step 1.2L3choose

On $β X$ , the family ${\hat{A} : A \in U} \cup {ξ^{- 1} [{x}]}$ has the finite-intersection property by step 1.2. Extend it by [L3] to an ultrafilter $W$ on $β X$ .

3.1step 2.1algebra

The inclusions forced by step 2.1 and maximality give $μ_{X} (W) = U$ and $β (ξ) (W) = η_{X} (x)$ . The algebra laws therefore give $ξ (U) = ξ μ_{X} (W) = ξ β (ξ) (W) = ξ η_{X} (x) = x$ .

4.1step 1.1step 3.1∎

Step 1.1 supplies the limit $ξ (U)$ and step 3.1 identifies every other limit with it, proving existence and uniqueness. If $X = \emptyset$ , no ultrafilter exists and the assertion is vacuous.

Depends on

Used by

Dependency tree · two levels

18 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

J. Goubault-Larrecq, Algebras of filter-related monads: I. Ultrafilters and Manes' theorem, Lemmas B and C (standard reference, not scraped)