LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-17

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 principal-ultrafilter and ultrafilter-flattening formulas are well-defined and natural

Statement

Write $β X$ for the set of ultrafilters on $X$ and $\hat{A} = {U \in β X : A \in U}$ for $A \subseteq X$ . The formulas

$η_{X} (x) = {A \subseteq X : x \in A}, μ_{X} (W) = {A \subseteq X : \hat{A} \in W}$

define natural transformations $η : 1_{S e t} \Rightarrow β$ and $μ : β^{2} \Rightarrow β$ .

Facts & Assumptions

Given: A set $X$ , a point $x \in X$ , and an ultrafilter $W$ on $β X$ .

[L1]

Pushforward makes $X \mapsto β X$ functorial (Pushforward sends ultrafilters to ultrafilters and is functorial).

[L2]

The complement-decision property characterises ultrafilters (Characterisation of ultrafilters: every set or its complement).

[L3]

In an ultrafilter, a finite union belongs exactly when one of its members belongs (Ultrafilters are prime: a union in $U$ has a member in $U$ ).

Proof

technique · direct

1.1L1L2

The subsets containing $x$ form a proper filter and decide every $A \subseteq X$ according as $x \in A$ or $x \in X ∖ A$ , so $η_{X} (x)$ is an ultrafilter by [L2]. For $f : X \to Y$ , the equivalence $B \in f_{*} η_{X} (x) ⟺ x \in f^{- 1} [B] ⟺ f (x) \in B$ proves naturality of $η$ .

1.2L2L3

The identities $\hat{X} = β X$ , $\hat{\emptyset} = \emptyset$ , $\hat{A \cap B} = \hat{A} \cap \hat{B}$ , and $A \subseteq B \Rightarrow \hat{A} \subseteq \hat{B}$ give the filter axioms for $μ_{X} (W)$ . Inner complement decision gives $\hat{X ∖ A} = β X ∖ \hat{A}$ ; their union belongs to the outer ultrafilter, so [L3] puts one of them in it. By [L2] the resulting filter is an ultrafilter.

2.1L1step 1.2∎

For $f : X \to Y$ and $B \subseteq Y$ , expanding definitions gives $B \in f_{*} μ_{X} (W)$ exactly when $\hat{f^{- 1} [B]} \in W$ . This is equivalent to $\hat{B} \in (β f)_{*} W$ , because $(β f)^{- 1} [\hat{B}] = \hat{f^{- 1} [B]}$ by [L1]. Hence $f_{*} μ_{X} = μ_{Y} β (f_{*})$ .

Depends on

Used by

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

E. Riehl, Category Theory in Context, 2nd ed., Exercise 5.1.ii (standard reference, not scraped)