LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)verified 2026-07-26 (claude-opus-5)

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Ultrafilters are prime: a union in $\mathcal{U}$ has a member in $\mathcal{U}$

Statement

Let $\mathcal{U}$ be an ultrafilter on a set $X$ (Ultrafilter) and let $A, B \subseteq X$ . If $A \cup B \in \mathcal{U}$ then $A \in \mathcal{U}$ or $B \in \mathcal{U}$ .

More generally, for every $n \in \mathbb{N}$ and every finite list $s : n \to \mathcal{P}(X)$ (The natural numbers $\mathbb{N}$ (von Neumann)), writing $\bigcup_{i \in n} s(i) = \{\, x \in X : x \in s(i) \text{ for some } i \in n \,\}$ , which is $\emptyset$ when $n = 0$ : if $\bigcup_{i \in n} s(i) \in \mathcal{U}$ then $s(i) \in \mathcal{U}$ for some $i \in n$ .

Facts & Assumptions

Given: A set $X$ , an ultrafilter $\mathcal{U}$ on $X$ , and subsets $A, B \subseteq X$ with $A \cup B \in \mathcal{U}$ .

[A1]

$\mathcal{U}$ is a filter on $X$ : $X \in \mathcal{U}$ , $\emptyset \notin \mathcal{U}$ , $\mathcal{U}$ is closed under pairwise intersection, and closed upward in $X$ (Filter on a set, Ultrafilter).

[L1]

For every $C \subseteq X$ , exactly one of $C \in \mathcal{U}$ and $X \setminus C \in \mathcal{U}$ holds (Characterisation of ultrafilters: every set or its complement).

[L2]

Induction on $\mathbb{N}$ : a property holding at $0$ and passing from $n$ to $\sigma(n)$ holds for every natural number; and $\sigma(n) = n \cup \{n\}$ (The principle of mathematical induction, The natural numbers $\mathbb{N}$ (von Neumann)).

Proof

technique · direct

1.1

Suppose $A \notin \mathcal{U}$ ; then $X \setminus A \in \mathcal{U}$ .

L1assume-hyp

1.2

As sets, $(A \cup B) \cap (X \setminus A) = B \cap (X \setminus A) \subseteq B \subseteq X$ .

given

2.1

Both $A \cup B$ and $X \setminus A$ lie in $\mathcal{U}$ , hence so does their intersection.

step 1.1givenA1

3.1

Upward closure applied to $(A \cup B) \cap (X \setminus A) \subseteq B$ gives $B \in \mathcal{U}$ ; so $A \in \mathcal{U}$ or $B \in \mathcal{U}$ .

step 2.1step 1.2A1

4.1

The finite case follows by induction on $n$ : for $n = 0$ the union is empty and $\emptyset \notin \mathcal{U}$ , so the hypothesis never holds and the claim is vacuous; and if the claim holds for lists of length $n$ , then for $s : \sigma(n) \to \mathcal{P}(X)$ one has $\bigcup_{i \in \sigma(n)} s(i) = \left(\bigcup_{i \in n} s(i)\right) \cup s(n)$ , so the binary case puts either $s(n)$ or $\bigcup_{i \in n} s(i)$ in $\mathcal{U}$ , and in the second alternative the claim for $n$ supplies an $i \in n$ with $s(i) \in \mathcal{U}$ .

step 3.1A1L2

5.1

So an ultrafilter is prime: it contains a member of every finite union it contains.

step 3.1step 4.1∎

Remarks

The converse holds too, and is worth stating even though it is not needed below: a filter $\mathcal{U}$ with the property that $A \cup B \in \mathcal{U}$ implies $A \in \mathcal{U}$ or $B \in \mathcal{U}$ is an ultrafilter, because $A \cup (X \setminus A) = X \in \mathcal{U}$ forces one of $A$ and $X \setminus A$ into $\mathcal{U}$ , which is Characterisation of ultrafilters: every set or its complement. Primeness and maximality are therefore the same condition, which is why the ultrafilter lemma is a form of the Boolean prime ideal theorem (What the ultrafilter lemma costs: a choice principle strictly weaker than AC).
The finite case does not extend to infinite unions. On $\mathbb{N}$ an ultrafilter containing every tail contains $\mathbb{N} = \bigcup_{x \in \mathbb{N}} \{x\}$ but no singleton (FALSE, once the ultrafilter lemma is available: every ultrafilter is principal), so a union of infinitely many sets can lie in $\mathcal{U}$ with no single member of the union doing so.
Read through the two-valued measure of Characterisation of ultrafilters: every set or its complement, primeness says that a union of finitely many null sets is null, the complementary form of closure under finite intersection.

Depends on

Used by

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Direct dependencies and their dependencies through the next three levels: 29 results over 10 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)
Boolean prime ideal theorem (Wikipedia) (standard reference, not scraped)
Ultrafilter (Wikipedia) (standard reference, not scraped)

Ultrafilters are prime: a union in U\mathcal{U}U has a member in U\mathcal{U}U