ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-07-31

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An ultrafilter extending the Fréchet filter on $N$ is free, and its existence uses the ultrafilter lemma

Example

There exists an ultrafilter $U$ on $N$ containing the Fréchet filter $F_{F r}$ , and every such extension is free. Its existence here is obtained from the ultrafilter lemma. That lemma is proved from Zorn's lemma and therefore from the Axiom of Choice; it does not construct or distinguish the resulting ultrafilter.

Facts & Assumptions

Given: The Fréchet filter $F_{F r}$ on $N$ .

[L1]

The Fréchet filter is a proper filter, and $N ∖ {n} \in F_{F r}$ for every $n \in N$ because this complement contains a tail (The subsets of $N$ containing a tail form the Fréchet filter, and it is proper and not an ultrafilter).

[L2]

Every filter on a set is contained in an ultrafilter. The proof uses Zorn's lemma and hence the Axiom of Choice (The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter).

[F1]

An ultrafilter is principal if it has the form ${A \subseteq X : x \in A}$ for some $x \in X$ , and it is free otherwise (Ultrafilter).

[L3]

An ultrafilter contains exactly one of $A$ and $X ∖ A$ for every $A \subseteq X$ (Characterisation of ultrafilters: every set or its complement).

Verification

technique · constructive

1.1

Let $U$ be an arbitrary ultrafilter on $N$ with $F_{F r} \subseteq U$ ; at least one such $U$ exists by applying [L2] to $F_{F r}$ .

L1L2construct

2.1

For every $n \in N$ , the set $N ∖ {n}$ lies in $U$ .

step 1.1L1

3.1

By [L3], step 2.1 forces ${n} \notin U$ for every $n \in N$ .

step 2.1L3

4.1

A principal ultrafilter at $n$ contains ${n}$ , so step 3.1 shows that $U$ is not principal and hence is free.

step 3.1F1

5.1

Since $U$ was arbitrary, every ultrafilter extending the Fréchet filter is free, and step 1.1 supplies one by the ultrafilter lemma, with the choice cost stated in [L2].

step 1.1step 4.1L2discharge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 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

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

An ultrafilter extending the Fréchet filter on NN is free, and its existence uses the ultrafilter lemma