ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Assuming the ultrafilter lemma, a free ultrafilter on $N$ converges to the added point in the one-point convergent-sequence space

Example

Assume the ultrafilter lemma. Let $X = N \cup {\infty}$ , make every natural isolated, and give $\infty$ the neighbourhood base $U_{N} = {\infty} \cup {n : n \geq N}$ . A free ultrafilter on $N$ , extended along the inclusion $N ↪ X$ , converges to $\infty$ .

Facts & Assumptions

Given: The identity net $n \mapsto n$ on the directed natural numbers.

[L1]

Its tail filter contains every tail $T_{N} = {n : n \geq N}$ (The tail filter of a net).

[L2]

The ultrafilter lemma extends that filter to an ultrafilter (The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter).

[L3]

A filter contains its whole set, omits the empty set, and is closed under intersections and supersets (Filter on a set).

[L4]

A filter converges to a point exactly when it contains every neighbourhood of that point (Convergence and cluster points of a filter on a topological space).

[L5]

A filter is an ultrafilter exactly when for every subset it contains that subset or its complement (Ultrafilter, Characterisation of ultrafilters: every set or its complement).

Verification

technique · direct

1.1

Choose an ultrafilter $U$ extending the tail filter. It contains every $T_{N}$ and contains no singleton, since ${k} \cap T_{k + 1} = \emptyset$ ; thus it is free.

L1L2

2.1

Put $U^{X} = {B \subseteq X : B \cap N \in U}$ . The filter axioms transfer through intersection with $N$ , so this is a filter on $X$ . For every $B \subseteq X$ , [L5] applied to $B \cap N$ shows that $U^{X}$ contains $B$ or $X ∖ B$ ; hence $U^{X}$ is an ultrafilter.

step 1.1L3L5

3.1

Every basic neighbourhood $U_{N}$ has $U_{N} \cap N = T_{N} \in U$ , hence $U_{N} \in U^{X}$ . Every neighbourhood of $\infty$ contains some $U_{N}$ , so upward closure gives $U^{X} \to \infty$ .

step 2.1L4

4.1

It is free: if ${x} \in U^{X}$ , then either $x = \infty$ and its intersection with $N$ is empty, or $x \in N$ and ${x} \in U$ , both impossible. Thus this supplies the claimed free ultrafilter and its convergence.

step 1.1step 2.1step 3.1L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 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 (set theory) (Wikipedia) (standard reference, not scraped)

Assuming the ultrafilter lemma, a free ultrafilter on NN converges to the added point in the one-point convergent-sequence space