ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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Assuming the ultrafilter lemma, a free ultrafilter on $\mathbb{N}$ converges to the added point in the one-point convergent-sequence space

Example

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

Facts & Assumptions

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

[L1]

Its tail filter contains every tail $T_N=\{n:n\ge N\}$ (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 $\mathcal U$ extending the tail filter. It contains every $T_N$ and contains no singleton, since $\{k\}\cap T_{k+1}=\varnothing$ ; thus it is free.

L1L2

2.1

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

step 1.1L3L5

3.1

Every basic neighbourhood $U_N$ has $U_N\cap\mathbb N=T_N\in\mathcal U$ , hence $U_N\in\mathcal U^X$ . Every neighbourhood of $\infty$ contains some $U_N$ , so upward closure gives $\mathcal U^X\to\infty$ .

step 2.1L4

4.1

It is free: if $\{x\}\in\mathcal U^X$ , then either $x=\infty$ and its intersection with $\mathbb N$ is empty, or $x\in\mathbb N$ and $\{x\}\in\mathcal U$ , 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 36 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)

Assuming the ultrafilter lemma, a free ultrafilter on N\mathbb{N}N converges to the added point in the one-point convergent-sequence space