LemmaStatement: AI-adaptedProof: 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, every net has a universal subnet

Statement

Assume the ultrafilter lemma. Every net has a universal subnet.

Facts & Assumptions

Given: A net $x : D \to X$ and its tail filter $F_{x}$ .

[A1]

$F_{x}$ contains every tail $T_{d}$ , and its members contain a tail (The tail filter of a net).

[L1]

The ultrafilter lemma extends $F_{x}$ to an ultrafilter $U$ (The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter).

[L2]

An ultrafilter contains every subset or its complement (Characterisation of ultrafilters: every set or its complement).

[A2]

A subnet uses an eventually cofinal index map (Subnet via an eventually cofinal index map).

[A3]

A universal net is eventually in every set or its complement (Universal net: eventually in every subset or eventually in its complement).

Proof

technique · constructive

1.1

Choose an ultrafilter $U \supseteq F_{x}$ by [L1]. Let $E = {(d, A) : d \in D, A \in U, x_{d} \in A}$ , ordered by $(d, A) ⪯ (e, B)$ when $d \leq e$ and $B \subseteq A$ , and put $y_{(d, A)} = x_{d}$ .

L1construct

2.1

The set $E$ is directed. Given $(d, A), (e, B)$ , choose $h \geq d, e$ . Since $A \cap B$ and the tail $T_{h}$ belong to $U$ , their intersection is nonempty; choose an index $k \geq h$ with $x_{k} \in A \cap B$ . Then $(k, A \cap B)$ is above both pairs.

step 1.1A1choose

2.2

The map $ϕ (d, A) = d$ is eventually cofinal: $(d_{0}, X)$ is an index for every $d_{0}$ , and every later index has first coordinate at least $d_{0}$ . Thus $y$ is a subnet of $x$ .

step 1.1A2

2.3

For $S \subseteq X$ , [L2] gives $S \in U$ or $X ∖ S \in U$ . In the first case choose any $d_{0} \in D$ . Since $S \cap T_{d_{0}} \in U$ , choose $j \geq d_{0}$ with $x_{j} \in S$ . Then $(j, S) \in E$ , and every later value lies in $S$ . The complementary case is identical. Thus $y$ is universal.

step 1.1A1A3L2choose

3.1

The constructed $y$ is a universal subnet of $x$ .

step 2.2step 2.3discharge-construct∎

Depends on

Used by

Assuming the ultrafilter lemma, a space is compact if and only if every universal net converges Corollary

Dependency tree · two levels

19 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

WVU Math 581 Topology I (standard reference, not scraped)
Net (mathematics) (Wikipedia) (standard reference, not scraped)
Boolean prime ideal theorem (Wikipedia) (standard reference, not scraped)