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TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31
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A net is universal exactly when its tail filter is an ultrafilter, and the canonical net of an ultrafilter is universal

Statement

A net is universal if and only if its tail filter is an ultrafilter. Moreover, the net derived from an ultrafilter is universal.

Facts & Assumptions

Given: A net xx in XX and a filter U\mathcal U on XX.

[A1]

SS belongs to the tail filter of xx exactly when xx is eventually in SS (The tail filter of a net).

[A2]

A filter is an ultrafilter exactly when, for every SXS\subseteq X, it contains SS or XSX\setminus S (Characterisation of ultrafilters: every set or its complement).

[A3]

The derived net of U\mathcal U is indexed by (A,a)(A,a) and later indices have first coordinate contained in AA (The canonical net indexed by the pairs (A,x)(A,x) with AA in a filter and xAx\in A).

Proof

technique · direct
1.1

By [A1], universality of xx says exactly that its tail filter contains SS or XSX\setminus S for every SXS\subseteq X. By [A2], this is exactly ultrafilterhood.

A1A2
1.2

If U\mathcal U is an ultrafilter and SXS\subseteq X, [A2] gives SUS\in\mathcal U or XSUX\setminus S\in\mathcal U. In the first case an index (S,a)(S,a) exists and every later value lies in SS by [A3]; the second case is identical.

A2A3
2.1

Thus the derived net of an ultrafilter is universal, completing both assertions.

step 1.1step 1.2

Depends on

Used by

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Dependency tree · next 3 levels

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Sources