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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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For a finite-sheeted covering, the total space is compact exactly when the base is compact

Statement

If p:EB is a finite-sheeted covering, then E is compact if and only if B is compact.

Facts & Assumptions

Given: The objects, hypotheses, and choice principles stated above.

[F1]

A covering map is a continuous surjection p:EB such that every bB has an open neighbourhood U for which p1(U) is a disjoint union of open sets Vj, called sheets, and each restriction pVj:VjU is a homeomorphism (def-continuous-map-top, def-homeomorphism-and-open-maps, def-disjoint-union-topology). Such a U is evenly covered, and p1(b) is the fibre over b. A covering is trivial when it is isomorphic over B to a product projection B×FB with F discrete. (Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings).

[F2]

Let (X,T) be a topological space (def-topological-space). An open cover of (X,T) is a family UT of open sets with X=U; a subcover of U is a subfamily that is itself an open cover; and (X,T) is compact when every open cover of it has a finite subcover. (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

[F3]

Let (X,TX) and (Y,TY) be topological spaces (def-topological-space), and let R carry its usual topology, the metric topology of dR(s,t)=st (lem-real-line-is-a-metric-space, def-metric-topology, def-metrizable-space). Then: 1. Continuous images. If f:XY is continuous (def-continuous-map-top) and (X,TX) is compact (def-compact-space), then f[X] is a compact subset of Y. More generally, if KX is a compact subset of X then f[K] is a compact subset of Y. 2. Extreme values. If (X,TX) is compact and nonempty and g:XR is continuous, then g[X] has a maximum and a minimum (def-max-min): there are xmax,xminX with g(xmin)    g(x)    g(xmax)for every xX. 3. Compact to Hausdorff. If (X,TX) is compact, (Y,TY) is Hausdorff (def-hausdorff-space) and f:XY is a continuous bijection, then f is a homeomorphism (def-homeomorphism-and-open-maps). Nonemptiness in claim 2 is a hypothesis and not an oversight: for X= the image is empty and has neither a maximum nor a minimum. No choice principle is used: the one selection made below is over a finite index set, where lem-finite-choice is a theorem of ZF. (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism).

Proof

technique · direct
1.1

The forward direction is the continuous image theorem and uses surjectivity.

givenF3F1F2
2.1

For the reverse direction, call an open VB adapted to a given open cover upstairs when V is evenly covered and every sheet above V lies in a single member of that cover. Because each fibre is finite, every point of B lies in some adapted V: take an evenly covered neighbourhood, and shrink it finitely many times, once per sheet. Let V be the set of all adapted open sets, formed outright rather than by selecting one per basepoint, so no choice principle is used.

step 1.1F1F3
3.1

V covers B, so the finite-subcover clause of [F2] supplies finitely many members of V covering B; each contributes finitely many sheets, each inside one cover member, so finitely many cover members exhaust the total space.

step 2.1F1F2F3
4.1

The preceding construction and implications establish the assertion.

step 3.1

Depends on

Used by

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