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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • 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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For a finite-sheeted covering, the total space is compact exactly when the base is compact

Statement

If p:E→B 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:E→B such that every b∈B has an open neighbourhood U for which p−1(U) is a disjoint union of open sets Vj, called sheets, and each restriction p∣Vj:Vj→U is a homeomorphism (def-continuous-map-top, def-homeomorphism-and-open-maps, def-disjoint-union-topology). Such a U is evenly covered, and p−1(b) is the fibre over b. A covering is trivial when it is isomorphic over B to a product projection B×F→B 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 U⊆T 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)=∣s−t∣ (lem-real-line-is-a-metric-space, def-metric-topology, def-metrizable-space). Then: 1. Continuous images. If f:X→Y 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 K⊆X 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:X→R is continuous, then g[X] has a maximum and a minimum (def-max-min): there are xmax⁡,xmin⁡∈X with g(xmin⁡)  ≤  g(x)  ≤  g(xmax⁡)for every x∈X. 3. Compact to Hausdorff. If (X,TX) is compact, (Y,TY) is Hausdorff (def-hausdorff-space) and f:X→Y 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
1.1givenF3F1F2

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

2.1step 1.1F1F3

For the reverse direction, call an open V⊆B 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.

3.1step 2.1F1F2F3

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.

4.1step 3.1∎

The preceding construction and implications establish the assertion.

Depends on

Used by

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