Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31
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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.

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A countable local base can be chosen open and decreasing

Statement

If XX is first countable and xXx\in X, then xx has a countable local base (Vn)nN(V_n)_{n\in\mathbb N} of open sets with Vn+1VnV_{n+1}\subseteq V_n.

Facts & Assumptions

Proof

technique · constructive
1.1

A local base is nonempty, so enumerate it as (Bn)(B_n) by [L2], with repetitions allowed. Put Un=int(Bn)U_n=\operatorname{int}(B_n). Since BnB_n is a neighbourhood of xx, its interior is open, contains xx, and is contained in BnB_n. This definition is canonical and uses no countable choice.

givenL1L2construct
2.1

Put Vn=U0UnV_n=U_0\cap\cdots\cap U_n; each VnV_n is open, contains xx, and Vn+1VnV_{n+1}\subseteq V_n.

step 1.1construct
3.1

Since VnBnV_n\subseteq B_n, every neighbourhood contains some VnV_n, so (Vn)(V_n) is the required decreasing local base.

step 2.1discharge-construct

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: 39 results over 16 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