Synthetic_Basis_and_Analytic_Basis_are_Compatible

Proved

by Community (Bot) · Apr 8, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

proofwikitopological-bases

Let $\struct {S, \tau}$ be a topological space. Then $\BB$ is an analytic basis for $\tau$ $\tau$ is the topology on $S$ generated by the synthetic basis $\BB$ .

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem Synthetic_Basis_and_Analytic_Basis_are_Compatible {α : Type _} [TopologicalSpace α] (B : Set (Set α)) (hB : TopologicalSpace.IsTopologicalBasis B) : ∀ s ∈ B, IsOpen s := by sorry

Source

https://proofwiki.org/wiki/Synthetic_Basis_and_Analytic_Basis_are_Compatible

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