The neighbourhood basis of a submodules basis, and when two of them agree #
Mathlib's SubmodulesBasis B turns a family B : ι → Submodule R M into a topology on M,
SubmodulesBasis.topology, by routing the family through
SubmodulesBasis.toModuleFilterBasis. That route leaves the neighbourhoods of 0 indexed by
sets U carrying a proof ∃ i, U = B i rather than by ι. This file adds the ι-indexed
form, which Mathlib already has one level down for subgroups as
RingSubgroupsBasis.hasBasis_nhds_zero, and the comparison lemma it makes routine: two
submodule bases on the same module that are mutually cofinal induce the same topology.
Neither statement mentions anything beyond SubmodulesBasis, so both live in that namespace
rather than in a TauCeti one.
Main results #
SubmodulesBasis.hasBasis_nhds_zero: the family is itself a neighbourhood basis at0for the topology it induces — a set is a neighbourhood of0exactly when it contains someB i.SubmodulesBasis.topology_eq: mutually cofinal submodule bases induce the same topology.
References #
- Mathlib's
Mathlib/Topology/Algebra/Nonarchimedean/Bases.lean, whoseRingSubgroupsBasis.hasBasis_nhds_zerothe first result mirrors.
The family is a neighbourhood basis at 0, indexed by ι: a set is a neighbourhood of
0 for the induced topology exactly when it contains some B i.
Mathlib reaches the same filter only through SubmodulesBasis.toModuleFilterBasis, whose basis
is indexed by the sets U with ∃ i, U = B i; this is the ι-indexed form, mirroring
RingSubgroupsBasis.hasBasis_nhds_zero.
Mutually cofinal submodule bases induce the same topology. If every B i contains some
B' j and every B' j contains some B i, the two families are neighbourhood bases at 0 for
the same filter, and both topologies are additive group topologies, so they agree everywhere.
No relation between the two index types is needed, and neither family need be antitone: cofinality in both directions is the whole hypothesis.