Lift filters along filter and set functions #
If (p : ι → Prop, s : ι → Set α)
is a basis of a filter f
, g
is a monotone function
Set α → Filter γ
, and for each i
, (pg : β i → Prop, sg : β i → Set α)
is a basis
of the filter g (s i)
, then
(fun (i : ι) (x : β i) ↦ p i ∧ pg i x, fun (i : ι) (x : β i) ↦ sg i x)
is a basis
of the filter f.lift g
.
This basis is parametrized by i : ι
and x : β i
, so in order to formulate this fact using
Filter.HasBasis
one has to use Σ i, β i
as the index type, see Filter.HasBasis.lift
.
This lemma states the corresponding mem_iff
statement without using a sigma type.
If (p : ι → Prop, s : ι → Set α)
is a basis of a filter f
, g
is a monotone function
Set α → Filter γ
, and for each i
, (pg : β i → Prop, sg : β i → Set α)
is a basis
of the filter g (s i)
, then
(fun (i : ι) (x : β i) ↦ p i ∧ pg i x, fun (i : ι) (x : β i) ↦ sg i x)
is a basis of the filter f.lift g
.
This basis is parametrized by i : ι
and x : β i
, so in order to formulate this fact using
has_basis
one has to use Σ i, β i
as the index type. See also Filter.HasBasis.mem_lift_iff
for the corresponding mem_iff
statement formulated without using a sigma type.