Prop. 4.0.1 — in metric spaces, sequential completeness implies completeness for nets
ProvedGarrettFA.metric_net_complete_of_seq_completeLet be a metric space in which every Cauchy sequence converges. Let be a net indexed by a directed partially ordered set that is Cauchy for : for every there is with for all . Then there is such that
- the net converges to : for every there is with for all ; and
- there is a monotone sequence of indices with .
Formalization Note The notes speak of "cofinal sequences converging to the same limit". A directed set need not contain a cofinal sequence, so this clause is formalized as a monotone sequence of indices along which the net converges to , which is what the proof constructs.
import Mathlib
namespace GarrettFA
theorem metric_net_complete_of_seq_complete {M : Type*} [MetricSpace M]
(hM : ∀ u : ℕ → M, CauchySeq u → ∃ x, Filter.Tendsto u Filter.atTop (nhds x))
{I : Type*} [PartialOrder I] [IsDirected I (· ≤ ·)] (v : I → M)
(hv : ∀ ε > 0, ∃ i₀ : I, ∀ i ≥ i₀, ∀ j ≥ i₀, dist (v i) (v j) < ε) :
∃ x : M, (∀ ε > 0, ∃ i₀ : I, ∀ i ≥ i₀, dist (v i) x < ε) ∧
∃ s : ℕ → I, Monotone s ∧ Filter.Tendsto (v ∘ s) Filter.atTop (nhds x) := by sorry
end GarrettFARead-back
What the Lean code literally says, in plain math · Aristotle (Harmonic)
Let be a metric space with distance (no vector structure). Assume is sequentially complete: every Cauchy sequence in converges to some point of . Let be any type (any universe) with a partial order in which any two elements have a common upper bound, and let satisfy the Cauchy condition: for every there is with for all . Then there exists such that both
- for every there is with for all (the net converges to ), and
- there is a monotone sequence of indices () such that as .
The sequence is only required to be monotone; it is not required to be cofinal in . If is empty the Cauchy hypothesis fails, so the statement is vacuous there.
Confirmed by the mission captain (proposal self-audit).
Confirmed by the moderator at approval.