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LeanPool.DirectedTopologyLean4.CoverLemma

Auxiliary lemmas #

theorem mid_point_Icc {i n : } (hn : n > 0) :
(2 * i + 1) / (2 * n) Set.Icc (i / n) ((i + 1) / n)

For any two natural numbers i n : ℕ with n > 0, we have that (2i+1)/(2n) is contained in the interval [i/n, (i+1)/n].

theorem mid_point_I {i n : } (hi : i < n) :
(2 * i + 1) / (2 * n) unitInterval

For any two natural numbers i n : ℕ with i < n, we have that (2i+1)/(2n) is contained in unit interval

theorem UnitIntervalSub.mem_I_of_mem_interval {t : } {n i : } (hi : i < n.succ) (h : t Set.Icc (i / n.succ) (↑(i + 1) / n.succ)) :
theorem UnitIntervalSub.mem_I_of_mem_interval_coed {t : } {n i : } (hi : i < n.succ) (h : t Set.Icc (i / (n + 1)) ((i + 1) / (n + 1))) :

Covering lemma for the unit interval #

theorem lebesgue_number_lemma_unit_interval {ι : Sort u} {c : ιSet } (hc₁ : ∀ (i : ι), IsOpen (c i)) (hc₂ : unitInterval ⋃ (i : ι), c i) :
n > 0, i < n, ∃ (j : ι), Set.Icc (i / n) ((i + 1) / n) c j

Covering lemma for the unit square #

The full unit square as a subset of I × I.

Equations
Instances For
    def UnitSubrectangle {n m i j : } (hi : i < n.succ) (hj : j < m.succ) :

    For any four natural numbers n m i j : ℕ such that i < n + 1 and j < m + 1, we have the rectangle [i/(n+1), (i+1)/(n+1)] × [j/(m+1), (j+1)/(m+1)] in the unit square.

    Equations
    Instances For
      theorem UnitSubrectangle.mem_unitSubrectangle {t₀ t₁ : } {n m i j : } (hi : i < n.succ) (hj : j < m.succ) (ht₀ : t₀ Set.Icc (i / n.succ) (↑(i + 1) / n.succ)) (ht₁ : t₁ Set.Icc (j / m.succ) (↑(j + 1) / m.succ)) :
      (t₀, , t₁, ) UnitSubrectangle hi hj
      theorem lebesgue_number_lemma_unitSquare {ι : Sort u} {c : ιSet (unitInterval × unitInterval)} (hc₁ : ∀ (i : ι), IsOpen (c i)) (hc₂ : UnitSquare ⋃ (i : ι), c i) :
      ∃ (n : ), ∀ (i j : ) (hi : i < n.succ) (hj : j < n.succ), ∃ (a : ι), UnitSubrectangle hi hj c a