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Created November 17, 2023 05:47
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Chapter 1 Section 5 Q1 of Introduction to Topology by Mendelson
/-
The original problem is incorrect. Instead, we prove that the statement is true
if and only if either A or B is the universe
-/
example (h0 : X ⊆ A) (h1 : Y ⊆ B) :
A = Set.univ ∨ B = Set.univ ↔ (X ×ˢ Y)ᶜ = A ×ˢ Yᶜ ∪ Xᶜ ×ˢ B := by
apply Iff.intro
{
intro h2
rw [Set.compl_prod_eq_union, Set.union_comm]
nth_rewrite 1 [← Set.union_compl_self A]
rw [← Set.union_compl_self B]
rw [Set.union_prod, Set.prod_union, ← Set.union_assoc]
rcases h2 with h2.l | h2.r
{
rw [h2.l]
simp
apply Set.subset_union_of_subset_left
apply Set.prod_mono
· simp
· simp; apply h1
}
{
rw [h2.r]
simp
rw [Set.union_assoc, Set.union_left_comm]
simp
apply Set.subset_union_of_subset_right
apply Set.prod_mono
· simp; apply h0
· simp
}
}
{
rw [Set.compl_prod_eq_union, Set.union_comm]
nth_rewrite 1 [← Set.union_compl_self A]
nth_rewrite 1 [← Set.union_compl_self B]
rw [Set.union_prod, Set.prod_union, ← Set.union_assoc,
Set.union_right_comm (A ×ˢ Yᶜ) (Aᶜ ×ˢ Yᶜ) (Xᶜ ×ˢ B),
Set.union_assoc]
simp
intro h2 h3
have h4 : Disjoint (Aᶜ ×ˢ Yᶜ) (A ×ˢ Yᶜ) := by
simp; left; exact disjoint_compl_left
have h5 : (Aᶜ ×ˢ Yᶜ) ⊆ (Xᶜ ×ˢ B) := by
exact Disjoint.subset_right_of_subset_union h2 h4
have h6 : (Aᶜ ⊆ Xᶜ ∧ Yᶜ ⊆ B) ∨ Aᶜ = ∅ ∨ Yᶜ = ∅ := by
exact Set.prod_subset_prod_iff.mp h5
simp_all
rcases h6 with h6.l | h6.r
{
have h7 : Bᶜ ⊆ Y := by
exact Set.compl_subset_comm.mp h6.l
have h8 : Bᶜ ⊆ B := by
tauto
have h9 : B ∩ Bᶜ = Bᶜ := by
exact Set.inter_eq_right.mpr h8
simp_all
right
exact Set.compl_empty_iff.mp (id h9.symm)
}
{
rcases h6.r with h6.rl | h6.rr
· left; exact h6.rl
· right; rw [h6.rr] at h1; apply Set.eq_univ_of_univ_subset h1
}
}
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