l3lab/ntp-mathlib huggingface dataset:
/-- In a real vector space of dimension `> 1`, the complement of any countable set is path
connected. -/
theorem Set.Countable.isPathConnected_compl_of_one_lt_rank
    (h : 1 < Module.rank ℝ E) {s : Set E} (hs : s.Countable) :
    IsPathConnected sᶜ := by
  have : Nontrivial E := (rank_pos_iff_nontrivial (R := ℝ)).1 (zero_lt_one.trans h)
  -- the set `sᶜ` is dense, therefore nonempty. Pick `a ∈ sᶜ`. We have to show that any
  -- `b ∈ sᶜ` can be joined to `a`.
  obtain ⟨a, ha⟩ : sᶜ.Nonempty := (hs.dense_compl ℝ).nonempty
  refine ⟨a, ha, ?_⟩
  intro b hb
  rcases eq_or_ne a b with rfl|hab
  · exact JoinedIn.refl ha
  /- Assume `b ≠ a`. Write `a = c - x` and `b = c + x` for some nonzero `x`. Choose `y` which
  is linearly independent from `x`. Then the segments joining `a = c - x` to `c + ty` are pairwise
  disjoint for varying `t` (except for the endpoint `a`) so only countably many of them can
  intersect `s`. In the same way, there are countably many `t`s for which the segment
  from `b = c + x` to `c + ty` intersects `s`. Choosing `t` outside of these countable exceptions,
  one gets a path in the complement of `s` from `a` to `z = c + ty` and then to `b`.
  -/
  let c := (2 : ℝ)⁻¹ • (a + b)
  let x := (2 : ℝ)⁻¹ • (b - a)
  have Ia : c - x = a := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have Ib : c + x = b := by
    

Proof state:
E : Type u_1
inst✝⁴ : AddCommGroup E
inst✝³ : Module ℝ E
inst✝² : TopologicalSpace E
inst✝¹ : ContinuousAdd E
inst✝ : ContinuousSMul ℝ E
h : 1 < Module.rank ℝ E
s : Set E
hs : Set.Countable s
this : Nontrivial E
a : E
ha : a ∈ sᶜ
b : E
hb : b ∈ sᶜ
hab : a ≠ b
c : E := 2⁻¹ • (a + b)
x : E := 2⁻¹ • (b - a)
Ia : c - x = a
⊢ c + x = b

Next tactic:
simp only [smul_add, smul_sub]
/-- In a real vector space of dimension `> 1`, the complement of any countable set is path
connected. -/
theorem Set.Countable.isPathConnected_compl_of_one_lt_rank
    (h : 1 < Module.rank ℝ E) {s : Set E} (hs : s.Countable) :
    IsPathConnected sᶜ := by
  have : Nontrivial E := (rank_pos_iff_nontrivial (R := ℝ)).1 (zero_lt_one.trans h)
  -- the set `sᶜ` is dense, therefore nonempty. Pick `a ∈ sᶜ`. We have to show that any
  -- `b ∈ sᶜ` can be joined to `a`.
  obtain ⟨a, ha⟩ : sᶜ.Nonempty := (hs.dense_compl ℝ).nonempty
  refine ⟨a, ha, ?_⟩
  intro b hb
  rcases eq_or_ne a b with rfl|hab
  · exact JoinedIn.refl ha
  /- Assume `b ≠ a`. Write `a = c - x` and `b = c + x` for some nonzero `x`. Choose `y` which
  is linearly independent from `x`. Then the segments joining `a = c - x` to `c + ty` are pairwise
  disjoint for varying `t` (except for the endpoint `a`) so only countably many of them can
  intersect `s`. In the same way, there are countably many `t`s for which the segment
  from `b = c + x` to `c + ty` intersects `s`. Choosing `t` outside of these countable exceptions,
  one gets a path in the complement of `s` from `a` to `z = c + ty` and then to `b`.
  -/
  let c := (2 : ℝ)⁻¹ • (a + b)
  let x := (2 : ℝ)⁻¹ • (b - a)
  have Ia : c - x = a := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have Ib : c + x = b := by
    simp only [smul_add, smul_sub]
    

Proof state:
E : Type u_1
inst✝⁴ : AddCommGroup E
inst✝³ : Module ℝ E
inst✝² : TopologicalSpace E
inst✝¹ : ContinuousAdd E
inst✝ : ContinuousSMul ℝ E
h : 1 < Module.rank ℝ E
s : Set E
hs : Set.Countable s
this : Nontrivial E
a : E
ha : a ∈ sᶜ
b : E
hb : b ∈ sᶜ
hab : a ≠ b
c : E := 2⁻¹ • (a + b)
x : E := 2⁻¹ • (b - a)
Ia : c - x = a
⊢ 2⁻¹ • a + 2⁻¹ • b + (2⁻¹ • b - 2⁻¹ • a) = b

Next tactic:
abel_nf
/-- In a real vector space of dimension `> 1`, the complement of any countable set is path
connected. -/
theorem Set.Countable.isPathConnected_compl_of_one_lt_rank
    (h : 1 < Module.rank ℝ E) {s : Set E} (hs : s.Countable) :
    IsPathConnected sᶜ := by
  have : Nontrivial E := (rank_pos_iff_nontrivial (R := ℝ)).1 (zero_lt_one.trans h)
  -- the set `sᶜ` is dense, therefore nonempty. Pick `a ∈ sᶜ`. We have to show that any
  -- `b ∈ sᶜ` can be joined to `a`.
  obtain ⟨a, ha⟩ : sᶜ.Nonempty := (hs.dense_compl ℝ).nonempty
  refine ⟨a, ha, ?_⟩
  intro b hb
  rcases eq_or_ne a b with rfl|hab
  · exact JoinedIn.refl ha
  /- Assume `b ≠ a`. Write `a = c - x` and `b = c + x` for some nonzero `x`. Choose `y` which
  is linearly independent from `x`. Then the segments joining `a = c - x` to `c + ty` are pairwise
  disjoint for varying `t` (except for the endpoint `a`) so only countably many of them can
  intersect `s`. In the same way, there are countably many `t`s for which the segment
  from `b = c + x` to `c + ty` intersects `s`. Choosing `t` outside of these countable exceptions,
  one gets a path in the complement of `s` from `a` to `z = c + ty` and then to `b`.
  -/
  let c := (2 : ℝ)⁻¹ • (a + b)
  let x := (2 : ℝ)⁻¹ • (b - a)
  have Ia : c - x = a := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have Ib : c + x = b := by
    simp only [smul_add, smul_sub]
    abel_nf
    

Proof state:
E : Type u_1
inst✝⁴ : AddCommGroup E
inst✝³ : Module ℝ E
inst✝² : TopologicalSpace E
inst✝¹ : ContinuousAdd E
inst✝ : ContinuousSMul ℝ E
h : 1 < Module.rank ℝ E
s : Set E
hs : Set.Countable s
this : Nontrivial E
a : E
ha : a ∈ sᶜ
b : E
hb : b ∈ sᶜ
hab : a ≠ b
c : E := 2⁻¹ • (a + b)
x : E := 2⁻¹ • (b - a)
Ia : c - x = a
⊢ 2 • 2⁻¹ • b = b

Next tactic:
simp [zsmul_eq_smul_cast ℝ 2]
/-- In a real vector space of dimension `> 1`, the complement of any countable set is path
connected. -/
theorem Set.Countable.isPathConnected_compl_of_one_lt_rank
    (h : 1 < Module.rank ℝ E) {s : Set E} (hs : s.Countable) :
    IsPathConnected sᶜ := by
  have : Nontrivial E := (rank_pos_iff_nontrivial (R := ℝ)).1 (zero_lt_one.trans h)
  -- the set `sᶜ` is dense, therefore nonempty. Pick `a ∈ sᶜ`. We have to show that any
  -- `b ∈ sᶜ` can be joined to `a`.
  obtain ⟨a, ha⟩ : sᶜ.Nonempty := (hs.dense_compl ℝ).nonempty
  refine ⟨a, ha, ?_⟩
  intro b hb
  rcases eq_or_ne a b with rfl|hab
  · exact JoinedIn.refl ha
  /- Assume `b ≠ a`. Write `a = c - x` and `b = c + x` for some nonzero `x`. Choose `y` which
  is linearly independent from `x`. Then the segments joining `a = c - x` to `c + ty` are pairwise
  disjoint for varying `t` (except for the endpoint `a`) so only countably many of them can
  intersect `s`. In the same way, there are countably many `t`s for which the segment
  from `b = c + x` to `c + ty` intersects `s`. Choosing `t` outside of these countable exceptions,
  one gets a path in the complement of `s` from `a` to `z = c + ty` and then to `b`.
  -/
  let c := (2 : ℝ)⁻¹ • (a + b)
  let x := (2 : ℝ)⁻¹ • (b - a)
  have Ia : c - x = a := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have Ib : c + x = b := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  

Proof state:
case intro.inr
E : Type u_1
inst✝⁴ : AddCommGroup E
inst✝³ : Module ℝ E
inst✝² : TopologicalSpace E
inst✝¹ : ContinuousAdd E
inst✝ : ContinuousSMul ℝ E
h : 1 < Module.rank ℝ E
s : Set E
hs : Set.Countable s
this : Nontrivial E
a : E
ha : a ∈ sᶜ
b : E
hb : b ∈ sᶜ
hab : a ≠ b
c : E := 2⁻¹ • (a + b)
x : E := 2⁻¹ • (b - a)
Ia : c - x = a
Ib : c + x = b
⊢ JoinedIn sᶜ a b

Next tactic:
have x_ne_zero : x ≠ 0 := by simpa using sub_ne_zero.2 hab.symm
/-- In a real vector space of dimension `> 1`, the complement of any countable set is path
connected. -/
theorem Set.Countable.isPathConnected_compl_of_one_lt_rank
    (h : 1 < Module.rank ℝ E) {s : Set E} (hs : s.Countable) :
    IsPathConnected sᶜ := by
  have : Nontrivial E := (rank_pos_iff_nontrivial (R := ℝ)).1 (zero_lt_one.trans h)
  -- the set `sᶜ` is dense, therefore nonempty. Pick `a ∈ sᶜ`. We have to show that any
  -- `b ∈ sᶜ` can be joined to `a`.
  obtain ⟨a, ha⟩ : sᶜ.Nonempty := (hs.dense_compl ℝ).nonempty
  refine ⟨a, ha, ?_⟩
  intro b hb
  rcases eq_or_ne a b with rfl|hab
  · exact JoinedIn.refl ha
  /- Assume `b ≠ a`. Write `a = c - x` and `b = c + x` for some nonzero `x`. Choose `y` which
  is linearly independent from `x`. Then the segments joining `a = c - x` to `c + ty` are pairwise
  disjoint for varying `t` (except for the endpoint `a`) so only countably many of them can
  intersect `s`. In the same way, there are countably many `t`s for which the segment
  from `b = c + x` to `c + ty` intersects `s`. Choosing `t` outside of these countable exceptions,
  one gets a path in the complement of `s` from `a` to `z = c + ty` and then to `b`.
  -/
  let c := (2 : ℝ)⁻¹ • (a + b)
  let x := (2 : ℝ)⁻¹ • (b - a)
  have Ia : c - x = a := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have Ib : c + x = b := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have x_ne_zero : x ≠ 0 := by 

Proof state:
E : Type u_1
inst✝⁴ : AddCommGroup E
inst✝³ : Module ℝ E
inst✝² : TopologicalSpace E
inst✝¹ : ContinuousAdd E
inst✝ : ContinuousSMul ℝ E
h : 1 < Module.rank ℝ E
s : Set E
hs : Set.Countable s
this : Nontrivial E
a : E
ha : a ∈ sᶜ
b : E
hb : b ∈ sᶜ
hab : a ≠ b
c : E := 2⁻¹ • (a + b)
x : E := 2⁻¹ • (b - a)
Ia : c - x = a
Ib : c + x = b
⊢ x ≠ 0

Next tactic:
simpa using sub_ne_zero.2 hab.symm
/-- In a real vector space of dimension `> 1`, the complement of any countable set is path
connected. -/
theorem Set.Countable.isPathConnected_compl_of_one_lt_rank
    (h : 1 < Module.rank ℝ E) {s : Set E} (hs : s.Countable) :
    IsPathConnected sᶜ := by
  have : Nontrivial E := (rank_pos_iff_nontrivial (R := ℝ)).1 (zero_lt_one.trans h)
  -- the set `sᶜ` is dense, therefore nonempty. Pick `a ∈ sᶜ`. We have to show that any
  -- `b ∈ sᶜ` can be joined to `a`.
  obtain ⟨a, ha⟩ : sᶜ.Nonempty := (hs.dense_compl ℝ).nonempty
  refine ⟨a, ha, ?_⟩
  intro b hb
  rcases eq_or_ne a b with rfl|hab
  · exact JoinedIn.refl ha
  /- Assume `b ≠ a`. Write `a = c - x` and `b = c + x` for some nonzero `x`. Choose `y` which
  is linearly independent from `x`. Then the segments joining `a = c - x` to `c + ty` are pairwise
  disjoint for varying `t` (except for the endpoint `a`) so only countably many of them can
  intersect `s`. In the same way, there are countably many `t`s for which the segment
  from `b = c + x` to `c + ty` intersects `s`. Choosing `t` outside of these countable exceptions,
  one gets a path in the complement of `s` from `a` to `z = c + ty` and then to `b`.
  -/
  let c := (2 : ℝ)⁻¹ • (a + b)
  let x := (2 : ℝ)⁻¹ • (b - a)
  have Ia : c - x = a := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have Ib : c + x = b := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have x_ne_zero : x ≠ 0 := by simpa using sub_ne_zero.2 hab.symm
  

Proof state:
case intro.inr
E : Type u_1
inst✝⁴ : AddCommGroup E
inst✝³ : Module ℝ E
inst✝² : TopologicalSpace E
inst✝¹ : ContinuousAdd E
inst✝ : ContinuousSMul ℝ E
h : 1 < Module.rank ℝ E
s : Set E
hs : Set.Countable s
this : Nontrivial E
a : E
ha : a ∈ sᶜ
b : E
hb : b ∈ sᶜ
hab : a ≠ b
c : E := 2⁻¹ • (a + b)
x : E := 2⁻¹ • (b - a)
Ia : c - x = a
Ib : c + x = b
x_ne_zero : x ≠ 0
⊢ JoinedIn sᶜ a b

Next tactic:
obtain ⟨y, hy⟩ : ∃ y, LinearIndependent ℝ ![x, y] :=
    exists_linearIndependent_pair_of_one_lt_rank h x_ne_zero
/-- In a real vector space of dimension `> 1`, the complement of any countable set is path
connected. -/
theorem Set.Countable.isPathConnected_compl_of_one_lt_rank
    (h : 1 < Module.rank ℝ E) {s : Set E} (hs : s.Countable) :
    IsPathConnected sᶜ := by
  have : Nontrivial E := (rank_pos_iff_nontrivial (R := ℝ)).1 (zero_lt_one.trans h)
  -- the set `sᶜ` is dense, therefore nonempty. Pick `a ∈ sᶜ`. We have to show that any
  -- `b ∈ sᶜ` can be joined to `a`.
  obtain ⟨a, ha⟩ : sᶜ.Nonempty := (hs.dense_compl ℝ).nonempty
  refine ⟨a, ha, ?_⟩
  intro b hb
  rcases eq_or_ne a b with rfl|hab
  · exact JoinedIn.refl ha
  /- Assume `b ≠ a`. Write `a = c - x` and `b = c + x` for some nonzero `x`. Choose `y` which
  is linearly independent from `x`. Then the segments joining `a = c - x` to `c + ty` are pairwise
  disjoint for varying `t` (except for the endpoint `a`) so only countably many of them can
  intersect `s`. In the same way, there are countably many `t`s for which the segment
  from `b = c + x` to `c + ty` intersects `s`. Choosing `t` outside of these countable exceptions,
  one gets a path in the complement of `s` from `a` to `z = c + ty` and then to `b`.
  -/
  let c := (2 : ℝ)⁻¹ • (a + b)
  let x := (2 : ℝ)⁻¹ • (b - a)
  have Ia : c - x = a := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have Ib : c + x = b := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have x_ne_zero : x ≠ 0 := by simpa using sub_ne_zero.2 hab.symm
  obtain ⟨y, hy⟩ : ∃ y, LinearIndependent ℝ ![x, y] :=
    exists_linearIndependent_pair_of_one_lt_rank h x_ne_zero
  

Proof state:
case intro.inr.intro
E : Type u_1
inst✝⁴ : AddCommGroup E
inst✝³ : Module ℝ E
inst✝² : TopologicalSpace E
inst✝¹ : ContinuousAdd E
inst✝ : ContinuousSMul ℝ E
h : 1 < Module.rank ℝ E
s : Set E
hs : Set.Countable s
this : Nontrivial E
a : E
ha : a ∈ sᶜ
b : E
hb : b ∈ sᶜ
hab : a ≠ b
c : E := 2⁻¹ • (a + b)
x : E := 2⁻¹ • (b - a)
Ia : c - x = a
Ib : c + x = b
x_ne_zero : x ≠ 0
y : E
hy : LinearIndependent ℝ ![x, y]
⊢ JoinedIn sᶜ a b

Next tactic:
have A : Set.Countable {t : ℝ | ([c + x -[ℝ] c + t • y] ∩ s).Nonempty} := by
    apply countable_setOf_nonempty_of_disjoint _ (fun t ↦ inter_subset_right _ _) hs
    intro t t' htt'
    apply disjoint_iff_inter_eq_empty.2
    have N : {c + x} ∩ s = ∅ := by
      simpa only [singleton_inter_eq_empty, mem_compl_iff, Ib] using hb
    rw [inter_assoc, inter_comm s, inter_assoc, inter_self, ← inter_assoc, ← subset_empty_iff, ← N]
    apply inter_subset_inter_left
    apply Eq.subset
    apply segment_inter_eq_endpoint_of_linearIndependent_of_ne hy htt'.symm
/-- In a real vector space of dimension `> 1`, the complement of any countable set is path
connected. -/
theorem Set.Countable.isPathConnected_compl_of_one_lt_rank
    (h : 1 < Module.rank ℝ E) {s : Set E} (hs : s.Countable) :
    IsPathConnected sᶜ := by
  have : Nontrivial E := (rank_pos_iff_nontrivial (R := ℝ)).1 (zero_lt_one.trans h)
  -- the set `sᶜ` is dense, therefore nonempty. Pick `a ∈ sᶜ`. We have to show that any
  -- `b ∈ sᶜ` can be joined to `a`.
  obtain ⟨a, ha⟩ : sᶜ.Nonempty := (hs.dense_compl ℝ).nonempty
  refine ⟨a, ha, ?_⟩
  intro b hb
  rcases eq_or_ne a b with rfl|hab
  · exact JoinedIn.refl ha
  /- Assume `b ≠ a`. Write `a = c - x` and `b = c + x` for some nonzero `x`. Choose `y` which
  is linearly independent from `x`. Then the segments joining `a = c - x` to `c + ty` are pairwise
  disjoint for varying `t` (except for the endpoint `a`) so only countably many of them can
  intersect `s`. In the same way, there are countably many `t`s for which the segment
  from `b = c + x` to `c + ty` intersects `s`. Choosing `t` outside of these countable exceptions,
  one gets a path in the complement of `s` from `a` to `z = c + ty` and then to `b`.
  -/
  let c := (2 : ℝ)⁻¹ • (a + b)
  let x := (2 : ℝ)⁻¹ • (b - a)
  have Ia : c - x = a := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have Ib : c + x = b := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have x_ne_zero : x ≠ 0 := by simpa using sub_ne_zero.2 hab.symm
  obtain ⟨y, hy⟩ : ∃ y, LinearIndependent ℝ ![x, y] :=
    exists_linearIndependent_pair_of_one_lt_rank h x_ne_zero
  have A : Set.Countable {t : ℝ | ([c + x -[ℝ] c + t • y] ∩ s).Nonempty} := by
    

Proof state:
E : Type u_1
inst✝⁴ : AddCommGroup E
inst✝³ : Module ℝ E
inst✝² : TopologicalSpace E
inst✝¹ : ContinuousAdd E
inst✝ : ContinuousSMul ℝ E
h : 1 < Module.rank ℝ E
s : Set E
hs : Set.Countable s
this : Nontrivial E
a : E
ha : a ∈ sᶜ
b : E
hb : b ∈ sᶜ
hab : a ≠ b
c : E := 2⁻¹ • (a + b)
x : E := 2⁻¹ • (b - a)
Ia : c - x = a
Ib : c + x = b
x_ne_zero : x ≠ 0
y : E
hy : LinearIndependent ℝ ![x, y]
⊢ Set.Countable {t | Set.Nonempty ([c + x-[ℝ]c + t • y] ∩ s)}

Next tactic:
apply countable_setOf_nonempty_of_disjoint _ (fun t ↦ inter_subset_right _ _) hs
/-- In a real vector space of dimension `> 1`, the complement of any countable set is path
connected. -/
theorem Set.Countable.isPathConnected_compl_of_one_lt_rank
    (h : 1 < Module.rank ℝ E) {s : Set E} (hs : s.Countable) :
    IsPathConnected sᶜ := by
  have : Nontrivial E := (rank_pos_iff_nontrivial (R := ℝ)).1 (zero_lt_one.trans h)
  -- the set `sᶜ` is dense, therefore nonempty. Pick `a ∈ sᶜ`. We have to show that any
  -- `b ∈ sᶜ` can be joined to `a`.
  obtain ⟨a, ha⟩ : sᶜ.Nonempty := (hs.dense_compl ℝ).nonempty
  refine ⟨a, ha, ?_⟩
  intro b hb
  rcases eq_or_ne a b with rfl|hab
  · exact JoinedIn.refl ha
  /- Assume `b ≠ a`. Write `a = c - x` and `b = c + x` for some nonzero `x`. Choose `y` which
  is linearly independent from `x`. Then the segments joining `a = c - x` to `c + ty` are pairwise
  disjoint for varying `t` (except for the endpoint `a`) so only countably many of them can
  intersect `s`. In the same way, there are countably many `t`s for which the segment
  from `b = c + x` to `c + ty` intersects `s`. Choosing `t` outside of these countable exceptions,
  one gets a path in the complement of `s` from `a` to `z = c + ty` and then to `b`.
  -/
  let c := (2 : ℝ)⁻¹ • (a + b)
  let x := (2 : ℝ)⁻¹ • (b - a)
  have Ia : c - x = a := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have Ib : c + x = b := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have x_ne_zero : x ≠ 0 := by simpa using sub_ne_zero.2 hab.symm
  obtain ⟨y, hy⟩ : ∃ y, LinearIndependent ℝ ![x, y] :=
    exists_linearIndependent_pair_of_one_lt_rank h x_ne_zero
  have A : Set.Countable {t : ℝ | ([c + x -[ℝ] c + t • y] ∩ s).Nonempty} := by
    apply countable_setOf_nonempty_of_disjoint _ (fun t ↦ inter_subset_right _ _) hs
    

Proof state:
E : Type u_1
inst✝⁴ : AddCommGroup E
inst✝³ : Module ℝ E
inst✝² : TopologicalSpace E
inst✝¹ : ContinuousAdd E
inst✝ : ContinuousSMul ℝ E
h : 1 < Module.rank ℝ E
s : Set E
hs : Set.Countable s
this : Nontrivial E
a : E
ha : a ∈ sᶜ
b : E
hb : b ∈ sᶜ
hab : a ≠ b
c : E := 2⁻¹ • (a + b)
x : E := 2⁻¹ • (b - a)
Ia : c - x = a
Ib : c + x = b
x_ne_zero : x ≠ 0
y : E
hy : LinearIndependent ℝ ![x, y]
⊢ Pairwise (Disjoint on fun t => [c + x-[ℝ]c + t • y] ∩ s)

Next tactic:
intro t t' htt'
/-- In a real vector space of dimension `> 1`, the complement of any countable set is path
connected. -/
theorem Set.Countable.isPathConnected_compl_of_one_lt_rank
    (h : 1 < Module.rank ℝ E) {s : Set E} (hs : s.Countable) :
    IsPathConnected sᶜ := by
  have : Nontrivial E := (rank_pos_iff_nontrivial (R := ℝ)).1 (zero_lt_one.trans h)
  -- the set `sᶜ` is dense, therefore nonempty. Pick `a ∈ sᶜ`. We have to show that any
  -- `b ∈ sᶜ` can be joined to `a`.
  obtain ⟨a, ha⟩ : sᶜ.Nonempty := (hs.dense_compl ℝ).nonempty
  refine ⟨a, ha, ?_⟩
  intro b hb
  rcases eq_or_ne a b with rfl|hab
  · exact JoinedIn.refl ha
  /- Assume `b ≠ a`. Write `a = c - x` and `b = c + x` for some nonzero `x`. Choose `y` which
  is linearly independent from `x`. Then the segments joining `a = c - x` to `c + ty` are pairwise
  disjoint for varying `t` (except for the endpoint `a`) so only countably many of them can
  intersect `s`. In the same way, there are countably many `t`s for which the segment
  from `b = c + x` to `c + ty` intersects `s`. Choosing `t` outside of these countable exceptions,
  one gets a path in the complement of `s` from `a` to `z = c + ty` and then to `b`.
  -/
  let c := (2 : ℝ)⁻¹ • (a + b)
  let x := (2 : ℝ)⁻¹ • (b - a)
  have Ia : c - x = a := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have Ib : c + x = b := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have x_ne_zero : x ≠ 0 := by simpa using sub_ne_zero.2 hab.symm
  obtain ⟨y, hy⟩ : ∃ y, LinearIndependent ℝ ![x, y] :=
    exists_linearIndependent_pair_of_one_lt_rank h x_ne_zero
  have A : Set.Countable {t : ℝ | ([c + x -[ℝ] c + t • y] ∩ s).Nonempty} := by
    apply countable_setOf_nonempty_of_disjoint _ (fun t ↦ inter_subset_right _ _) hs
    intro t t' htt'
    

Proof state:
E : Type u_1
inst✝⁴ : AddCommGroup E
inst✝³ : Module ℝ E
inst✝² : TopologicalSpace E
inst✝¹ : ContinuousAdd E
inst✝ : ContinuousSMul ℝ E
h : 1 < Module.rank ℝ E
s : Set E
hs : Set.Countable s
this : Nontrivial E
a : E
ha : a ∈ sᶜ
b : E
hb : b ∈ sᶜ
hab : a ≠ b
c : E := 2⁻¹ • (a + b)
x : E := 2⁻¹ • (b - a)
Ia : c - x = a
Ib : c + x = b
x_ne_zero : x ≠ 0
y : E
hy : LinearIndependent ℝ ![x, y]
t t' : ℝ
htt' : t ≠ t'
⊢ (Disjoint on fun t => [c + x-[ℝ]c + t • y] ∩ s) t t'

Next tactic:
apply disjoint_iff_inter_eq_empty.2
/-- In a real vector space of dimension `> 1`, the complement of any countable set is path
connected. -/
theorem Set.Countable.isPathConnected_compl_of_one_lt_rank
    (h : 1 < Module.rank ℝ E) {s : Set E} (hs : s.Countable) :
    IsPathConnected sᶜ := by
  have : Nontrivial E := (rank_pos_iff_nontrivial (R := ℝ)).1 (zero_lt_one.trans h)
  -- the set `sᶜ` is dense, therefore nonempty. Pick `a ∈ sᶜ`. We have to show that any
  -- `b ∈ sᶜ` can be joined to `a`.
  obtain ⟨a, ha⟩ : sᶜ.Nonempty := (hs.dense_compl ℝ).nonempty
  refine ⟨a, ha, ?_⟩
  intro b hb
  rcases eq_or_ne a b with rfl|hab
  · exact JoinedIn.refl ha
  /- Assume `b ≠ a`. Write `a = c - x` and `b = c + x` for some nonzero `x`. Choose `y` which
  is linearly independent from `x`. Then the segments joining `a = c - x` to `c + ty` are pairwise
  disjoint for varying `t` (except for the endpoint `a`) so only countably many of them can
  intersect `s`. In the same way, there are countably many `t`s for which the segment
  from `b = c + x` to `c + ty` intersects `s`. Choosing `t` outside of these countable exceptions,
  one gets a path in the complement of `s` from `a` to `z = c + ty` and then to `b`.
  -/
  let c := (2 : ℝ)⁻¹ • (a + b)
  let x := (2 : ℝ)⁻¹ • (b - a)
  have Ia : c - x = a := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have Ib : c + x = b := by
    simp only [smul_add, smul_sub]
    abel_nf
    simp [zsmul_eq_smul_cast ℝ 2]
  have x_ne_zero : x ≠ 0 := by simpa using sub_ne_zero.2 hab.symm
  obtain ⟨y, hy⟩ : ∃ y, LinearIndependent ℝ ![x, y] :=
    exists_linearIndependent_pair_of_one_lt_rank h x_ne_zero
  have A : Set.Countable {t : ℝ | ([c + x -[ℝ] c + t • y] ∩ s).Nonempty} := by
    apply countable_setOf_nonempty_of_disjoint _ (fun t ↦ inter_subset_right _ _) hs
    intro t t' htt'
    apply disjoint_iff_inter_eq_empty.2
    

Proof state:
E : Type u_1
inst✝⁴ : AddCommGroup E
inst✝³ : Module ℝ E
inst✝² : TopologicalSpace E
inst✝¹ : ContinuousAdd E
inst✝ : ContinuousSMul ℝ E
h : 1 < Module.rank ℝ E
s : Set E
hs : Set.Countable s
this : Nontrivial E
a : E
ha : a ∈ sᶜ
b : E
hb : b ∈ sᶜ
hab : a ≠ b
c : E := 2⁻¹ • (a + b)
x : E := 2⁻¹ • (b - a)
Ia : c - x = a
Ib : c + x = b
x_ne_zero : x ≠ 0
y : E
hy : LinearIndependent ℝ ![x, y]
t t' : ℝ
htt' : t ≠ t'
⊢ (fun t => [c + x-[ℝ]c + t • y] ∩ s) t ∩ (fun t => [c + x-[ℝ]c + t • y] ∩ s) t' = ∅

Next tactic:
have N : {c + x} ∩ s = ∅ := by
      simpa only [singleton_inter_eq_empty, mem_compl_iff, Ib] using hb