l3lab/ntp-mathlib huggingface dataset:
/-- Given a closed nonempty subset `s` of `Π (n : ℕ), E n`, there exists a Lipschitz retraction
onto this set, i.e., a Lipschitz map with range equal to `s`, equal to the identity on `s`. -/
theorem exists_lipschitz_retraction_of_isClosed {s : Set (∀ n, E n)} (hs : IsClosed s)
    (hne : s.Nonempty) :
    ∃ f : (∀ n, E n) → ∀ n, E n, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f := by
  /- The map `f` is defined as follows. For `x ∈ s`, let `f x = x`. Otherwise, consider the longest
    prefix `w` that `x` shares with an element of `s`, and let `f x = z_w` where `z_w` is an element
    of `s` starting with `w`. All the desired properties are clear, except the fact that `f` is
    `1`-Lipschitz: if two points `x, y` belong to a common cylinder of length `n`, one should show
    that their images also belong to a common cylinder of length `n`. This is a case analysis:
    * if both `x, y ∈ s`, then this is clear.
    * if `x ∈ s` but `y ∉ s`, then the longest prefix `w` of `y` shared by an element of `s` is of
    length at least `n` (because of `x`), and then `f y` starts with `w` and therefore stays in the
    same length `n` cylinder.
    * if `x ∉ s`, `y ∉ s`, let `w` be the longest prefix of `x` shared by an element of `s`. If its
    length is `< n`, then it is also the longest prefix of `y`, and we get `f x = f y = z_w`.
    Otherwise, `f x` remains in the same `n`-cylinder as `x`. Similarly for `y`. Finally, `f x` and
    `f y` are again in the same `n`-cylinder, as desired. -/
  set f := fun x => if x ∈ s then x else (inter_cylinder_longestPrefix_nonempty hs hne x).some
  

Proof state:
E : ℕ → Type u_1
inst✝¹ : (n : ℕ) → TopologicalSpace (E n)
inst✝ : ∀ (n : ℕ), DiscreteTopology (E n)
s : Set ((n : ℕ) → E n)
hs : IsClosed s
hne : Set.Nonempty s
f : ((n : ℕ) → E n) → (n : ℕ) → E n :=
  fun x => if x ∈ s then x else Set.Nonempty.some (_ : Set.Nonempty (s ∩ cylinder x (longestPrefix x s)))
⊢ ∃ f, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f

Next tactic:
have fs : ∀ x ∈ s, f x = x := fun x xs => by simp [xs]
/-- Given a closed nonempty subset `s` of `Π (n : ℕ), E n`, there exists a Lipschitz retraction
onto this set, i.e., a Lipschitz map with range equal to `s`, equal to the identity on `s`. -/
theorem exists_lipschitz_retraction_of_isClosed {s : Set (∀ n, E n)} (hs : IsClosed s)
    (hne : s.Nonempty) :
    ∃ f : (∀ n, E n) → ∀ n, E n, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f := by
  /- The map `f` is defined as follows. For `x ∈ s`, let `f x = x`. Otherwise, consider the longest
    prefix `w` that `x` shares with an element of `s`, and let `f x = z_w` where `z_w` is an element
    of `s` starting with `w`. All the desired properties are clear, except the fact that `f` is
    `1`-Lipschitz: if two points `x, y` belong to a common cylinder of length `n`, one should show
    that their images also belong to a common cylinder of length `n`. This is a case analysis:
    * if both `x, y ∈ s`, then this is clear.
    * if `x ∈ s` but `y ∉ s`, then the longest prefix `w` of `y` shared by an element of `s` is of
    length at least `n` (because of `x`), and then `f y` starts with `w` and therefore stays in the
    same length `n` cylinder.
    * if `x ∉ s`, `y ∉ s`, let `w` be the longest prefix of `x` shared by an element of `s`. If its
    length is `< n`, then it is also the longest prefix of `y`, and we get `f x = f y = z_w`.
    Otherwise, `f x` remains in the same `n`-cylinder as `x`. Similarly for `y`. Finally, `f x` and
    `f y` are again in the same `n`-cylinder, as desired. -/
  set f := fun x => if x ∈ s then x else (inter_cylinder_longestPrefix_nonempty hs hne x).some
  have fs : ∀ x ∈ s, f x = x := fun x xs => by 

Proof state:
E : ℕ → Type u_1
inst✝¹ : (n : ℕ) → TopologicalSpace (E n)
inst✝ : ∀ (n : ℕ), DiscreteTopology (E n)
s : Set ((n : ℕ) → E n)
hs : IsClosed s
hne : Set.Nonempty s
f : ((n : ℕ) → E n) → (n : ℕ) → E n :=
  fun x => if x ∈ s then x else Set.Nonempty.some (_ : Set.Nonempty (s ∩ cylinder x (longestPrefix x s)))
x : (n : ℕ) → E n
xs : x ∈ s
⊢ f x = x

Next tactic:
simp [xs]
/-- Given a closed nonempty subset `s` of `Π (n : ℕ), E n`, there exists a Lipschitz retraction
onto this set, i.e., a Lipschitz map with range equal to `s`, equal to the identity on `s`. -/
theorem exists_lipschitz_retraction_of_isClosed {s : Set (∀ n, E n)} (hs : IsClosed s)
    (hne : s.Nonempty) :
    ∃ f : (∀ n, E n) → ∀ n, E n, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f := by
  /- The map `f` is defined as follows. For `x ∈ s`, let `f x = x`. Otherwise, consider the longest
    prefix `w` that `x` shares with an element of `s`, and let `f x = z_w` where `z_w` is an element
    of `s` starting with `w`. All the desired properties are clear, except the fact that `f` is
    `1`-Lipschitz: if two points `x, y` belong to a common cylinder of length `n`, one should show
    that their images also belong to a common cylinder of length `n`. This is a case analysis:
    * if both `x, y ∈ s`, then this is clear.
    * if `x ∈ s` but `y ∉ s`, then the longest prefix `w` of `y` shared by an element of `s` is of
    length at least `n` (because of `x`), and then `f y` starts with `w` and therefore stays in the
    same length `n` cylinder.
    * if `x ∉ s`, `y ∉ s`, let `w` be the longest prefix of `x` shared by an element of `s`. If its
    length is `< n`, then it is also the longest prefix of `y`, and we get `f x = f y = z_w`.
    Otherwise, `f x` remains in the same `n`-cylinder as `x`. Similarly for `y`. Finally, `f x` and
    `f y` are again in the same `n`-cylinder, as desired. -/
  set f := fun x => if x ∈ s then x else (inter_cylinder_longestPrefix_nonempty hs hne x).some
  have fs : ∀ x ∈ s, f x = x := fun x xs => by simp [xs]
  

Proof state:
E : ℕ → Type u_1
inst✝¹ : (n : ℕ) → TopologicalSpace (E n)
inst✝ : ∀ (n : ℕ), DiscreteTopology (E n)
s : Set ((n : ℕ) → E n)
hs : IsClosed s
hne : Set.Nonempty s
f : ((n : ℕ) → E n) → (n : ℕ) → E n :=
  fun x => if x ∈ s then x else Set.Nonempty.some (_ : Set.Nonempty (s ∩ cylinder x (longestPrefix x s)))
fs : ∀ x ∈ s, f x = x
⊢ ∃ f, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f

Next tactic:
refine' ⟨f, fs, _, _⟩
/-- Given a closed nonempty subset `s` of `Π (n : ℕ), E n`, there exists a Lipschitz retraction
onto this set, i.e., a Lipschitz map with range equal to `s`, equal to the identity on `s`. -/
theorem exists_lipschitz_retraction_of_isClosed {s : Set (∀ n, E n)} (hs : IsClosed s)
    (hne : s.Nonempty) :
    ∃ f : (∀ n, E n) → ∀ n, E n, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f := by
  /- The map `f` is defined as follows. For `x ∈ s`, let `f x = x`. Otherwise, consider the longest
    prefix `w` that `x` shares with an element of `s`, and let `f x = z_w` where `z_w` is an element
    of `s` starting with `w`. All the desired properties are clear, except the fact that `f` is
    `1`-Lipschitz: if two points `x, y` belong to a common cylinder of length `n`, one should show
    that their images also belong to a common cylinder of length `n`. This is a case analysis:
    * if both `x, y ∈ s`, then this is clear.
    * if `x ∈ s` but `y ∉ s`, then the longest prefix `w` of `y` shared by an element of `s` is of
    length at least `n` (because of `x`), and then `f y` starts with `w` and therefore stays in the
    same length `n` cylinder.
    * if `x ∉ s`, `y ∉ s`, let `w` be the longest prefix of `x` shared by an element of `s`. If its
    length is `< n`, then it is also the longest prefix of `y`, and we get `f x = f y = z_w`.
    Otherwise, `f x` remains in the same `n`-cylinder as `x`. Similarly for `y`. Finally, `f x` and
    `f y` are again in the same `n`-cylinder, as desired. -/
  set f := fun x => if x ∈ s then x else (inter_cylinder_longestPrefix_nonempty hs hne x).some
  have fs : ∀ x ∈ s, f x = x := fun x xs => by simp [xs]
  refine' ⟨f, fs, _, _⟩
  -- check that the range of `f` is `s`.
  · 

Proof state:
case refine'_1
E : ℕ → Type u_1
inst✝¹ : (n : ℕ) → TopologicalSpace (E n)
inst✝ : ∀ (n : ℕ), DiscreteTopology (E n)
s : Set ((n : ℕ) → E n)
hs : IsClosed s
hne : Set.Nonempty s
f : ((n : ℕ) → E n) → (n : ℕ) → E n :=
  fun x => if x ∈ s then x else Set.Nonempty.some (_ : Set.Nonempty (s ∩ cylinder x (longestPrefix x s)))
fs : ∀ x ∈ s, f x = x
⊢ range f = s

Next tactic:
apply Subset.antisymm
/-- Given a closed nonempty subset `s` of `Π (n : ℕ), E n`, there exists a Lipschitz retraction
onto this set, i.e., a Lipschitz map with range equal to `s`, equal to the identity on `s`. -/
theorem exists_lipschitz_retraction_of_isClosed {s : Set (∀ n, E n)} (hs : IsClosed s)
    (hne : s.Nonempty) :
    ∃ f : (∀ n, E n) → ∀ n, E n, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f := by
  /- The map `f` is defined as follows. For `x ∈ s`, let `f x = x`. Otherwise, consider the longest
    prefix `w` that `x` shares with an element of `s`, and let `f x = z_w` where `z_w` is an element
    of `s` starting with `w`. All the desired properties are clear, except the fact that `f` is
    `1`-Lipschitz: if two points `x, y` belong to a common cylinder of length `n`, one should show
    that their images also belong to a common cylinder of length `n`. This is a case analysis:
    * if both `x, y ∈ s`, then this is clear.
    * if `x ∈ s` but `y ∉ s`, then the longest prefix `w` of `y` shared by an element of `s` is of
    length at least `n` (because of `x`), and then `f y` starts with `w` and therefore stays in the
    same length `n` cylinder.
    * if `x ∉ s`, `y ∉ s`, let `w` be the longest prefix of `x` shared by an element of `s`. If its
    length is `< n`, then it is also the longest prefix of `y`, and we get `f x = f y = z_w`.
    Otherwise, `f x` remains in the same `n`-cylinder as `x`. Similarly for `y`. Finally, `f x` and
    `f y` are again in the same `n`-cylinder, as desired. -/
  set f := fun x => if x ∈ s then x else (inter_cylinder_longestPrefix_nonempty hs hne x).some
  have fs : ∀ x ∈ s, f x = x := fun x xs => by simp [xs]
  refine' ⟨f, fs, _, _⟩
  -- check that the range of `f` is `s`.
  · apply Subset.antisymm
    · 

Proof state:
case refine'_1.h₁
E : ℕ → Type u_1
inst✝¹ : (n : ℕ) → TopologicalSpace (E n)
inst✝ : ∀ (n : ℕ), DiscreteTopology (E n)
s : Set ((n : ℕ) → E n)
hs : IsClosed s
hne : Set.Nonempty s
f : ((n : ℕ) → E n) → (n : ℕ) → E n :=
  fun x => if x ∈ s then x else Set.Nonempty.some (_ : Set.Nonempty (s ∩ cylinder x (longestPrefix x s)))
fs : ∀ x ∈ s, f x = x
⊢ range f ⊆ s

Next tactic:
rintro x ⟨y, rfl⟩
/-- Given a closed nonempty subset `s` of `Π (n : ℕ), E n`, there exists a Lipschitz retraction
onto this set, i.e., a Lipschitz map with range equal to `s`, equal to the identity on `s`. -/
theorem exists_lipschitz_retraction_of_isClosed {s : Set (∀ n, E n)} (hs : IsClosed s)
    (hne : s.Nonempty) :
    ∃ f : (∀ n, E n) → ∀ n, E n, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f := by
  /- The map `f` is defined as follows. For `x ∈ s`, let `f x = x`. Otherwise, consider the longest
    prefix `w` that `x` shares with an element of `s`, and let `f x = z_w` where `z_w` is an element
    of `s` starting with `w`. All the desired properties are clear, except the fact that `f` is
    `1`-Lipschitz: if two points `x, y` belong to a common cylinder of length `n`, one should show
    that their images also belong to a common cylinder of length `n`. This is a case analysis:
    * if both `x, y ∈ s`, then this is clear.
    * if `x ∈ s` but `y ∉ s`, then the longest prefix `w` of `y` shared by an element of `s` is of
    length at least `n` (because of `x`), and then `f y` starts with `w` and therefore stays in the
    same length `n` cylinder.
    * if `x ∉ s`, `y ∉ s`, let `w` be the longest prefix of `x` shared by an element of `s`. If its
    length is `< n`, then it is also the longest prefix of `y`, and we get `f x = f y = z_w`.
    Otherwise, `f x` remains in the same `n`-cylinder as `x`. Similarly for `y`. Finally, `f x` and
    `f y` are again in the same `n`-cylinder, as desired. -/
  set f := fun x => if x ∈ s then x else (inter_cylinder_longestPrefix_nonempty hs hne x).some
  have fs : ∀ x ∈ s, f x = x := fun x xs => by simp [xs]
  refine' ⟨f, fs, _, _⟩
  -- check that the range of `f` is `s`.
  · apply Subset.antisymm
    · rintro x ⟨y, rfl⟩
      

Proof state:
case refine'_1.h₁.intro
E : ℕ → Type u_1
inst✝¹ : (n : ℕ) → TopologicalSpace (E n)
inst✝ : ∀ (n : ℕ), DiscreteTopology (E n)
s : Set ((n : ℕ) → E n)
hs : IsClosed s
hne : Set.Nonempty s
f : ((n : ℕ) → E n) → (n : ℕ) → E n :=
  fun x => if x ∈ s then x else Set.Nonempty.some (_ : Set.Nonempty (s ∩ cylinder x (longestPrefix x s)))
fs : ∀ x ∈ s, f x = x
y : (n : ℕ) → E n
⊢ f y ∈ s

Next tactic:
by_cases hy : y ∈ s
/-- Given a closed nonempty subset `s` of `Π (n : ℕ), E n`, there exists a Lipschitz retraction
onto this set, i.e., a Lipschitz map with range equal to `s`, equal to the identity on `s`. -/
theorem exists_lipschitz_retraction_of_isClosed {s : Set (∀ n, E n)} (hs : IsClosed s)
    (hne : s.Nonempty) :
    ∃ f : (∀ n, E n) → ∀ n, E n, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f := by
  /- The map `f` is defined as follows. For `x ∈ s`, let `f x = x`. Otherwise, consider the longest
    prefix `w` that `x` shares with an element of `s`, and let `f x = z_w` where `z_w` is an element
    of `s` starting with `w`. All the desired properties are clear, except the fact that `f` is
    `1`-Lipschitz: if two points `x, y` belong to a common cylinder of length `n`, one should show
    that their images also belong to a common cylinder of length `n`. This is a case analysis:
    * if both `x, y ∈ s`, then this is clear.
    * if `x ∈ s` but `y ∉ s`, then the longest prefix `w` of `y` shared by an element of `s` is of
    length at least `n` (because of `x`), and then `f y` starts with `w` and therefore stays in the
    same length `n` cylinder.
    * if `x ∉ s`, `y ∉ s`, let `w` be the longest prefix of `x` shared by an element of `s`. If its
    length is `< n`, then it is also the longest prefix of `y`, and we get `f x = f y = z_w`.
    Otherwise, `f x` remains in the same `n`-cylinder as `x`. Similarly for `y`. Finally, `f x` and
    `f y` are again in the same `n`-cylinder, as desired. -/
  set f := fun x => if x ∈ s then x else (inter_cylinder_longestPrefix_nonempty hs hne x).some
  have fs : ∀ x ∈ s, f x = x := fun x xs => by simp [xs]
  refine' ⟨f, fs, _, _⟩
  -- check that the range of `f` is `s`.
  · apply Subset.antisymm
    · rintro x ⟨y, rfl⟩
      by_cases hy : y ∈ s
      · 

Proof state:
case pos
E : ℕ → Type u_1
inst✝¹ : (n : ℕ) → TopologicalSpace (E n)
inst✝ : ∀ (n : ℕ), DiscreteTopology (E n)
s : Set ((n : ℕ) → E n)
hs : IsClosed s
hne : Set.Nonempty s
f : ((n : ℕ) → E n) → (n : ℕ) → E n :=
  fun x => if x ∈ s then x else Set.Nonempty.some (_ : Set.Nonempty (s ∩ cylinder x (longestPrefix x s)))
fs : ∀ x ∈ s, f x = x
y : (n : ℕ) → E n
hy : y ∈ s
⊢ f y ∈ s

Next tactic:
rwa [fs y hy]
/-- Given a closed nonempty subset `s` of `Π (n : ℕ), E n`, there exists a Lipschitz retraction
onto this set, i.e., a Lipschitz map with range equal to `s`, equal to the identity on `s`. -/
theorem exists_lipschitz_retraction_of_isClosed {s : Set (∀ n, E n)} (hs : IsClosed s)
    (hne : s.Nonempty) :
    ∃ f : (∀ n, E n) → ∀ n, E n, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f := by
  /- The map `f` is defined as follows. For `x ∈ s`, let `f x = x`. Otherwise, consider the longest
    prefix `w` that `x` shares with an element of `s`, and let `f x = z_w` where `z_w` is an element
    of `s` starting with `w`. All the desired properties are clear, except the fact that `f` is
    `1`-Lipschitz: if two points `x, y` belong to a common cylinder of length `n`, one should show
    that their images also belong to a common cylinder of length `n`. This is a case analysis:
    * if both `x, y ∈ s`, then this is clear.
    * if `x ∈ s` but `y ∉ s`, then the longest prefix `w` of `y` shared by an element of `s` is of
    length at least `n` (because of `x`), and then `f y` starts with `w` and therefore stays in the
    same length `n` cylinder.
    * if `x ∉ s`, `y ∉ s`, let `w` be the longest prefix of `x` shared by an element of `s`. If its
    length is `< n`, then it is also the longest prefix of `y`, and we get `f x = f y = z_w`.
    Otherwise, `f x` remains in the same `n`-cylinder as `x`. Similarly for `y`. Finally, `f x` and
    `f y` are again in the same `n`-cylinder, as desired. -/
  set f := fun x => if x ∈ s then x else (inter_cylinder_longestPrefix_nonempty hs hne x).some
  have fs : ∀ x ∈ s, f x = x := fun x xs => by simp [xs]
  refine' ⟨f, fs, _, _⟩
  -- check that the range of `f` is `s`.
  · apply Subset.antisymm
    · rintro x ⟨y, rfl⟩
      by_cases hy : y ∈ s
      · rwa [fs y hy]
      

Proof state:
case neg
E : ℕ → Type u_1
inst✝¹ : (n : ℕ) → TopologicalSpace (E n)
inst✝ : ∀ (n : ℕ), DiscreteTopology (E n)
s : Set ((n : ℕ) → E n)
hs : IsClosed s
hne : Set.Nonempty s
f : ((n : ℕ) → E n) → (n : ℕ) → E n :=
  fun x => if x ∈ s then x else Set.Nonempty.some (_ : Set.Nonempty (s ∩ cylinder x (longestPrefix x s)))
fs : ∀ x ∈ s, f x = x
y : (n : ℕ) → E n
hy : y ∉ s
⊢ f y ∈ s

Next tactic:
simpa [if_neg hy] using (inter_cylinder_longestPrefix_nonempty hs hne y).choose_spec.1
/-- Given a closed nonempty subset `s` of `Π (n : ℕ), E n`, there exists a Lipschitz retraction
onto this set, i.e., a Lipschitz map with range equal to `s`, equal to the identity on `s`. -/
theorem exists_lipschitz_retraction_of_isClosed {s : Set (∀ n, E n)} (hs : IsClosed s)
    (hne : s.Nonempty) :
    ∃ f : (∀ n, E n) → ∀ n, E n, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f := by
  /- The map `f` is defined as follows. For `x ∈ s`, let `f x = x`. Otherwise, consider the longest
    prefix `w` that `x` shares with an element of `s`, and let `f x = z_w` where `z_w` is an element
    of `s` starting with `w`. All the desired properties are clear, except the fact that `f` is
    `1`-Lipschitz: if two points `x, y` belong to a common cylinder of length `n`, one should show
    that their images also belong to a common cylinder of length `n`. This is a case analysis:
    * if both `x, y ∈ s`, then this is clear.
    * if `x ∈ s` but `y ∉ s`, then the longest prefix `w` of `y` shared by an element of `s` is of
    length at least `n` (because of `x`), and then `f y` starts with `w` and therefore stays in the
    same length `n` cylinder.
    * if `x ∉ s`, `y ∉ s`, let `w` be the longest prefix of `x` shared by an element of `s`. If its
    length is `< n`, then it is also the longest prefix of `y`, and we get `f x = f y = z_w`.
    Otherwise, `f x` remains in the same `n`-cylinder as `x`. Similarly for `y`. Finally, `f x` and
    `f y` are again in the same `n`-cylinder, as desired. -/
  set f := fun x => if x ∈ s then x else (inter_cylinder_longestPrefix_nonempty hs hne x).some
  have fs : ∀ x ∈ s, f x = x := fun x xs => by simp [xs]
  refine' ⟨f, fs, _, _⟩
  -- check that the range of `f` is `s`.
  · apply Subset.antisymm
    · rintro x ⟨y, rfl⟩
      by_cases hy : y ∈ s
      · rwa [fs y hy]
      simpa [if_neg hy] using (inter_cylinder_longestPrefix_nonempty hs hne y).choose_spec.1
    · 

Proof state:
case refine'_1.h₂
E : ℕ → Type u_1
inst✝¹ : (n : ℕ) → TopologicalSpace (E n)
inst✝ : ∀ (n : ℕ), DiscreteTopology (E n)
s : Set ((n : ℕ) → E n)
hs : IsClosed s
hne : Set.Nonempty s
f : ((n : ℕ) → E n) → (n : ℕ) → E n :=
  fun x => if x ∈ s then x else Set.Nonempty.some (_ : Set.Nonempty (s ∩ cylinder x (longestPrefix x s)))
fs : ∀ x ∈ s, f x = x
⊢ s ⊆ range f

Next tactic:
intro x hx
/-- Given a closed nonempty subset `s` of `Π (n : ℕ), E n`, there exists a Lipschitz retraction
onto this set, i.e., a Lipschitz map with range equal to `s`, equal to the identity on `s`. -/
theorem exists_lipschitz_retraction_of_isClosed {s : Set (∀ n, E n)} (hs : IsClosed s)
    (hne : s.Nonempty) :
    ∃ f : (∀ n, E n) → ∀ n, E n, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f := by
  /- The map `f` is defined as follows. For `x ∈ s`, let `f x = x`. Otherwise, consider the longest
    prefix `w` that `x` shares with an element of `s`, and let `f x = z_w` where `z_w` is an element
    of `s` starting with `w`. All the desired properties are clear, except the fact that `f` is
    `1`-Lipschitz: if two points `x, y` belong to a common cylinder of length `n`, one should show
    that their images also belong to a common cylinder of length `n`. This is a case analysis:
    * if both `x, y ∈ s`, then this is clear.
    * if `x ∈ s` but `y ∉ s`, then the longest prefix `w` of `y` shared by an element of `s` is of
    length at least `n` (because of `x`), and then `f y` starts with `w` and therefore stays in the
    same length `n` cylinder.
    * if `x ∉ s`, `y ∉ s`, let `w` be the longest prefix of `x` shared by an element of `s`. If its
    length is `< n`, then it is also the longest prefix of `y`, and we get `f x = f y = z_w`.
    Otherwise, `f x` remains in the same `n`-cylinder as `x`. Similarly for `y`. Finally, `f x` and
    `f y` are again in the same `n`-cylinder, as desired. -/
  set f := fun x => if x ∈ s then x else (inter_cylinder_longestPrefix_nonempty hs hne x).some
  have fs : ∀ x ∈ s, f x = x := fun x xs => by simp [xs]
  refine' ⟨f, fs, _, _⟩
  -- check that the range of `f` is `s`.
  · apply Subset.antisymm
    · rintro x ⟨y, rfl⟩
      by_cases hy : y ∈ s
      · rwa [fs y hy]
      simpa [if_neg hy] using (inter_cylinder_longestPrefix_nonempty hs hne y).choose_spec.1
    · intro x hx
      

Proof state:
case refine'_1.h₂
E : ℕ → Type u_1
inst✝¹ : (n : ℕ) → TopologicalSpace (E n)
inst✝ : ∀ (n : ℕ), DiscreteTopology (E n)
s : Set ((n : ℕ) → E n)
hs : IsClosed s
hne : Set.Nonempty s
f : ((n : ℕ) → E n) → (n : ℕ) → E n :=
  fun x => if x ∈ s then x else Set.Nonempty.some (_ : Set.Nonempty (s ∩ cylinder x (longestPrefix x s)))
fs : ∀ x ∈ s, f x = x
x : (n : ℕ) → E n
hx : x ∈ s
⊢ x ∈ range f

Next tactic:
rw [← fs x hx]
/-- Given a closed nonempty subset `s` of `Π (n : ℕ), E n`, there exists a Lipschitz retraction
onto this set, i.e., a Lipschitz map with range equal to `s`, equal to the identity on `s`. -/
theorem exists_lipschitz_retraction_of_isClosed {s : Set (∀ n, E n)} (hs : IsClosed s)
    (hne : s.Nonempty) :
    ∃ f : (∀ n, E n) → ∀ n, E n, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f := by
  /- The map `f` is defined as follows. For `x ∈ s`, let `f x = x`. Otherwise, consider the longest
    prefix `w` that `x` shares with an element of `s`, and let `f x = z_w` where `z_w` is an element
    of `s` starting with `w`. All the desired properties are clear, except the fact that `f` is
    `1`-Lipschitz: if two points `x, y` belong to a common cylinder of length `n`, one should show
    that their images also belong to a common cylinder of length `n`. This is a case analysis:
    * if both `x, y ∈ s`, then this is clear.
    * if `x ∈ s` but `y ∉ s`, then the longest prefix `w` of `y` shared by an element of `s` is of
    length at least `n` (because of `x`), and then `f y` starts with `w` and therefore stays in the
    same length `n` cylinder.
    * if `x ∉ s`, `y ∉ s`, let `w` be the longest prefix of `x` shared by an element of `s`. If its
    length is `< n`, then it is also the longest prefix of `y`, and we get `f x = f y = z_w`.
    Otherwise, `f x` remains in the same `n`-cylinder as `x`. Similarly for `y`. Finally, `f x` and
    `f y` are again in the same `n`-cylinder, as desired. -/
  set f := fun x => if x ∈ s then x else (inter_cylinder_longestPrefix_nonempty hs hne x).some
  have fs : ∀ x ∈ s, f x = x := fun x xs => by simp [xs]
  refine' ⟨f, fs, _, _⟩
  -- check that the range of `f` is `s`.
  · apply Subset.antisymm
    · rintro x ⟨y, rfl⟩
      by_cases hy : y ∈ s
      · rwa [fs y hy]
      simpa [if_neg hy] using (inter_cylinder_longestPrefix_nonempty hs hne y).choose_spec.1
    · intro x hx
      rw [← fs x hx]
      

Proof state:
case refine'_1.h₂
E : ℕ → Type u_1
inst✝¹ : (n : ℕ) → TopologicalSpace (E n)
inst✝ : ∀ (n : ℕ), DiscreteTopology (E n)
s : Set ((n : ℕ) → E n)
hs : IsClosed s
hne : Set.Nonempty s
f : ((n : ℕ) → E n) → (n : ℕ) → E n :=
  fun x => if x ∈ s then x else Set.Nonempty.some (_ : Set.Nonempty (s ∩ cylinder x (longestPrefix x s)))
fs : ∀ x ∈ s, f x = x
x : (n : ℕ) → E n
hx : x ∈ s
⊢ f x ∈ range f

Next tactic:
exact mem_range_self _