l3lab/ntp-mathlib huggingface dataset:
/-- Given a Polish space, and countably many finer Polish topologies, there exists another Polish
topology which is finer than all of them.

Porting note: TODO: the topology `t'` is `t ⊓ ⨅ i, m i`. -/
theorem exists_polishSpace_forall_le {ι : Type*} [Countable ι] [t : TopologicalSpace α]
    [p : PolishSpace α] (m : ι → TopologicalSpace α) (hm : ∀ n, m n ≤ t)
    (h'm : ∀ n, @PolishSpace α (m n)) :
    ∃ t' : TopologicalSpace α, (∀ n, t' ≤ m n) ∧ t' ≤ t ∧ @PolishSpace α t' := by
  rcases isEmpty_or_nonempty ι with (hι | hι)
  · exact ⟨t, fun i => (IsEmpty.elim hι i : _), le_rfl, p⟩
  inhabit ι
  /- Consider the product of infinitely many copies of `α`, each endowed with the topology `m n`.
    This is a Polish space, as a product of Polish spaces. Pulling back this topology under the
    diagonal embedding of `α`, one gets a Polish topology which is finer than all the `m n`. -/
  letI : ∀ n : ι, TopologicalSpace (AuxCopy α n) := fun n => m n
  haveI : ∀ n : ι, PolishSpace (AuxCopy α n) := fun n => h'm n
  letI T : TopologicalSpace (∀ n : ι, AuxCopy α n) := inferInstance
  let f : α → ∀ n : ι, AuxCopy α n := fun x _ => x
  -- show that the induced topology is finer than all the `m n`.
  have T_le_m : ∀ n, T.induced f ≤ m n := fun n ↦ by
    rw [induced_to_pi]
    exact iInf_le_of_le n (@induced_id _ (m n)).le
  

Proof state:
case inr
α : Type u_1
β : Type u_2
ι : Type u_3
inst✝ : Countable ι
t : TopologicalSpace α
p : PolishSpace α
m : ι → TopologicalSpace α
hm : ∀ (n : ι), m n ≤ t
h'm : ∀ (n : ι), PolishSpace α
hι : Nonempty ι
inhabited_h : Inhabited ι
this✝ : (n : ι) → TopologicalSpace (AuxCopy α n) := fun n => m n
this : ∀ (n : ι), PolishSpace (AuxCopy α n)
T : TopologicalSpace ((n : ι) → AuxCopy α n) := inferInstance
f : α → (n : ι) → AuxCopy α n := fun x x_1 => x
T_le_m : ∀ (n : ι), induced f T ≤ m n
⊢ ∃ t', (∀ (n : ι), t' ≤ m n) ∧ t' ≤ t ∧ PolishSpace α

Next tactic:
refine' ⟨T.induced f, fun n => T_le_m n, (T_le_m default).trans (hm default), _⟩
/-- Given a Polish space, and countably many finer Polish topologies, there exists another Polish
topology which is finer than all of them.

Porting note: TODO: the topology `t'` is `t ⊓ ⨅ i, m i`. -/
theorem exists_polishSpace_forall_le {ι : Type*} [Countable ι] [t : TopologicalSpace α]
    [p : PolishSpace α] (m : ι → TopologicalSpace α) (hm : ∀ n, m n ≤ t)
    (h'm : ∀ n, @PolishSpace α (m n)) :
    ∃ t' : TopologicalSpace α, (∀ n, t' ≤ m n) ∧ t' ≤ t ∧ @PolishSpace α t' := by
  rcases isEmpty_or_nonempty ι with (hι | hι)
  · exact ⟨t, fun i => (IsEmpty.elim hι i : _), le_rfl, p⟩
  inhabit ι
  /- Consider the product of infinitely many copies of `α`, each endowed with the topology `m n`.
    This is a Polish space, as a product of Polish spaces. Pulling back this topology under the
    diagonal embedding of `α`, one gets a Polish topology which is finer than all the `m n`. -/
  letI : ∀ n : ι, TopologicalSpace (AuxCopy α n) := fun n => m n
  haveI : ∀ n : ι, PolishSpace (AuxCopy α n) := fun n => h'm n
  letI T : TopologicalSpace (∀ n : ι, AuxCopy α n) := inferInstance
  let f : α → ∀ n : ι, AuxCopy α n := fun x _ => x
  -- show that the induced topology is finer than all the `m n`.
  have T_le_m : ∀ n, T.induced f ≤ m n := fun n ↦ by
    rw [induced_to_pi]
    exact iInf_le_of_le n (@induced_id _ (m n)).le
  refine' ⟨T.induced f, fun n => T_le_m n, (T_le_m default).trans (hm default), _⟩
  -- show that the new topology is Polish, as the pullback of a Polish topology under a closed
  -- embedding.
  

Proof state:
case inr
α : Type u_1
β : Type u_2
ι : Type u_3
inst✝ : Countable ι
t : TopologicalSpace α
p : PolishSpace α
m : ι → TopologicalSpace α
hm : ∀ (n : ι), m n ≤ t
h'm : ∀ (n : ι), PolishSpace α
hι : Nonempty ι
inhabited_h : Inhabited ι
this✝ : (n : ι) → TopologicalSpace (AuxCopy α n) := fun n => m n
this : ∀ (n : ι), PolishSpace (AuxCopy α n)
T : TopologicalSpace ((n : ι) → AuxCopy α n) := inferInstance
f : α → (n : ι) → AuxCopy α n := fun x x_1 => x
T_le_m : ∀ (n : ι), induced f T ≤ m n
⊢ PolishSpace α

Next tactic:
have A : range f = ⋂ n, { x | x n = x default } := by
    ext x
    constructor
    · rintro ⟨y, rfl⟩
      exact mem_iInter.2 fun n => by simp only [mem_setOf_eq]
    · refine fun hx ↦ ⟨x default, ?_⟩
      ext1 n
      symm
      exact mem_iInter.1 hx n
/-- Given a Polish space, and countably many finer Polish topologies, there exists another Polish
topology which is finer than all of them.

Porting note: TODO: the topology `t'` is `t ⊓ ⨅ i, m i`. -/
theorem exists_polishSpace_forall_le {ι : Type*} [Countable ι] [t : TopologicalSpace α]
    [p : PolishSpace α] (m : ι → TopologicalSpace α) (hm : ∀ n, m n ≤ t)
    (h'm : ∀ n, @PolishSpace α (m n)) :
    ∃ t' : TopologicalSpace α, (∀ n, t' ≤ m n) ∧ t' ≤ t ∧ @PolishSpace α t' := by
  rcases isEmpty_or_nonempty ι with (hι | hι)
  · exact ⟨t, fun i => (IsEmpty.elim hι i : _), le_rfl, p⟩
  inhabit ι
  /- Consider the product of infinitely many copies of `α`, each endowed with the topology `m n`.
    This is a Polish space, as a product of Polish spaces. Pulling back this topology under the
    diagonal embedding of `α`, one gets a Polish topology which is finer than all the `m n`. -/
  letI : ∀ n : ι, TopologicalSpace (AuxCopy α n) := fun n => m n
  haveI : ∀ n : ι, PolishSpace (AuxCopy α n) := fun n => h'm n
  letI T : TopologicalSpace (∀ n : ι, AuxCopy α n) := inferInstance
  let f : α → ∀ n : ι, AuxCopy α n := fun x _ => x
  -- show that the induced topology is finer than all the `m n`.
  have T_le_m : ∀ n, T.induced f ≤ m n := fun n ↦ by
    rw [induced_to_pi]
    exact iInf_le_of_le n (@induced_id _ (m n)).le
  refine' ⟨T.induced f, fun n => T_le_m n, (T_le_m default).trans (hm default), _⟩
  -- show that the new topology is Polish, as the pullback of a Polish topology under a closed
  -- embedding.
  have A : range f = ⋂ n, { x | x n = x default } := by
    

Proof state:
α : Type u_1
β : Type u_2
ι : Type u_3
inst✝ : Countable ι
t : TopologicalSpace α
p : PolishSpace α
m : ι → TopologicalSpace α
hm : ∀ (n : ι), m n ≤ t
h'm : ∀ (n : ι), PolishSpace α
hι : Nonempty ι
inhabited_h : Inhabited ι
this✝ : (n : ι) → TopologicalSpace (AuxCopy α n) := fun n => m n
this : ∀ (n : ι), PolishSpace (AuxCopy α n)
T : TopologicalSpace ((n : ι) → AuxCopy α n) := inferInstance
f : α → (n : ι) → AuxCopy α n := fun x x_1 => x
T_le_m : ∀ (n : ι), induced f T ≤ m n
⊢ range f = ⋂ n, {x | x n = x default}

Next tactic:
ext x
/-- Given a Polish space, and countably many finer Polish topologies, there exists another Polish
topology which is finer than all of them.

Porting note: TODO: the topology `t'` is `t ⊓ ⨅ i, m i`. -/
theorem exists_polishSpace_forall_le {ι : Type*} [Countable ι] [t : TopologicalSpace α]
    [p : PolishSpace α] (m : ι → TopologicalSpace α) (hm : ∀ n, m n ≤ t)
    (h'm : ∀ n, @PolishSpace α (m n)) :
    ∃ t' : TopologicalSpace α, (∀ n, t' ≤ m n) ∧ t' ≤ t ∧ @PolishSpace α t' := by
  rcases isEmpty_or_nonempty ι with (hι | hι)
  · exact ⟨t, fun i => (IsEmpty.elim hι i : _), le_rfl, p⟩
  inhabit ι
  /- Consider the product of infinitely many copies of `α`, each endowed with the topology `m n`.
    This is a Polish space, as a product of Polish spaces. Pulling back this topology under the
    diagonal embedding of `α`, one gets a Polish topology which is finer than all the `m n`. -/
  letI : ∀ n : ι, TopologicalSpace (AuxCopy α n) := fun n => m n
  haveI : ∀ n : ι, PolishSpace (AuxCopy α n) := fun n => h'm n
  letI T : TopologicalSpace (∀ n : ι, AuxCopy α n) := inferInstance
  let f : α → ∀ n : ι, AuxCopy α n := fun x _ => x
  -- show that the induced topology is finer than all the `m n`.
  have T_le_m : ∀ n, T.induced f ≤ m n := fun n ↦ by
    rw [induced_to_pi]
    exact iInf_le_of_le n (@induced_id _ (m n)).le
  refine' ⟨T.induced f, fun n => T_le_m n, (T_le_m default).trans (hm default), _⟩
  -- show that the new topology is Polish, as the pullback of a Polish topology under a closed
  -- embedding.
  have A : range f = ⋂ n, { x | x n = x default } := by
    ext x
    

Proof state:
case h
α : Type u_1
β : Type u_2
ι : Type u_3
inst✝ : Countable ι
t : TopologicalSpace α
p : PolishSpace α
m : ι → TopologicalSpace α
hm : ∀ (n : ι), m n ≤ t
h'm : ∀ (n : ι), PolishSpace α
hι : Nonempty ι
inhabited_h : Inhabited ι
this✝ : (n : ι) → TopologicalSpace (AuxCopy α n) := fun n => m n
this : ∀ (n : ι), PolishSpace (AuxCopy α n)
T : TopologicalSpace ((n : ι) → AuxCopy α n) := inferInstance
f : α → (n : ι) → AuxCopy α n := fun x x_1 => x
T_le_m : ∀ (n : ι), induced f T ≤ m n
x : (n : ι) → AuxCopy α n
⊢ x ∈ range f ↔ x ∈ ⋂ n, {x | x n = x default}

Next tactic:
constructor
/-- Given a Polish space, and countably many finer Polish topologies, there exists another Polish
topology which is finer than all of them.

Porting note: TODO: the topology `t'` is `t ⊓ ⨅ i, m i`. -/
theorem exists_polishSpace_forall_le {ι : Type*} [Countable ι] [t : TopologicalSpace α]
    [p : PolishSpace α] (m : ι → TopologicalSpace α) (hm : ∀ n, m n ≤ t)
    (h'm : ∀ n, @PolishSpace α (m n)) :
    ∃ t' : TopologicalSpace α, (∀ n, t' ≤ m n) ∧ t' ≤ t ∧ @PolishSpace α t' := by
  rcases isEmpty_or_nonempty ι with (hι | hι)
  · exact ⟨t, fun i => (IsEmpty.elim hι i : _), le_rfl, p⟩
  inhabit ι
  /- Consider the product of infinitely many copies of `α`, each endowed with the topology `m n`.
    This is a Polish space, as a product of Polish spaces. Pulling back this topology under the
    diagonal embedding of `α`, one gets a Polish topology which is finer than all the `m n`. -/
  letI : ∀ n : ι, TopologicalSpace (AuxCopy α n) := fun n => m n
  haveI : ∀ n : ι, PolishSpace (AuxCopy α n) := fun n => h'm n
  letI T : TopologicalSpace (∀ n : ι, AuxCopy α n) := inferInstance
  let f : α → ∀ n : ι, AuxCopy α n := fun x _ => x
  -- show that the induced topology is finer than all the `m n`.
  have T_le_m : ∀ n, T.induced f ≤ m n := fun n ↦ by
    rw [induced_to_pi]
    exact iInf_le_of_le n (@induced_id _ (m n)).le
  refine' ⟨T.induced f, fun n => T_le_m n, (T_le_m default).trans (hm default), _⟩
  -- show that the new topology is Polish, as the pullback of a Polish topology under a closed
  -- embedding.
  have A : range f = ⋂ n, { x | x n = x default } := by
    ext x
    constructor
    · 

Proof state:
case h.mp
α : Type u_1
β : Type u_2
ι : Type u_3
inst✝ : Countable ι
t : TopologicalSpace α
p : PolishSpace α
m : ι → TopologicalSpace α
hm : ∀ (n : ι), m n ≤ t
h'm : ∀ (n : ι), PolishSpace α
hι : Nonempty ι
inhabited_h : Inhabited ι
this✝ : (n : ι) → TopologicalSpace (AuxCopy α n) := fun n => m n
this : ∀ (n : ι), PolishSpace (AuxCopy α n)
T : TopologicalSpace ((n : ι) → AuxCopy α n) := inferInstance
f : α → (n : ι) → AuxCopy α n := fun x x_1 => x
T_le_m : ∀ (n : ι), induced f T ≤ m n
x : (n : ι) → AuxCopy α n
⊢ x ∈ range f → x ∈ ⋂ n, {x | x n = x default}

Next tactic:
rintro ⟨y, rfl⟩
/-- Given a Polish space, and countably many finer Polish topologies, there exists another Polish
topology which is finer than all of them.

Porting note: TODO: the topology `t'` is `t ⊓ ⨅ i, m i`. -/
theorem exists_polishSpace_forall_le {ι : Type*} [Countable ι] [t : TopologicalSpace α]
    [p : PolishSpace α] (m : ι → TopologicalSpace α) (hm : ∀ n, m n ≤ t)
    (h'm : ∀ n, @PolishSpace α (m n)) :
    ∃ t' : TopologicalSpace α, (∀ n, t' ≤ m n) ∧ t' ≤ t ∧ @PolishSpace α t' := by
  rcases isEmpty_or_nonempty ι with (hι | hι)
  · exact ⟨t, fun i => (IsEmpty.elim hι i : _), le_rfl, p⟩
  inhabit ι
  /- Consider the product of infinitely many copies of `α`, each endowed with the topology `m n`.
    This is a Polish space, as a product of Polish spaces. Pulling back this topology under the
    diagonal embedding of `α`, one gets a Polish topology which is finer than all the `m n`. -/
  letI : ∀ n : ι, TopologicalSpace (AuxCopy α n) := fun n => m n
  haveI : ∀ n : ι, PolishSpace (AuxCopy α n) := fun n => h'm n
  letI T : TopologicalSpace (∀ n : ι, AuxCopy α n) := inferInstance
  let f : α → ∀ n : ι, AuxCopy α n := fun x _ => x
  -- show that the induced topology is finer than all the `m n`.
  have T_le_m : ∀ n, T.induced f ≤ m n := fun n ↦ by
    rw [induced_to_pi]
    exact iInf_le_of_le n (@induced_id _ (m n)).le
  refine' ⟨T.induced f, fun n => T_le_m n, (T_le_m default).trans (hm default), _⟩
  -- show that the new topology is Polish, as the pullback of a Polish topology under a closed
  -- embedding.
  have A : range f = ⋂ n, { x | x n = x default } := by
    ext x
    constructor
    · rintro ⟨y, rfl⟩
      

Proof state:
case h.mp.intro
α : Type u_1
β : Type u_2
ι : Type u_3
inst✝ : Countable ι
t : TopologicalSpace α
p : PolishSpace α
m : ι → TopologicalSpace α
hm : ∀ (n : ι), m n ≤ t
h'm : ∀ (n : ι), PolishSpace α
hι : Nonempty ι
inhabited_h : Inhabited ι
this✝ : (n : ι) → TopologicalSpace (AuxCopy α n) := fun n => m n
this : ∀ (n : ι), PolishSpace (AuxCopy α n)
T : TopologicalSpace ((n : ι) → AuxCopy α n) := inferInstance
f : α → (n : ι) → AuxCopy α n := fun x x_1 => x
T_le_m : ∀ (n : ι), induced f T ≤ m n
y : α
⊢ f y ∈ ⋂ n, {x | x n = x default}

Next tactic:
exact mem_iInter.2 fun n => by simp only [mem_setOf_eq]
/-- Given a Polish space, and countably many finer Polish topologies, there exists another Polish
topology which is finer than all of them.

Porting note: TODO: the topology `t'` is `t ⊓ ⨅ i, m i`. -/
theorem exists_polishSpace_forall_le {ι : Type*} [Countable ι] [t : TopologicalSpace α]
    [p : PolishSpace α] (m : ι → TopologicalSpace α) (hm : ∀ n, m n ≤ t)
    (h'm : ∀ n, @PolishSpace α (m n)) :
    ∃ t' : TopologicalSpace α, (∀ n, t' ≤ m n) ∧ t' ≤ t ∧ @PolishSpace α t' := by
  rcases isEmpty_or_nonempty ι with (hι | hι)
  · exact ⟨t, fun i => (IsEmpty.elim hι i : _), le_rfl, p⟩
  inhabit ι
  /- Consider the product of infinitely many copies of `α`, each endowed with the topology `m n`.
    This is a Polish space, as a product of Polish spaces. Pulling back this topology under the
    diagonal embedding of `α`, one gets a Polish topology which is finer than all the `m n`. -/
  letI : ∀ n : ι, TopologicalSpace (AuxCopy α n) := fun n => m n
  haveI : ∀ n : ι, PolishSpace (AuxCopy α n) := fun n => h'm n
  letI T : TopologicalSpace (∀ n : ι, AuxCopy α n) := inferInstance
  let f : α → ∀ n : ι, AuxCopy α n := fun x _ => x
  -- show that the induced topology is finer than all the `m n`.
  have T_le_m : ∀ n, T.induced f ≤ m n := fun n ↦ by
    rw [induced_to_pi]
    exact iInf_le_of_le n (@induced_id _ (m n)).le
  refine' ⟨T.induced f, fun n => T_le_m n, (T_le_m default).trans (hm default), _⟩
  -- show that the new topology is Polish, as the pullback of a Polish topology under a closed
  -- embedding.
  have A : range f = ⋂ n, { x | x n = x default } := by
    ext x
    constructor
    · rintro ⟨y, rfl⟩
      exact mem_iInter.2 fun n => by 

Proof state:
α : Type u_1
β : Type u_2
ι : Type u_3
inst✝ : Countable ι
t : TopologicalSpace α
p : PolishSpace α
m : ι → TopologicalSpace α
hm : ∀ (n : ι), m n ≤ t
h'm : ∀ (n : ι), PolishSpace α
hι : Nonempty ι
inhabited_h : Inhabited ι
this✝ : (n : ι) → TopologicalSpace (AuxCopy α n) := fun n => m n
this : ∀ (n : ι), PolishSpace (AuxCopy α n)
T : TopologicalSpace ((n : ι) → AuxCopy α n) := inferInstance
f : α → (n : ι) → AuxCopy α n := fun x x_1 => x
T_le_m : ∀ (n : ι), induced f T ≤ m n
y : α
n : ι
⊢ f y ∈ {x | x n = x default}

Next tactic:
simp only [mem_setOf_eq]
/-- Given a Polish space, and countably many finer Polish topologies, there exists another Polish
topology which is finer than all of them.

Porting note: TODO: the topology `t'` is `t ⊓ ⨅ i, m i`. -/
theorem exists_polishSpace_forall_le {ι : Type*} [Countable ι] [t : TopologicalSpace α]
    [p : PolishSpace α] (m : ι → TopologicalSpace α) (hm : ∀ n, m n ≤ t)
    (h'm : ∀ n, @PolishSpace α (m n)) :
    ∃ t' : TopologicalSpace α, (∀ n, t' ≤ m n) ∧ t' ≤ t ∧ @PolishSpace α t' := by
  rcases isEmpty_or_nonempty ι with (hι | hι)
  · exact ⟨t, fun i => (IsEmpty.elim hι i : _), le_rfl, p⟩
  inhabit ι
  /- Consider the product of infinitely many copies of `α`, each endowed with the topology `m n`.
    This is a Polish space, as a product of Polish spaces. Pulling back this topology under the
    diagonal embedding of `α`, one gets a Polish topology which is finer than all the `m n`. -/
  letI : ∀ n : ι, TopologicalSpace (AuxCopy α n) := fun n => m n
  haveI : ∀ n : ι, PolishSpace (AuxCopy α n) := fun n => h'm n
  letI T : TopologicalSpace (∀ n : ι, AuxCopy α n) := inferInstance
  let f : α → ∀ n : ι, AuxCopy α n := fun x _ => x
  -- show that the induced topology is finer than all the `m n`.
  have T_le_m : ∀ n, T.induced f ≤ m n := fun n ↦ by
    rw [induced_to_pi]
    exact iInf_le_of_le n (@induced_id _ (m n)).le
  refine' ⟨T.induced f, fun n => T_le_m n, (T_le_m default).trans (hm default), _⟩
  -- show that the new topology is Polish, as the pullback of a Polish topology under a closed
  -- embedding.
  have A : range f = ⋂ n, { x | x n = x default } := by
    ext x
    constructor
    · rintro ⟨y, rfl⟩
      exact mem_iInter.2 fun n => by simp only [mem_setOf_eq]
    · 

Proof state:
case h.mpr
α : Type u_1
β : Type u_2
ι : Type u_3
inst✝ : Countable ι
t : TopologicalSpace α
p : PolishSpace α
m : ι → TopologicalSpace α
hm : ∀ (n : ι), m n ≤ t
h'm : ∀ (n : ι), PolishSpace α
hι : Nonempty ι
inhabited_h : Inhabited ι
this✝ : (n : ι) → TopologicalSpace (AuxCopy α n) := fun n => m n
this : ∀ (n : ι), PolishSpace (AuxCopy α n)
T : TopologicalSpace ((n : ι) → AuxCopy α n) := inferInstance
f : α → (n : ι) → AuxCopy α n := fun x x_1 => x
T_le_m : ∀ (n : ι), induced f T ≤ m n
x : (n : ι) → AuxCopy α n
⊢ x ∈ ⋂ n, {x | x n = x default} → x ∈ range f

Next tactic:
refine fun hx ↦ ⟨x default, ?_⟩
/-- Given a Polish space, and countably many finer Polish topologies, there exists another Polish
topology which is finer than all of them.

Porting note: TODO: the topology `t'` is `t ⊓ ⨅ i, m i`. -/
theorem exists_polishSpace_forall_le {ι : Type*} [Countable ι] [t : TopologicalSpace α]
    [p : PolishSpace α] (m : ι → TopologicalSpace α) (hm : ∀ n, m n ≤ t)
    (h'm : ∀ n, @PolishSpace α (m n)) :
    ∃ t' : TopologicalSpace α, (∀ n, t' ≤ m n) ∧ t' ≤ t ∧ @PolishSpace α t' := by
  rcases isEmpty_or_nonempty ι with (hι | hι)
  · exact ⟨t, fun i => (IsEmpty.elim hι i : _), le_rfl, p⟩
  inhabit ι
  /- Consider the product of infinitely many copies of `α`, each endowed with the topology `m n`.
    This is a Polish space, as a product of Polish spaces. Pulling back this topology under the
    diagonal embedding of `α`, one gets a Polish topology which is finer than all the `m n`. -/
  letI : ∀ n : ι, TopologicalSpace (AuxCopy α n) := fun n => m n
  haveI : ∀ n : ι, PolishSpace (AuxCopy α n) := fun n => h'm n
  letI T : TopologicalSpace (∀ n : ι, AuxCopy α n) := inferInstance
  let f : α → ∀ n : ι, AuxCopy α n := fun x _ => x
  -- show that the induced topology is finer than all the `m n`.
  have T_le_m : ∀ n, T.induced f ≤ m n := fun n ↦ by
    rw [induced_to_pi]
    exact iInf_le_of_le n (@induced_id _ (m n)).le
  refine' ⟨T.induced f, fun n => T_le_m n, (T_le_m default).trans (hm default), _⟩
  -- show that the new topology is Polish, as the pullback of a Polish topology under a closed
  -- embedding.
  have A : range f = ⋂ n, { x | x n = x default } := by
    ext x
    constructor
    · rintro ⟨y, rfl⟩
      exact mem_iInter.2 fun n => by simp only [mem_setOf_eq]
    · refine fun hx ↦ ⟨x default, ?_⟩
      

Proof state:
case h.mpr
α : Type u_1
β : Type u_2
ι : Type u_3
inst✝ : Countable ι
t : TopologicalSpace α
p : PolishSpace α
m : ι → TopologicalSpace α
hm : ∀ (n : ι), m n ≤ t
h'm : ∀ (n : ι), PolishSpace α
hι : Nonempty ι
inhabited_h : Inhabited ι
this✝ : (n : ι) → TopologicalSpace (AuxCopy α n) := fun n => m n
this : ∀ (n : ι), PolishSpace (AuxCopy α n)
T : TopologicalSpace ((n : ι) → AuxCopy α n) := inferInstance
f : α → (n : ι) → AuxCopy α n := fun x x_1 => x
T_le_m : ∀ (n : ι), induced f T ≤ m n
x : (n : ι) → AuxCopy α n
hx : x ∈ ⋂ n, {x | x n = x default}
⊢ f (x default) = x

Next tactic:
ext1 n
/-- Given a Polish space, and countably many finer Polish topologies, there exists another Polish
topology which is finer than all of them.

Porting note: TODO: the topology `t'` is `t ⊓ ⨅ i, m i`. -/
theorem exists_polishSpace_forall_le {ι : Type*} [Countable ι] [t : TopologicalSpace α]
    [p : PolishSpace α] (m : ι → TopologicalSpace α) (hm : ∀ n, m n ≤ t)
    (h'm : ∀ n, @PolishSpace α (m n)) :
    ∃ t' : TopologicalSpace α, (∀ n, t' ≤ m n) ∧ t' ≤ t ∧ @PolishSpace α t' := by
  rcases isEmpty_or_nonempty ι with (hι | hι)
  · exact ⟨t, fun i => (IsEmpty.elim hι i : _), le_rfl, p⟩
  inhabit ι
  /- Consider the product of infinitely many copies of `α`, each endowed with the topology `m n`.
    This is a Polish space, as a product of Polish spaces. Pulling back this topology under the
    diagonal embedding of `α`, one gets a Polish topology which is finer than all the `m n`. -/
  letI : ∀ n : ι, TopologicalSpace (AuxCopy α n) := fun n => m n
  haveI : ∀ n : ι, PolishSpace (AuxCopy α n) := fun n => h'm n
  letI T : TopologicalSpace (∀ n : ι, AuxCopy α n) := inferInstance
  let f : α → ∀ n : ι, AuxCopy α n := fun x _ => x
  -- show that the induced topology is finer than all the `m n`.
  have T_le_m : ∀ n, T.induced f ≤ m n := fun n ↦ by
    rw [induced_to_pi]
    exact iInf_le_of_le n (@induced_id _ (m n)).le
  refine' ⟨T.induced f, fun n => T_le_m n, (T_le_m default).trans (hm default), _⟩
  -- show that the new topology is Polish, as the pullback of a Polish topology under a closed
  -- embedding.
  have A : range f = ⋂ n, { x | x n = x default } := by
    ext x
    constructor
    · rintro ⟨y, rfl⟩
      exact mem_iInter.2 fun n => by simp only [mem_setOf_eq]
    · refine fun hx ↦ ⟨x default, ?_⟩
      ext1 n
      

Proof state:
case h.mpr.h
α : Type u_1
β : Type u_2
ι : Type u_3
inst✝ : Countable ι
t : TopologicalSpace α
p : PolishSpace α
m : ι → TopologicalSpace α
hm : ∀ (n : ι), m n ≤ t
h'm : ∀ (n : ι), PolishSpace α
hι : Nonempty ι
inhabited_h : Inhabited ι
this✝ : (n : ι) → TopologicalSpace (AuxCopy α n) := fun n => m n
this : ∀ (n : ι), PolishSpace (AuxCopy α n)
T : TopologicalSpace ((n : ι) → AuxCopy α n) := inferInstance
f : α → (n : ι) → AuxCopy α n := fun x x_1 => x
T_le_m : ∀ (n : ι), induced f T ≤ m n
x : (n : ι) → AuxCopy α n
hx : x ∈ ⋂ n, {x | x n = x default}
n : ι
⊢ f (x default) n = x n

Next tactic:
symm