l3lab/ntp-mathlib huggingface dataset:
/-- In a topological vector space over a nontrivially normed field, any neighborhood of zero has
the same cardinality as the whole space.

See also `cardinal_eq_of_mem_nhds`. -/
lemma cardinal_eq_of_mem_nhds_zero
    {E : Type*} (𝕜 : Type*) [NontriviallyNormedField 𝕜] [AddCommGroup E] [Module 𝕜 E]
    [TopologicalSpace E] [ContinuousSMul 𝕜 E] {s : Set E} (hs : s ∈ 𝓝 (0 : E)) : #s = #E := by
  /- As `s` is a neighborhood of `0`, the space is covered by the rescaled sets `c^n • s`,
  where `c` is any element of `𝕜` with norm `> 1`. All these sets are in bijection and have
  therefore the same cardinality. The conclusion follows. -/
  obtain ⟨c, hc⟩ : ∃ x : 𝕜 , 1 < ‖x‖ := NormedField.exists_lt_norm 𝕜 1
  have cn_ne : ∀ n, c^n ≠ 0 := by
    intro n
    apply pow_ne_zero
    rintro rfl
    simp only [norm_zero] at hc
    exact lt_irrefl _ (hc.trans zero_lt_one)
  have A : ∀ (x : E), ∀ᶠ n in (atTop : Filter ℕ), x ∈ c^n • s := by
    intro x
    have : Tendsto (fun n ↦ (c^n) ⁻¹ • x) atTop (𝓝 ((0 : 𝕜) • x)) := by
      

Proof state:
E : Type u_1
𝕜 : Type u_2
inst✝⁴ : NontriviallyNormedField 𝕜
inst✝³ : AddCommGroup E
inst✝² : Module 𝕜 E
inst✝¹ : TopologicalSpace E
inst✝ : ContinuousSMul 𝕜 E
s : Set E
hs : s ∈ 𝓝 0
c : 𝕜
hc : 1 < ‖c‖
cn_ne : ∀ (n : ℕ), c ^ n ≠ 0
x : E
⊢ Tendsto (fun n => (c ^ n)⁻¹ • x) atTop (𝓝 (0 • x))

Next tactic:
have : Tendsto (fun n ↦ (c^n)⁻¹) atTop (𝓝 0) := by
        simp_rw [← inv_pow]
        apply tendsto_pow_atTop_nhds_0_of_norm_lt_1
        rw [norm_inv]
        exact inv_lt_one hc
/-- In a topological vector space over a nontrivially normed field, any neighborhood of zero has
the same cardinality as the whole space.

See also `cardinal_eq_of_mem_nhds`. -/
lemma cardinal_eq_of_mem_nhds_zero
    {E : Type*} (𝕜 : Type*) [NontriviallyNormedField 𝕜] [AddCommGroup E] [Module 𝕜 E]
    [TopologicalSpace E] [ContinuousSMul 𝕜 E] {s : Set E} (hs : s ∈ 𝓝 (0 : E)) : #s = #E := by
  /- As `s` is a neighborhood of `0`, the space is covered by the rescaled sets `c^n • s`,
  where `c` is any element of `𝕜` with norm `> 1`. All these sets are in bijection and have
  therefore the same cardinality. The conclusion follows. -/
  obtain ⟨c, hc⟩ : ∃ x : 𝕜 , 1 < ‖x‖ := NormedField.exists_lt_norm 𝕜 1
  have cn_ne : ∀ n, c^n ≠ 0 := by
    intro n
    apply pow_ne_zero
    rintro rfl
    simp only [norm_zero] at hc
    exact lt_irrefl _ (hc.trans zero_lt_one)
  have A : ∀ (x : E), ∀ᶠ n in (atTop : Filter ℕ), x ∈ c^n • s := by
    intro x
    have : Tendsto (fun n ↦ (c^n) ⁻¹ • x) atTop (𝓝 ((0 : 𝕜) • x)) := by
      have : Tendsto (fun n ↦ (c^n)⁻¹) atTop (𝓝 0) := by
        

Proof state:
E : Type u_1
𝕜 : Type u_2
inst✝⁴ : NontriviallyNormedField 𝕜
inst✝³ : AddCommGroup E
inst✝² : Module 𝕜 E
inst✝¹ : TopologicalSpace E
inst✝ : ContinuousSMul 𝕜 E
s : Set E
hs : s ∈ 𝓝 0
c : 𝕜
hc : 1 < ‖c‖
cn_ne : ∀ (n : ℕ), c ^ n ≠ 0
x : E
⊢ Tendsto (fun n => (c ^ n)⁻¹) atTop (𝓝 0)

Next tactic:
simp_rw [← inv_pow]
/-- In a topological vector space over a nontrivially normed field, any neighborhood of zero has
the same cardinality as the whole space.

See also `cardinal_eq_of_mem_nhds`. -/
lemma cardinal_eq_of_mem_nhds_zero
    {E : Type*} (𝕜 : Type*) [NontriviallyNormedField 𝕜] [AddCommGroup E] [Module 𝕜 E]
    [TopologicalSpace E] [ContinuousSMul 𝕜 E] {s : Set E} (hs : s ∈ 𝓝 (0 : E)) : #s = #E := by
  /- As `s` is a neighborhood of `0`, the space is covered by the rescaled sets `c^n • s`,
  where `c` is any element of `𝕜` with norm `> 1`. All these sets are in bijection and have
  therefore the same cardinality. The conclusion follows. -/
  obtain ⟨c, hc⟩ : ∃ x : 𝕜 , 1 < ‖x‖ := NormedField.exists_lt_norm 𝕜 1
  have cn_ne : ∀ n, c^n ≠ 0 := by
    intro n
    apply pow_ne_zero
    rintro rfl
    simp only [norm_zero] at hc
    exact lt_irrefl _ (hc.trans zero_lt_one)
  have A : ∀ (x : E), ∀ᶠ n in (atTop : Filter ℕ), x ∈ c^n • s := by
    intro x
    have : Tendsto (fun n ↦ (c^n) ⁻¹ • x) atTop (𝓝 ((0 : 𝕜) • x)) := by
      have : Tendsto (fun n ↦ (c^n)⁻¹) atTop (𝓝 0) := by
        simp_rw [← inv_pow]
        

Proof state:
E : Type u_1
𝕜 : Type u_2
inst✝⁴ : NontriviallyNormedField 𝕜
inst✝³ : AddCommGroup E
inst✝² : Module 𝕜 E
inst✝¹ : TopologicalSpace E
inst✝ : ContinuousSMul 𝕜 E
s : Set E
hs : s ∈ 𝓝 0
c : 𝕜
hc : 1 < ‖c‖
cn_ne : ∀ (n : ℕ), c ^ n ≠ 0
x : E
⊢ Tendsto (fun n => c⁻¹ ^ n) atTop (𝓝 0)

Next tactic:
apply tendsto_pow_atTop_nhds_0_of_norm_lt_1
/-- In a topological vector space over a nontrivially normed field, any neighborhood of zero has
the same cardinality as the whole space.

See also `cardinal_eq_of_mem_nhds`. -/
lemma cardinal_eq_of_mem_nhds_zero
    {E : Type*} (𝕜 : Type*) [NontriviallyNormedField 𝕜] [AddCommGroup E] [Module 𝕜 E]
    [TopologicalSpace E] [ContinuousSMul 𝕜 E] {s : Set E} (hs : s ∈ 𝓝 (0 : E)) : #s = #E := by
  /- As `s` is a neighborhood of `0`, the space is covered by the rescaled sets `c^n • s`,
  where `c` is any element of `𝕜` with norm `> 1`. All these sets are in bijection and have
  therefore the same cardinality. The conclusion follows. -/
  obtain ⟨c, hc⟩ : ∃ x : 𝕜 , 1 < ‖x‖ := NormedField.exists_lt_norm 𝕜 1
  have cn_ne : ∀ n, c^n ≠ 0 := by
    intro n
    apply pow_ne_zero
    rintro rfl
    simp only [norm_zero] at hc
    exact lt_irrefl _ (hc.trans zero_lt_one)
  have A : ∀ (x : E), ∀ᶠ n in (atTop : Filter ℕ), x ∈ c^n • s := by
    intro x
    have : Tendsto (fun n ↦ (c^n) ⁻¹ • x) atTop (𝓝 ((0 : 𝕜) • x)) := by
      have : Tendsto (fun n ↦ (c^n)⁻¹) atTop (𝓝 0) := by
        simp_rw [← inv_pow]
        apply tendsto_pow_atTop_nhds_0_of_norm_lt_1
        

Proof state:
case h
E : Type u_1
𝕜 : Type u_2
inst✝⁴ : NontriviallyNormedField 𝕜
inst✝³ : AddCommGroup E
inst✝² : Module 𝕜 E
inst✝¹ : TopologicalSpace E
inst✝ : ContinuousSMul 𝕜 E
s : Set E
hs : s ∈ 𝓝 0
c : 𝕜
hc : 1 < ‖c‖
cn_ne : ∀ (n : ℕ), c ^ n ≠ 0
x : E
⊢ ‖c⁻¹‖ < 1

Next tactic:
rw [norm_inv]
/-- In a topological vector space over a nontrivially normed field, any neighborhood of zero has
the same cardinality as the whole space.

See also `cardinal_eq_of_mem_nhds`. -/
lemma cardinal_eq_of_mem_nhds_zero
    {E : Type*} (𝕜 : Type*) [NontriviallyNormedField 𝕜] [AddCommGroup E] [Module 𝕜 E]
    [TopologicalSpace E] [ContinuousSMul 𝕜 E] {s : Set E} (hs : s ∈ 𝓝 (0 : E)) : #s = #E := by
  /- As `s` is a neighborhood of `0`, the space is covered by the rescaled sets `c^n • s`,
  where `c` is any element of `𝕜` with norm `> 1`. All these sets are in bijection and have
  therefore the same cardinality. The conclusion follows. -/
  obtain ⟨c, hc⟩ : ∃ x : 𝕜 , 1 < ‖x‖ := NormedField.exists_lt_norm 𝕜 1
  have cn_ne : ∀ n, c^n ≠ 0 := by
    intro n
    apply pow_ne_zero
    rintro rfl
    simp only [norm_zero] at hc
    exact lt_irrefl _ (hc.trans zero_lt_one)
  have A : ∀ (x : E), ∀ᶠ n in (atTop : Filter ℕ), x ∈ c^n • s := by
    intro x
    have : Tendsto (fun n ↦ (c^n) ⁻¹ • x) atTop (𝓝 ((0 : 𝕜) • x)) := by
      have : Tendsto (fun n ↦ (c^n)⁻¹) atTop (𝓝 0) := by
        simp_rw [← inv_pow]
        apply tendsto_pow_atTop_nhds_0_of_norm_lt_1
        rw [norm_inv]
        

Proof state:
case h
E : Type u_1
𝕜 : Type u_2
inst✝⁴ : NontriviallyNormedField 𝕜
inst✝³ : AddCommGroup E
inst✝² : Module 𝕜 E
inst✝¹ : TopologicalSpace E
inst✝ : ContinuousSMul 𝕜 E
s : Set E
hs : s ∈ 𝓝 0
c : 𝕜
hc : 1 < ‖c‖
cn_ne : ∀ (n : ℕ), c ^ n ≠ 0
x : E
⊢ ‖c‖⁻¹ < 1

Next tactic:
exact inv_lt_one hc
/-- In a topological vector space over a nontrivially normed field, any neighborhood of zero has
the same cardinality as the whole space.

See also `cardinal_eq_of_mem_nhds`. -/
lemma cardinal_eq_of_mem_nhds_zero
    {E : Type*} (𝕜 : Type*) [NontriviallyNormedField 𝕜] [AddCommGroup E] [Module 𝕜 E]
    [TopologicalSpace E] [ContinuousSMul 𝕜 E] {s : Set E} (hs : s ∈ 𝓝 (0 : E)) : #s = #E := by
  /- As `s` is a neighborhood of `0`, the space is covered by the rescaled sets `c^n • s`,
  where `c` is any element of `𝕜` with norm `> 1`. All these sets are in bijection and have
  therefore the same cardinality. The conclusion follows. -/
  obtain ⟨c, hc⟩ : ∃ x : 𝕜 , 1 < ‖x‖ := NormedField.exists_lt_norm 𝕜 1
  have cn_ne : ∀ n, c^n ≠ 0 := by
    intro n
    apply pow_ne_zero
    rintro rfl
    simp only [norm_zero] at hc
    exact lt_irrefl _ (hc.trans zero_lt_one)
  have A : ∀ (x : E), ∀ᶠ n in (atTop : Filter ℕ), x ∈ c^n • s := by
    intro x
    have : Tendsto (fun n ↦ (c^n) ⁻¹ • x) atTop (𝓝 ((0 : 𝕜) • x)) := by
      have : Tendsto (fun n ↦ (c^n)⁻¹) atTop (𝓝 0) := by
        simp_rw [← inv_pow]
        apply tendsto_pow_atTop_nhds_0_of_norm_lt_1
        rw [norm_inv]
        exact inv_lt_one hc
      

Proof state:
E : Type u_1
𝕜 : Type u_2
inst✝⁴ : NontriviallyNormedField 𝕜
inst✝³ : AddCommGroup E
inst✝² : Module 𝕜 E
inst✝¹ : TopologicalSpace E
inst✝ : ContinuousSMul 𝕜 E
s : Set E
hs : s ∈ 𝓝 0
c : 𝕜
hc : 1 < ‖c‖
cn_ne : ∀ (n : ℕ), c ^ n ≠ 0
x : E
this : Tendsto (fun n => (c ^ n)⁻¹) atTop (𝓝 0)
⊢ Tendsto (fun n => (c ^ n)⁻¹ • x) atTop (𝓝 (0 • x))

Next tactic:
exact Tendsto.smul_const this x
/-- In a topological vector space over a nontrivially normed field, any neighborhood of zero has
the same cardinality as the whole space.

See also `cardinal_eq_of_mem_nhds`. -/
lemma cardinal_eq_of_mem_nhds_zero
    {E : Type*} (𝕜 : Type*) [NontriviallyNormedField 𝕜] [AddCommGroup E] [Module 𝕜 E]
    [TopologicalSpace E] [ContinuousSMul 𝕜 E] {s : Set E} (hs : s ∈ 𝓝 (0 : E)) : #s = #E := by
  /- As `s` is a neighborhood of `0`, the space is covered by the rescaled sets `c^n • s`,
  where `c` is any element of `𝕜` with norm `> 1`. All these sets are in bijection and have
  therefore the same cardinality. The conclusion follows. -/
  obtain ⟨c, hc⟩ : ∃ x : 𝕜 , 1 < ‖x‖ := NormedField.exists_lt_norm 𝕜 1
  have cn_ne : ∀ n, c^n ≠ 0 := by
    intro n
    apply pow_ne_zero
    rintro rfl
    simp only [norm_zero] at hc
    exact lt_irrefl _ (hc.trans zero_lt_one)
  have A : ∀ (x : E), ∀ᶠ n in (atTop : Filter ℕ), x ∈ c^n • s := by
    intro x
    have : Tendsto (fun n ↦ (c^n) ⁻¹ • x) atTop (𝓝 ((0 : 𝕜) • x)) := by
      have : Tendsto (fun n ↦ (c^n)⁻¹) atTop (𝓝 0) := by
        simp_rw [← inv_pow]
        apply tendsto_pow_atTop_nhds_0_of_norm_lt_1
        rw [norm_inv]
        exact inv_lt_one hc
      exact Tendsto.smul_const this x
    

Proof state:
E : Type u_1
𝕜 : Type u_2
inst✝⁴ : NontriviallyNormedField 𝕜
inst✝³ : AddCommGroup E
inst✝² : Module 𝕜 E
inst✝¹ : TopologicalSpace E
inst✝ : ContinuousSMul 𝕜 E
s : Set E
hs : s ∈ 𝓝 0
c : 𝕜
hc : 1 < ‖c‖
cn_ne : ∀ (n : ℕ), c ^ n ≠ 0
x : E
this : Tendsto (fun n => (c ^ n)⁻¹ • x) atTop (𝓝 (0 • x))
⊢ ∀ᶠ (n : ℕ) in atTop, x ∈ c ^ n • s

Next tactic:
rw [zero_smul] at this
/-- In a topological vector space over a nontrivially normed field, any neighborhood of zero has
the same cardinality as the whole space.

See also `cardinal_eq_of_mem_nhds`. -/
lemma cardinal_eq_of_mem_nhds_zero
    {E : Type*} (𝕜 : Type*) [NontriviallyNormedField 𝕜] [AddCommGroup E] [Module 𝕜 E]
    [TopologicalSpace E] [ContinuousSMul 𝕜 E] {s : Set E} (hs : s ∈ 𝓝 (0 : E)) : #s = #E := by
  /- As `s` is a neighborhood of `0`, the space is covered by the rescaled sets `c^n • s`,
  where `c` is any element of `𝕜` with norm `> 1`. All these sets are in bijection and have
  therefore the same cardinality. The conclusion follows. -/
  obtain ⟨c, hc⟩ : ∃ x : 𝕜 , 1 < ‖x‖ := NormedField.exists_lt_norm 𝕜 1
  have cn_ne : ∀ n, c^n ≠ 0 := by
    intro n
    apply pow_ne_zero
    rintro rfl
    simp only [norm_zero] at hc
    exact lt_irrefl _ (hc.trans zero_lt_one)
  have A : ∀ (x : E), ∀ᶠ n in (atTop : Filter ℕ), x ∈ c^n • s := by
    intro x
    have : Tendsto (fun n ↦ (c^n) ⁻¹ • x) atTop (𝓝 ((0 : 𝕜) • x)) := by
      have : Tendsto (fun n ↦ (c^n)⁻¹) atTop (𝓝 0) := by
        simp_rw [← inv_pow]
        apply tendsto_pow_atTop_nhds_0_of_norm_lt_1
        rw [norm_inv]
        exact inv_lt_one hc
      exact Tendsto.smul_const this x
    rw [zero_smul] at this
    

Proof state:
E : Type u_1
𝕜 : Type u_2
inst✝⁴ : NontriviallyNormedField 𝕜
inst✝³ : AddCommGroup E
inst✝² : Module 𝕜 E
inst✝¹ : TopologicalSpace E
inst✝ : ContinuousSMul 𝕜 E
s : Set E
hs : s ∈ 𝓝 0
c : 𝕜
hc : 1 < ‖c‖
cn_ne : ∀ (n : ℕ), c ^ n ≠ 0
x : E
this : Tendsto (fun n => (c ^ n)⁻¹ • x) atTop (𝓝 0)
⊢ ∀ᶠ (n : ℕ) in atTop, x ∈ c ^ n • s

Next tactic:
filter_upwards [this hs] with n (hn : (c ^ n)⁻¹ • x ∈ s)
/-- In a topological vector space over a nontrivially normed field, any neighborhood of zero has
the same cardinality as the whole space.

See also `cardinal_eq_of_mem_nhds`. -/
lemma cardinal_eq_of_mem_nhds_zero
    {E : Type*} (𝕜 : Type*) [NontriviallyNormedField 𝕜] [AddCommGroup E] [Module 𝕜 E]
    [TopologicalSpace E] [ContinuousSMul 𝕜 E] {s : Set E} (hs : s ∈ 𝓝 (0 : E)) : #s = #E := by
  /- As `s` is a neighborhood of `0`, the space is covered by the rescaled sets `c^n • s`,
  where `c` is any element of `𝕜` with norm `> 1`. All these sets are in bijection and have
  therefore the same cardinality. The conclusion follows. -/
  obtain ⟨c, hc⟩ : ∃ x : 𝕜 , 1 < ‖x‖ := NormedField.exists_lt_norm 𝕜 1
  have cn_ne : ∀ n, c^n ≠ 0 := by
    intro n
    apply pow_ne_zero
    rintro rfl
    simp only [norm_zero] at hc
    exact lt_irrefl _ (hc.trans zero_lt_one)
  have A : ∀ (x : E), ∀ᶠ n in (atTop : Filter ℕ), x ∈ c^n • s := by
    intro x
    have : Tendsto (fun n ↦ (c^n) ⁻¹ • x) atTop (𝓝 ((0 : 𝕜) • x)) := by
      have : Tendsto (fun n ↦ (c^n)⁻¹) atTop (𝓝 0) := by
        simp_rw [← inv_pow]
        apply tendsto_pow_atTop_nhds_0_of_norm_lt_1
        rw [norm_inv]
        exact inv_lt_one hc
      exact Tendsto.smul_const this x
    rw [zero_smul] at this
    filter_upwards [this hs] with n (hn : (c ^ n)⁻¹ • x ∈ s)
    

Proof state:
case h
E : Type u_1
𝕜 : Type u_2
inst✝⁴ : NontriviallyNormedField 𝕜
inst✝³ : AddCommGroup E
inst✝² : Module 𝕜 E
inst✝¹ : TopologicalSpace E
inst✝ : ContinuousSMul 𝕜 E
s : Set E
hs : s ∈ 𝓝 0
c : 𝕜
hc : 1 < ‖c‖
cn_ne : ∀ (n : ℕ), c ^ n ≠ 0
x : E
this : Tendsto (fun n => (c ^ n)⁻¹ • x) atTop (𝓝 0)
n : ℕ
hn : (c ^ n)⁻¹ • x ∈ s
⊢ x ∈ c ^ n • s

Next tactic:
exact (mem_smul_set_iff_inv_smul_mem₀ (cn_ne n) _ _).2 hn
/-- In a topological vector space over a nontrivially normed field, any neighborhood of zero has
the same cardinality as the whole space.

See also `cardinal_eq_of_mem_nhds`. -/
lemma cardinal_eq_of_mem_nhds_zero
    {E : Type*} (𝕜 : Type*) [NontriviallyNormedField 𝕜] [AddCommGroup E] [Module 𝕜 E]
    [TopologicalSpace E] [ContinuousSMul 𝕜 E] {s : Set E} (hs : s ∈ 𝓝 (0 : E)) : #s = #E := by
  /- As `s` is a neighborhood of `0`, the space is covered by the rescaled sets `c^n • s`,
  where `c` is any element of `𝕜` with norm `> 1`. All these sets are in bijection and have
  therefore the same cardinality. The conclusion follows. -/
  obtain ⟨c, hc⟩ : ∃ x : 𝕜 , 1 < ‖x‖ := NormedField.exists_lt_norm 𝕜 1
  have cn_ne : ∀ n, c^n ≠ 0 := by
    intro n
    apply pow_ne_zero
    rintro rfl
    simp only [norm_zero] at hc
    exact lt_irrefl _ (hc.trans zero_lt_one)
  have A : ∀ (x : E), ∀ᶠ n in (atTop : Filter ℕ), x ∈ c^n • s := by
    intro x
    have : Tendsto (fun n ↦ (c^n) ⁻¹ • x) atTop (𝓝 ((0 : 𝕜) • x)) := by
      have : Tendsto (fun n ↦ (c^n)⁻¹) atTop (𝓝 0) := by
        simp_rw [← inv_pow]
        apply tendsto_pow_atTop_nhds_0_of_norm_lt_1
        rw [norm_inv]
        exact inv_lt_one hc
      exact Tendsto.smul_const this x
    rw [zero_smul] at this
    filter_upwards [this hs] with n (hn : (c ^ n)⁻¹ • x ∈ s)
    exact (mem_smul_set_iff_inv_smul_mem₀ (cn_ne n) _ _).2 hn
  

Proof state:
case intro
E : Type u_1
𝕜 : Type u_2
inst✝⁴ : NontriviallyNormedField 𝕜
inst✝³ : AddCommGroup E
inst✝² : Module 𝕜 E
inst✝¹ : TopologicalSpace E
inst✝ : ContinuousSMul 𝕜 E
s : Set E
hs : s ∈ 𝓝 0
c : 𝕜
hc : 1 < ‖c‖
cn_ne : ∀ (n : ℕ), c ^ n ≠ 0
A : ∀ (x : E), ∀ᶠ (n : ℕ) in atTop, x ∈ c ^ n • s
⊢ #↑s = #E

Next tactic:
have B : ∀ n, #(c^n • s :) = #s := by
    intro n
    have : (c^n • s :) ≃ s :=
    { toFun := fun x ↦ ⟨(c^n)⁻¹ • x.1, (mem_smul_set_iff_inv_smul_mem₀ (cn_ne n) _ _).1 x.2⟩
      invFun := fun x ↦ ⟨(c^n) • x.1, smul_mem_smul_set x.2⟩
      left_inv := fun x ↦ by simp [smul_smul, mul_inv_cancel (cn_ne n)]
      right_inv := fun x ↦ by simp [smul_smul, inv_mul_cancel (cn_ne n)] }
    exact Cardinal.mk_congr this
/-- In a topological vector space over a nontrivially normed field, any neighborhood of zero has
the same cardinality as the whole space.

See also `cardinal_eq_of_mem_nhds`. -/
lemma cardinal_eq_of_mem_nhds_zero
    {E : Type*} (𝕜 : Type*) [NontriviallyNormedField 𝕜] [AddCommGroup E] [Module 𝕜 E]
    [TopologicalSpace E] [ContinuousSMul 𝕜 E] {s : Set E} (hs : s ∈ 𝓝 (0 : E)) : #s = #E := by
  /- As `s` is a neighborhood of `0`, the space is covered by the rescaled sets `c^n • s`,
  where `c` is any element of `𝕜` with norm `> 1`. All these sets are in bijection and have
  therefore the same cardinality. The conclusion follows. -/
  obtain ⟨c, hc⟩ : ∃ x : 𝕜 , 1 < ‖x‖ := NormedField.exists_lt_norm 𝕜 1
  have cn_ne : ∀ n, c^n ≠ 0 := by
    intro n
    apply pow_ne_zero
    rintro rfl
    simp only [norm_zero] at hc
    exact lt_irrefl _ (hc.trans zero_lt_one)
  have A : ∀ (x : E), ∀ᶠ n in (atTop : Filter ℕ), x ∈ c^n • s := by
    intro x
    have : Tendsto (fun n ↦ (c^n) ⁻¹ • x) atTop (𝓝 ((0 : 𝕜) • x)) := by
      have : Tendsto (fun n ↦ (c^n)⁻¹) atTop (𝓝 0) := by
        simp_rw [← inv_pow]
        apply tendsto_pow_atTop_nhds_0_of_norm_lt_1
        rw [norm_inv]
        exact inv_lt_one hc
      exact Tendsto.smul_const this x
    rw [zero_smul] at this
    filter_upwards [this hs] with n (hn : (c ^ n)⁻¹ • x ∈ s)
    exact (mem_smul_set_iff_inv_smul_mem₀ (cn_ne n) _ _).2 hn
  have B : ∀ n, #(c^n • s :) = #s := by
    

Proof state:
E : Type u_1
𝕜 : Type u_2
inst✝⁴ : NontriviallyNormedField 𝕜
inst✝³ : AddCommGroup E
inst✝² : Module 𝕜 E
inst✝¹ : TopologicalSpace E
inst✝ : ContinuousSMul 𝕜 E
s : Set E
hs : s ∈ 𝓝 0
c : 𝕜
hc : 1 < ‖c‖
cn_ne : ∀ (n : ℕ), c ^ n ≠ 0
A : ∀ (x : E), ∀ᶠ (n : ℕ) in atTop, x ∈ c ^ n • s
⊢ ∀ (n : ℕ), #↑(c ^ n • s) = #↑s

Next tactic:
intro n
/-- In a topological vector space over a nontrivially normed field, any neighborhood of zero has
the same cardinality as the whole space.

See also `cardinal_eq_of_mem_nhds`. -/
lemma cardinal_eq_of_mem_nhds_zero
    {E : Type*} (𝕜 : Type*) [NontriviallyNormedField 𝕜] [AddCommGroup E] [Module 𝕜 E]
    [TopologicalSpace E] [ContinuousSMul 𝕜 E] {s : Set E} (hs : s ∈ 𝓝 (0 : E)) : #s = #E := by
  /- As `s` is a neighborhood of `0`, the space is covered by the rescaled sets `c^n • s`,
  where `c` is any element of `𝕜` with norm `> 1`. All these sets are in bijection and have
  therefore the same cardinality. The conclusion follows. -/
  obtain ⟨c, hc⟩ : ∃ x : 𝕜 , 1 < ‖x‖ := NormedField.exists_lt_norm 𝕜 1
  have cn_ne : ∀ n, c^n ≠ 0 := by
    intro n
    apply pow_ne_zero
    rintro rfl
    simp only [norm_zero] at hc
    exact lt_irrefl _ (hc.trans zero_lt_one)
  have A : ∀ (x : E), ∀ᶠ n in (atTop : Filter ℕ), x ∈ c^n • s := by
    intro x
    have : Tendsto (fun n ↦ (c^n) ⁻¹ • x) atTop (𝓝 ((0 : 𝕜) • x)) := by
      have : Tendsto (fun n ↦ (c^n)⁻¹) atTop (𝓝 0) := by
        simp_rw [← inv_pow]
        apply tendsto_pow_atTop_nhds_0_of_norm_lt_1
        rw [norm_inv]
        exact inv_lt_one hc
      exact Tendsto.smul_const this x
    rw [zero_smul] at this
    filter_upwards [this hs] with n (hn : (c ^ n)⁻¹ • x ∈ s)
    exact (mem_smul_set_iff_inv_smul_mem₀ (cn_ne n) _ _).2 hn
  have B : ∀ n, #(c^n • s :) = #s := by
    intro n
    

Proof state:
E : Type u_1
𝕜 : Type u_2
inst✝⁴ : NontriviallyNormedField 𝕜
inst✝³ : AddCommGroup E
inst✝² : Module 𝕜 E
inst✝¹ : TopologicalSpace E
inst✝ : ContinuousSMul 𝕜 E
s : Set E
hs : s ∈ 𝓝 0
c : 𝕜
hc : 1 < ‖c‖
cn_ne : ∀ (n : ℕ), c ^ n ≠ 0
A : ∀ (x : E), ∀ᶠ (n : ℕ) in atTop, x ∈ c ^ n • s
n : ℕ
⊢ #↑(c ^ n • s) = #↑s

Next tactic:
have : (c^n • s :) ≃ s :=
    { toFun := fun x ↦ ⟨(c^n)⁻¹ • x.1, (mem_smul_set_iff_inv_smul_mem₀ (cn_ne n) _ _).1 x.2⟩
      invFun := fun x ↦ ⟨(c^n) • x.1, smul_mem_smul_set x.2⟩
      left_inv := fun x ↦ by simp [smul_smul, mul_inv_cancel (cn_ne n)]
      right_inv := fun x ↦ by simp [smul_smul, inv_mul_cancel (cn_ne n)] }
/-- In a topological vector space over a nontrivially normed field, any neighborhood of zero has
the same cardinality as the whole space.

See also `cardinal_eq_of_mem_nhds`. -/
lemma cardinal_eq_of_mem_nhds_zero
    {E : Type*} (𝕜 : Type*) [NontriviallyNormedField 𝕜] [AddCommGroup E] [Module 𝕜 E]
    [TopologicalSpace E] [ContinuousSMul 𝕜 E] {s : Set E} (hs : s ∈ 𝓝 (0 : E)) : #s = #E := by
  /- As `s` is a neighborhood of `0`, the space is covered by the rescaled sets `c^n • s`,
  where `c` is any element of `𝕜` with norm `> 1`. All these sets are in bijection and have
  therefore the same cardinality. The conclusion follows. -/
  obtain ⟨c, hc⟩ : ∃ x : 𝕜 , 1 < ‖x‖ := NormedField.exists_lt_norm 𝕜 1
  have cn_ne : ∀ n, c^n ≠ 0 := by
    intro n
    apply pow_ne_zero
    rintro rfl
    simp only [norm_zero] at hc
    exact lt_irrefl _ (hc.trans zero_lt_one)
  have A : ∀ (x : E), ∀ᶠ n in (atTop : Filter ℕ), x ∈ c^n • s := by
    intro x
    have : Tendsto (fun n ↦ (c^n) ⁻¹ • x) atTop (𝓝 ((0 : 𝕜) • x)) := by
      have : Tendsto (fun n ↦ (c^n)⁻¹) atTop (𝓝 0) := by
        simp_rw [← inv_pow]
        apply tendsto_pow_atTop_nhds_0_of_norm_lt_1
        rw [norm_inv]
        exact inv_lt_one hc
      exact Tendsto.smul_const this x
    rw [zero_smul] at this
    filter_upwards [this hs] with n (hn : (c ^ n)⁻¹ • x ∈ s)
    exact (mem_smul_set_iff_inv_smul_mem₀ (cn_ne n) _ _).2 hn
  have B : ∀ n, #(c^n • s :) = #s := by
    intro n
    have : (c^n • s :) ≃ s :=
    { toFun := fun x ↦ ⟨(c^n)⁻¹ • x.1, (mem_smul_set_iff_inv_smul_mem₀ (cn_ne n) _ _).1 x.2⟩
      invFun := fun x ↦ ⟨(c^n) • x.1, smul_mem_smul_set x.2⟩
      left_inv := fun x ↦ by 

Proof state:
E : Type u_1
𝕜 : Type u_2
inst✝⁴ : NontriviallyNormedField 𝕜
inst✝³ : AddCommGroup E
inst✝² : Module 𝕜 E
inst✝¹ : TopologicalSpace E
inst✝ : ContinuousSMul 𝕜 E
s : Set E
hs : s ∈ 𝓝 0
c : 𝕜
hc : 1 < ‖c‖
cn_ne : ∀ (n : ℕ), c ^ n ≠ 0
A : ∀ (x : E), ∀ᶠ (n : ℕ) in atTop, x ∈ c ^ n • s
n : ℕ
x : ↑(c ^ n • s)
⊢ (fun x => { val := c ^ n • ↑x, property := (_ : c ^ n • ↑x ∈ c ^ n • s) })
      ((fun x => { val := (c ^ n)⁻¹ • ↑x, property := (_ : (c ^ n)⁻¹ • ↑x ∈ s) }) x) =
    x

Next tactic:
simp [smul_smul, mul_inv_cancel (cn_ne n)]