Metamath Proof Explorer


Theorem heiborlem7

Description: Lemma for heibor . Since the sizes of the balls decrease exponentially, the sequence converges to zero. (Contributed by Jeff Madsen, 23-Jan-2014)

Ref Expression
Hypotheses heibor.1 ⊢ J = MetOpen ⁡ D
heibor.3 ⊢ K = u | ¬ ∃ v ∈ 𝒫 U ∩ Fin u ⊆ ⋃ v
heibor.4 ⊢ G = y n | n ∈ ℕ 0 ∧ y ∈ F ⁡ n ∧ y B n ∈ K
heibor.5 ⊢ B = z ∈ X , m ∈ ℕ 0 ⟼ z ball ⁡ D 1 2 m
heibor.6 ⊢ φ → D ∈ CMet ⁡ X
heibor.7 ⊢ φ → F : ℕ 0 ⟶ 𝒫 X ∩ Fin
heibor.8 ⊢ φ → ∀ n ∈ ℕ 0 X = ⋃ y ∈ F ⁡ n y B n
heibor.9 ⊢ φ → ∀ x ∈ G T ⁡ x G 2 nd ⁡ x + 1 ∧ B ⁡ x ∩ T ⁡ x B 2 nd ⁡ x + 1 ∈ K
heibor.10 ⊢ φ → C G 0
heibor.11 ⊢ S = seq 0 T m ∈ ℕ 0 ⟼ if m = 0 C m − 1
heibor.12 ⊢ M = n ∈ ℕ ⟼ S ⁡ n 3 2 n
Assertion heiborlem7 ⊢ ∀ r ∈ ℝ + ∃ k ∈ ℕ 2 nd ⁡ M ⁡ k < r

Proof

Step Hyp Ref Expression
1 heibor.1 ⊢ J = MetOpen ⁡ D
2 heibor.3 ⊢ K = u | ¬ ∃ v ∈ 𝒫 U ∩ Fin u ⊆ ⋃ v
3 heibor.4 ⊢ G = y n | n ∈ ℕ 0 ∧ y ∈ F ⁡ n ∧ y B n ∈ K
4 heibor.5 ⊢ B = z ∈ X , m ∈ ℕ 0 ⟼ z ball ⁡ D 1 2 m
5 heibor.6 ⊢ φ → D ∈ CMet ⁡ X
6 heibor.7 ⊢ φ → F : ℕ 0 ⟶ 𝒫 X ∩ Fin
7 heibor.8 ⊢ φ → ∀ n ∈ ℕ 0 X = ⋃ y ∈ F ⁡ n y B n
8 heibor.9 ⊢ φ → ∀ x ∈ G T ⁡ x G 2 nd ⁡ x + 1 ∧ B ⁡ x ∩ T ⁡ x B 2 nd ⁡ x + 1 ∈ K
9 heibor.10 ⊢ φ → C G 0
10 heibor.11 ⊢ S = seq 0 T m ∈ ℕ 0 ⟼ if m = 0 C m − 1
11 heibor.12 ⊢ M = n ∈ ℕ ⟼ S ⁡ n 3 2 n
12 3re ⊢ 3 ∈ ℝ
13 3pos ⊢ 0 < 3
14 12 13 elrpii ⊢ 3 ∈ ℝ +
15 rpdivcl ⊢ r ∈ ℝ + ∧ 3 ∈ ℝ + → r 3 ∈ ℝ +
16 14 15 mpan2 ⊢ r ∈ ℝ + → r 3 ∈ ℝ +
17 2re ⊢ 2 ∈ ℝ
18 1lt2 ⊢ 1 < 2
19 expnlbnd ⊢ r 3 ∈ ℝ + ∧ 2 ∈ ℝ ∧ 1 < 2 → ∃ k ∈ ℕ 1 2 k < r 3
20 17 18 19 mp3an23 ⊢ r 3 ∈ ℝ + → ∃ k ∈ ℕ 1 2 k < r 3
21 16 20 syl ⊢ r ∈ ℝ + → ∃ k ∈ ℕ 1 2 k < r 3
22 2nn ⊢ 2 ∈ ℕ
23 nnnn0 ⊢ k ∈ ℕ → k ∈ ℕ 0
24 nnexpcl ⊢ 2 ∈ ℕ ∧ k ∈ ℕ 0 → 2 k ∈ ℕ
25 22 23 24 sylancr ⊢ k ∈ ℕ → 2 k ∈ ℕ
26 25 nnrpd ⊢ k ∈ ℕ → 2 k ∈ ℝ +
27 rpcn ⊢ 2 k ∈ ℝ + → 2 k ∈ ℂ
28 rpne0 ⊢ 2 k ∈ ℝ + → 2 k ≠ 0
29 3cn ⊢ 3 ∈ ℂ
30 divrec ⊢ 3 ∈ ℂ ∧ 2 k ∈ ℂ ∧ 2 k ≠ 0 → 3 2 k = 3 ⁢ 1 2 k
31 29 30 mp3an1 ⊢ 2 k ∈ ℂ ∧ 2 k ≠ 0 → 3 2 k = 3 ⁢ 1 2 k
32 27 28 31 syl2anc ⊢ 2 k ∈ ℝ + → 3 2 k = 3 ⁢ 1 2 k
33 26 32 syl ⊢ k ∈ ℕ → 3 2 k = 3 ⁢ 1 2 k
34 33 adantl ⊢ r ∈ ℝ + ∧ k ∈ ℕ → 3 2 k = 3 ⁢ 1 2 k
35 34 breq1d ⊢ r ∈ ℝ + ∧ k ∈ ℕ → 3 2 k < r ↔ 3 ⁢ 1 2 k < r
36 25 nnrecred ⊢ k ∈ ℕ → 1 2 k ∈ ℝ
37 rpre ⊢ r ∈ ℝ + → r ∈ ℝ
38 12 13 pm3.2i ⊢ 3 ∈ ℝ ∧ 0 < 3
39 ltmuldiv2 ⊢ 1 2 k ∈ ℝ ∧ r ∈ ℝ ∧ 3 ∈ ℝ ∧ 0 < 3 → 3 ⁢ 1 2 k < r ↔ 1 2 k < r 3
40 38 39 mp3an3 ⊢ 1 2 k ∈ ℝ ∧ r ∈ ℝ → 3 ⁢ 1 2 k < r ↔ 1 2 k < r 3
41 36 37 40 syl2anr ⊢ r ∈ ℝ + ∧ k ∈ ℕ → 3 ⁢ 1 2 k < r ↔ 1 2 k < r 3
42 35 41 bitrd ⊢ r ∈ ℝ + ∧ k ∈ ℕ → 3 2 k < r ↔ 1 2 k < r 3
43 42 rexbidva ⊢ r ∈ ℝ + → ∃ k ∈ ℕ 3 2 k < r ↔ ∃ k ∈ ℕ 1 2 k < r 3
44 21 43 mpbird ⊢ r ∈ ℝ + → ∃ k ∈ ℕ 3 2 k < r
45 fveq2 ⊢ n = k → S ⁡ n = S ⁡ k
46 oveq2 ⊢ n = k → 2 n = 2 k
47 46 oveq2d ⊢ n = k → 3 2 n = 3 2 k
48 45 47 opeq12d ⊢ n = k → S ⁡ n 3 2 n = S ⁡ k 3 2 k
49 opex ⊢ S ⁡ k 3 2 k ∈ V
50 48 11 49 fvmpt ⊢ k ∈ ℕ → M ⁡ k = S ⁡ k 3 2 k
51 50 fveq2d ⊢ k ∈ ℕ → 2 nd ⁡ M ⁡ k = 2 nd ⁡ S ⁡ k 3 2 k
52 fvex ⊢ S ⁡ k ∈ V
53 ovex ⊢ 3 2 k ∈ V
54 52 53 op2nd ⊢ 2 nd ⁡ S ⁡ k 3 2 k = 3 2 k
55 51 54 eqtrdi ⊢ k ∈ ℕ → 2 nd ⁡ M ⁡ k = 3 2 k
56 55 breq1d ⊢ k ∈ ℕ → 2 nd ⁡ M ⁡ k < r ↔ 3 2 k < r
57 56 rexbiia ⊢ ∃ k ∈ ℕ 2 nd ⁡ M ⁡ k < r ↔ ∃ k ∈ ℕ 3 2 k < r
58 44 57 sylibr ⊢ r ∈ ℝ + → ∃ k ∈ ℕ 2 nd ⁡ M ⁡ k < r
59 58 rgen ⊢ ∀ r ∈ ℝ + ∃ k ∈ ℕ 2 nd ⁡ M ⁡ k < r