Metamath Proof Explorer


Theorem ovolmge0

Description: The set M is composed of nonnegative extended real numbers. (Contributed by Mario Carneiro, 16-Mar-2014)

Ref Expression
Hypothesis elovolm.1 ⊢ M = y ∈ ℝ * | ∃ f ∈ ≤ ∩ ℝ 2 ℕ A ⊆ ⋃ ran ⁡ . ∘ f ∧ y = sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * <
Assertion ovolmge0 ⊢ B ∈ M → 0 ≤ B

Proof

Step Hyp Ref Expression
1 elovolm.1 ⊢ M = y ∈ ℝ * | ∃ f ∈ ≤ ∩ ℝ 2 ℕ A ⊆ ⋃ ran ⁡ . ∘ f ∧ y = sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * <
2 1 elovolm ⊢ B ∈ M ↔ ∃ f ∈ ≤ ∩ ℝ 2 ℕ A ⊆ ⋃ ran ⁡ . ∘ f ∧ B = sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * <
3 elovolmlem ⊢ f ∈ ≤ ∩ ℝ 2 ℕ ↔ f : ℕ ⟶ ≤ ∩ ℝ 2
4 eqid ⊢ abs ∘ − ∘ f = abs ∘ − ∘ f
5 eqid ⊢ seq 1 + abs ∘ − ∘ f = seq 1 + abs ∘ − ∘ f
6 4 5 ovolsf ⊢ f : ℕ ⟶ ≤ ∩ ℝ 2 → seq 1 + abs ∘ − ∘ f : ℕ ⟶ 0 +∞
7 1nn ⊢ 1 ∈ ℕ
8 ffvelcdm ⊢ seq 1 + abs ∘ − ∘ f : ℕ ⟶ 0 +∞ ∧ 1 ∈ ℕ → seq 1 + abs ∘ − ∘ f ⁡ 1 ∈ 0 +∞
9 6 7 8 sylancl ⊢ f : ℕ ⟶ ≤ ∩ ℝ 2 → seq 1 + abs ∘ − ∘ f ⁡ 1 ∈ 0 +∞
10 elrege0 ⊢ seq 1 + abs ∘ − ∘ f ⁡ 1 ∈ 0 +∞ ↔ seq 1 + abs ∘ − ∘ f ⁡ 1 ∈ ℝ ∧ 0 ≤ seq 1 + abs ∘ − ∘ f ⁡ 1
11 10 simprbi ⊢ seq 1 + abs ∘ − ∘ f ⁡ 1 ∈ 0 +∞ → 0 ≤ seq 1 + abs ∘ − ∘ f ⁡ 1
12 9 11 syl ⊢ f : ℕ ⟶ ≤ ∩ ℝ 2 → 0 ≤ seq 1 + abs ∘ − ∘ f ⁡ 1
13 6 frnd ⊢ f : ℕ ⟶ ≤ ∩ ℝ 2 → ran ⁡ seq 1 + abs ∘ − ∘ f ⊆ 0 +∞
14 icossxr ⊢ 0 +∞ ⊆ ℝ *
15 13 14 sstrdi ⊢ f : ℕ ⟶ ≤ ∩ ℝ 2 → ran ⁡ seq 1 + abs ∘ − ∘ f ⊆ ℝ *
16 6 ffnd ⊢ f : ℕ ⟶ ≤ ∩ ℝ 2 → seq 1 + abs ∘ − ∘ f Fn ℕ
17 fnfvelrn ⊢ seq 1 + abs ∘ − ∘ f Fn ℕ ∧ 1 ∈ ℕ → seq 1 + abs ∘ − ∘ f ⁡ 1 ∈ ran ⁡ seq 1 + abs ∘ − ∘ f
18 16 7 17 sylancl ⊢ f : ℕ ⟶ ≤ ∩ ℝ 2 → seq 1 + abs ∘ − ∘ f ⁡ 1 ∈ ran ⁡ seq 1 + abs ∘ − ∘ f
19 supxrub ⊢ ran ⁡ seq 1 + abs ∘ − ∘ f ⊆ ℝ * ∧ seq 1 + abs ∘ − ∘ f ⁡ 1 ∈ ran ⁡ seq 1 + abs ∘ − ∘ f → seq 1 + abs ∘ − ∘ f ⁡ 1 ≤ sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * <
20 15 18 19 syl2anc ⊢ f : ℕ ⟶ ≤ ∩ ℝ 2 → seq 1 + abs ∘ − ∘ f ⁡ 1 ≤ sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * <
21 0xr ⊢ 0 ∈ ℝ *
22 14 9 sselid ⊢ f : ℕ ⟶ ≤ ∩ ℝ 2 → seq 1 + abs ∘ − ∘ f ⁡ 1 ∈ ℝ *
23 supxrcl ⊢ ran ⁡ seq 1 + abs ∘ − ∘ f ⊆ ℝ * → sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * < ∈ ℝ *
24 15 23 syl ⊢ f : ℕ ⟶ ≤ ∩ ℝ 2 → sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * < ∈ ℝ *
25 xrletr ⊢ 0 ∈ ℝ * ∧ seq 1 + abs ∘ − ∘ f ⁡ 1 ∈ ℝ * ∧ sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * < ∈ ℝ * → 0 ≤ seq 1 + abs ∘ − ∘ f ⁡ 1 ∧ seq 1 + abs ∘ − ∘ f ⁡ 1 ≤ sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * < → 0 ≤ sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * <
26 21 22 24 25 mp3an2i ⊢ f : ℕ ⟶ ≤ ∩ ℝ 2 → 0 ≤ seq 1 + abs ∘ − ∘ f ⁡ 1 ∧ seq 1 + abs ∘ − ∘ f ⁡ 1 ≤ sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * < → 0 ≤ sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * <
27 12 20 26 mp2and ⊢ f : ℕ ⟶ ≤ ∩ ℝ 2 → 0 ≤ sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * <
28 3 27 sylbi ⊢ f ∈ ≤ ∩ ℝ 2 ℕ → 0 ≤ sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * <
29 breq2 ⊢ B = sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * < → 0 ≤ B ↔ 0 ≤ sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * <
30 28 29 syl5ibrcom ⊢ f ∈ ≤ ∩ ℝ 2 ℕ → B = sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * < → 0 ≤ B
31 30 adantld ⊢ f ∈ ≤ ∩ ℝ 2 ℕ → A ⊆ ⋃ ran ⁡ . ∘ f ∧ B = sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * < → 0 ≤ B
32 31 rexlimiv ⊢ ∃ f ∈ ≤ ∩ ℝ 2 ℕ A ⊆ ⋃ ran ⁡ . ∘ f ∧ B = sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * < → 0 ≤ B
33 2 32 sylbi ⊢ B ∈ M → 0 ≤ B