Metamath Proof Explorer


Theorem minveclem5

Description: Lemma for minvec . Discharge the assumptions in minveclem4 . (Contributed by Mario Carneiro, 9-May-2014) (Revised by Mario Carneiro, 15-Oct-2015)

Ref Expression
Hypotheses minvec.x ⊢ X = Base U
minvec.m ⊢ - ˙ = - U
minvec.n ⊢ N = norm ⁡ U
minvec.u ⊢ φ → U ∈ CPreHil
minvec.y ⊢ φ → Y ∈ LSubSp ⁡ U
minvec.w ⊢ φ → U ↾ 𝑠 Y ∈ CMetSp
minvec.a ⊢ φ → A ∈ X
minvec.j ⊢ J = TopOpen ⁡ U
minvec.r ⊢ R = ran ⁡ y ∈ Y ⟼ N ⁡ A - ˙ y
minvec.s ⊢ S = inf R ℝ <
minvec.d ⊢ D = dist ⁡ U ↾ X × X
Assertion minveclem5 ⊢ φ → ∃ x ∈ Y ∀ y ∈ Y N ⁡ A - ˙ x ≤ N ⁡ A - ˙ y

Proof

Step Hyp Ref Expression
1 minvec.x ⊢ X = Base U
2 minvec.m ⊢ - ˙ = - U
3 minvec.n ⊢ N = norm ⁡ U
4 minvec.u ⊢ φ → U ∈ CPreHil
5 minvec.y ⊢ φ → Y ∈ LSubSp ⁡ U
6 minvec.w ⊢ φ → U ↾ 𝑠 Y ∈ CMetSp
7 minvec.a ⊢ φ → A ∈ X
8 minvec.j ⊢ J = TopOpen ⁡ U
9 minvec.r ⊢ R = ran ⁡ y ∈ Y ⟼ N ⁡ A - ˙ y
10 minvec.s ⊢ S = inf R ℝ <
11 minvec.d ⊢ D = dist ⁡ U ↾ X × X
12 oveq2 ⊢ s = r → S 2 + s = S 2 + r
13 12 breq2d ⊢ s = r → A D z 2 ≤ S 2 + s ↔ A D z 2 ≤ S 2 + r
14 13 rabbidv ⊢ s = r → z ∈ Y | A D z 2 ≤ S 2 + s = z ∈ Y | A D z 2 ≤ S 2 + r
15 oveq2 ⊢ z = y → A D z = A D y
16 15 oveq1d ⊢ z = y → A D z 2 = A D y 2
17 16 breq1d ⊢ z = y → A D z 2 ≤ S 2 + r ↔ A D y 2 ≤ S 2 + r
18 17 cbvrabv ⊢ z ∈ Y | A D z 2 ≤ S 2 + r = y ∈ Y | A D y 2 ≤ S 2 + r
19 14 18 eqtrdi ⊢ s = r → z ∈ Y | A D z 2 ≤ S 2 + s = y ∈ Y | A D y 2 ≤ S 2 + r
20 19 cbvmptv ⊢ s ∈ ℝ + ⟼ z ∈ Y | A D z 2 ≤ S 2 + s = r ∈ ℝ + ⟼ y ∈ Y | A D y 2 ≤ S 2 + r
21 20 rneqi ⊢ ran ⁡ s ∈ ℝ + ⟼ z ∈ Y | A D z 2 ≤ S 2 + s = ran ⁡ r ∈ ℝ + ⟼ y ∈ Y | A D y 2 ≤ S 2 + r
22 eqid ⊢ ⋃ J fLim X filGen ran ⁡ s ∈ ℝ + ⟼ z ∈ Y | A D z 2 ≤ S 2 + s = ⋃ J fLim X filGen ran ⁡ s ∈ ℝ + ⟼ z ∈ Y | A D z 2 ≤ S 2 + s
23 eqid ⊢ A D ⋃ J fLim X filGen ran ⁡ s ∈ ℝ + ⟼ z ∈ Y | A D z 2 ≤ S 2 + s + S 2 2 − S 2 = A D ⋃ J fLim X filGen ran ⁡ s ∈ ℝ + ⟼ z ∈ Y | A D z 2 ≤ S 2 + s + S 2 2 − S 2
24 1 2 3 4 5 6 7 8 9 10 11 21 22 23 minveclem4 ⊢ φ → ∃ x ∈ Y ∀ y ∈ Y N ⁡ A - ˙ x ≤ N ⁡ A - ˙ y