Metamath Proof Explorer


Theorem lclkr

Description: The set of functionals with closed kernels is a subspace. Part of proof of Theorem 3.6 of Holland95 p. 218, line 20, stating "The f_M that arise this way generate a subspace F of E'". Our proof was suggested by Mario Carneiro, 5-Jan-2015. (Contributed by NM, 18-Jan-2015)

Ref Expression
Hypotheses lclkr.h ⊢ H = LHyp ⁡ K
lclkr.u ⊢ U = DVecH ⁡ K ⁡ W
lclkr.o ⊢ ⊥ ˙ = ocH ⁡ K ⁡ W
lclkr.f ⊢ F = LFnl ⁡ U
lclkr.l ⊢ L = LKer ⁡ U
lclkr.d ⊢ D = LDual ⁡ U
lclkr.s ⊢ S = LSubSp ⁡ D
lclkr.c ⊢ C = f ∈ F | ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ f = L ⁡ f
lclkr.k ⊢ φ → K ∈ HL ∧ W ∈ H
Assertion lclkr ⊢ φ → C ∈ S

Proof

Step Hyp Ref Expression
1 lclkr.h ⊢ H = LHyp ⁡ K
2 lclkr.u ⊢ U = DVecH ⁡ K ⁡ W
3 lclkr.o ⊢ ⊥ ˙ = ocH ⁡ K ⁡ W
4 lclkr.f ⊢ F = LFnl ⁡ U
5 lclkr.l ⊢ L = LKer ⁡ U
6 lclkr.d ⊢ D = LDual ⁡ U
7 lclkr.s ⊢ S = LSubSp ⁡ D
8 lclkr.c ⊢ C = f ∈ F | ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ f = L ⁡ f
9 lclkr.k ⊢ φ → K ∈ HL ∧ W ∈ H
10 ssrab2 ⊢ f ∈ F | ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ f = L ⁡ f ⊆ F
11 10 a1i ⊢ φ → f ∈ F | ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ f = L ⁡ f ⊆ F
12 8 a1i ⊢ φ → C = f ∈ F | ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ f = L ⁡ f
13 eqid ⊢ Base D = Base D
14 1 2 9 dvhlmod ⊢ φ → U ∈ LMod
15 4 6 13 14 ldualvbase ⊢ φ → Base D = F
16 11 12 15 3sstr4d ⊢ φ → C ⊆ Base D
17 eqid ⊢ Scalar ⁡ U = Scalar ⁡ U
18 eqid ⊢ 0 Scalar ⁡ U = 0 Scalar ⁡ U
19 eqid ⊢ Base U = Base U
20 17 18 19 4 lfl0f ⊢ U ∈ LMod → Base U × 0 Scalar ⁡ U ∈ F
21 14 20 syl ⊢ φ → Base U × 0 Scalar ⁡ U ∈ F
22 1 2 3 19 9 dochoc1 ⊢ φ → ⊥ ˙ ⁡ ⊥ ˙ ⁡ Base U = Base U
23 eqid ⊢ Base U × 0 Scalar ⁡ U = Base U × 0 Scalar ⁡ U
24 17 18 19 4 5 lkr0f ⊢ U ∈ LMod ∧ Base U × 0 Scalar ⁡ U ∈ F → L ⁡ Base U × 0 Scalar ⁡ U = Base U ↔ Base U × 0 Scalar ⁡ U = Base U × 0 Scalar ⁡ U
25 14 20 24 syl2anc2 ⊢ φ → L ⁡ Base U × 0 Scalar ⁡ U = Base U ↔ Base U × 0 Scalar ⁡ U = Base U × 0 Scalar ⁡ U
26 23 25 mpbiri ⊢ φ → L ⁡ Base U × 0 Scalar ⁡ U = Base U
27 26 fveq2d ⊢ φ → ⊥ ˙ ⁡ L ⁡ Base U × 0 Scalar ⁡ U = ⊥ ˙ ⁡ Base U
28 27 fveq2d ⊢ φ → ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ Base U × 0 Scalar ⁡ U = ⊥ ˙ ⁡ ⊥ ˙ ⁡ Base U
29 22 28 26 3eqtr4d ⊢ φ → ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ Base U × 0 Scalar ⁡ U = L ⁡ Base U × 0 Scalar ⁡ U
30 8 lcfl1lem ⊢ Base U × 0 Scalar ⁡ U ∈ C ↔ Base U × 0 Scalar ⁡ U ∈ F ∧ ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ Base U × 0 Scalar ⁡ U = L ⁡ Base U × 0 Scalar ⁡ U
31 21 29 30 sylanbrc ⊢ φ → Base U × 0 Scalar ⁡ U ∈ C
32 31 ne0d ⊢ φ → C ≠ ∅
33 eqid ⊢ + D = + D
34 9 adantr ⊢ φ ∧ x ∈ Base Scalar ⁡ D ∧ a ∈ C ∧ b ∈ C → K ∈ HL ∧ W ∈ H
35 eqid ⊢ Base Scalar ⁡ U = Base Scalar ⁡ U
36 eqid ⊢ ⋅ D = ⋅ D
37 simpr1 ⊢ φ ∧ x ∈ Base Scalar ⁡ D ∧ a ∈ C ∧ b ∈ C → x ∈ Base Scalar ⁡ D
38 eqid ⊢ Scalar ⁡ D = Scalar ⁡ D
39 eqid ⊢ Base Scalar ⁡ D = Base Scalar ⁡ D
40 17 35 6 38 39 14 ldualsbase ⊢ φ → Base Scalar ⁡ D = Base Scalar ⁡ U
41 40 adantr ⊢ φ ∧ x ∈ Base Scalar ⁡ D ∧ a ∈ C ∧ b ∈ C → Base Scalar ⁡ D = Base Scalar ⁡ U
42 37 41 eleqtrd ⊢ φ ∧ x ∈ Base Scalar ⁡ D ∧ a ∈ C ∧ b ∈ C → x ∈ Base Scalar ⁡ U
43 simpr2 ⊢ φ ∧ x ∈ Base Scalar ⁡ D ∧ a ∈ C ∧ b ∈ C → a ∈ C
44 1 3 2 4 5 6 17 35 36 8 34 42 43 lclkrlem1 ⊢ φ ∧ x ∈ Base Scalar ⁡ D ∧ a ∈ C ∧ b ∈ C → x ⋅ D a ∈ C
45 simpr3 ⊢ φ ∧ x ∈ Base Scalar ⁡ D ∧ a ∈ C ∧ b ∈ C → b ∈ C
46 1 3 2 4 5 6 33 8 34 44 45 lclkrlem2 ⊢ φ ∧ x ∈ Base Scalar ⁡ D ∧ a ∈ C ∧ b ∈ C → x ⋅ D a + D b ∈ C
47 46 ralrimivvva ⊢ φ → ∀ x ∈ Base Scalar ⁡ D ∀ a ∈ C ∀ b ∈ C x ⋅ D a + D b ∈ C
48 38 39 13 33 36 7 islss ⊢ C ∈ S ↔ C ⊆ Base D ∧ C ≠ ∅ ∧ ∀ x ∈ Base Scalar ⁡ D ∀ a ∈ C ∀ b ∈ C x ⋅ D a + D b ∈ C
49 16 32 47 48 syl3anbrc ⊢ φ → C ∈ S