Metamath Proof Explorer


Theorem occl

Description: Closure of complement of Hilbert subset. Part of Remark 3.12 of Beran p. 107. (Contributed by NM, 8-Aug-2000) (Proof shortened by Mario Carneiro, 14-May-2014) (New usage is discouraged.)

Ref Expression
Assertion occl ⊢ A ⊆ ℋ → ⊥ ⁡ A ∈ C ℋ

Proof

Step Hyp Ref Expression
1 ocsh ⊢ A ⊆ ℋ → ⊥ ⁡ A ∈ S ℋ
2 ax-hcompl ⊢ f ∈ Cauchy → ∃ x ∈ ℋ f ⇝v x
3 vex ⊢ f ∈ V
4 vex ⊢ x ∈ V
5 3 4 breldm ⊢ f ⇝v x → f ∈ dom ⁡ ⇝v
6 5 rexlimivw ⊢ ∃ x ∈ ℋ f ⇝v x → f ∈ dom ⁡ ⇝v
7 2 6 syl ⊢ f ∈ Cauchy → f ∈ dom ⁡ ⇝v
8 7 ad2antlr ⊢ A ⊆ ℋ ∧ f ∈ Cauchy ∧ f : ℕ ⟶ ⊥ ⁡ A → f ∈ dom ⁡ ⇝v
9 hlimf ⊢ ⇝v : dom ⁡ ⇝v ⟶ ℋ
10 9 ffvelcdmi ⊢ f ∈ dom ⁡ ⇝v → ⇝v ⁡ f ∈ ℋ
11 8 10 syl ⊢ A ⊆ ℋ ∧ f ∈ Cauchy ∧ f : ℕ ⟶ ⊥ ⁡ A → ⇝v ⁡ f ∈ ℋ
12 simplll ⊢ A ⊆ ℋ ∧ f ∈ Cauchy ∧ f : ℕ ⟶ ⊥ ⁡ A ∧ x ∈ A → A ⊆ ℋ
13 simpllr ⊢ A ⊆ ℋ ∧ f ∈ Cauchy ∧ f : ℕ ⟶ ⊥ ⁡ A ∧ x ∈ A → f ∈ Cauchy
14 simplr ⊢ A ⊆ ℋ ∧ f ∈ Cauchy ∧ f : ℕ ⟶ ⊥ ⁡ A ∧ x ∈ A → f : ℕ ⟶ ⊥ ⁡ A
15 simpr ⊢ A ⊆ ℋ ∧ f ∈ Cauchy ∧ f : ℕ ⟶ ⊥ ⁡ A ∧ x ∈ A → x ∈ A
16 12 13 14 15 occllem ⊢ A ⊆ ℋ ∧ f ∈ Cauchy ∧ f : ℕ ⟶ ⊥ ⁡ A ∧ x ∈ A → ⇝v ⁡ f ⋅ ih x = 0
17 16 ralrimiva ⊢ A ⊆ ℋ ∧ f ∈ Cauchy ∧ f : ℕ ⟶ ⊥ ⁡ A → ∀ x ∈ A ⇝v ⁡ f ⋅ ih x = 0
18 ocel ⊢ A ⊆ ℋ → ⇝v ⁡ f ∈ ⊥ ⁡ A ↔ ⇝v ⁡ f ∈ ℋ ∧ ∀ x ∈ A ⇝v ⁡ f ⋅ ih x = 0
19 18 ad2antrr ⊢ A ⊆ ℋ ∧ f ∈ Cauchy ∧ f : ℕ ⟶ ⊥ ⁡ A → ⇝v ⁡ f ∈ ⊥ ⁡ A ↔ ⇝v ⁡ f ∈ ℋ ∧ ∀ x ∈ A ⇝v ⁡ f ⋅ ih x = 0
20 11 17 19 mpbir2and ⊢ A ⊆ ℋ ∧ f ∈ Cauchy ∧ f : ℕ ⟶ ⊥ ⁡ A → ⇝v ⁡ f ∈ ⊥ ⁡ A
21 ffun ⊢ ⇝v : dom ⁡ ⇝v ⟶ ℋ → Fun ⁡ ⇝v
22 funfvbrb ⊢ Fun ⁡ ⇝v → f ∈ dom ⁡ ⇝v ↔ f ⇝v ⇝v ⁡ f
23 9 21 22 mp2b ⊢ f ∈ dom ⁡ ⇝v ↔ f ⇝v ⇝v ⁡ f
24 8 23 sylib ⊢ A ⊆ ℋ ∧ f ∈ Cauchy ∧ f : ℕ ⟶ ⊥ ⁡ A → f ⇝v ⇝v ⁡ f
25 breq2 ⊢ x = ⇝v ⁡ f → f ⇝v x ↔ f ⇝v ⇝v ⁡ f
26 25 rspcev ⊢ ⇝v ⁡ f ∈ ⊥ ⁡ A ∧ f ⇝v ⇝v ⁡ f → ∃ x ∈ ⊥ ⁡ A f ⇝v x
27 20 24 26 syl2anc ⊢ A ⊆ ℋ ∧ f ∈ Cauchy ∧ f : ℕ ⟶ ⊥ ⁡ A → ∃ x ∈ ⊥ ⁡ A f ⇝v x
28 27 ex ⊢ A ⊆ ℋ ∧ f ∈ Cauchy → f : ℕ ⟶ ⊥ ⁡ A → ∃ x ∈ ⊥ ⁡ A f ⇝v x
29 28 ralrimiva ⊢ A ⊆ ℋ → ∀ f ∈ Cauchy f : ℕ ⟶ ⊥ ⁡ A → ∃ x ∈ ⊥ ⁡ A f ⇝v x
30 isch3 ⊢ ⊥ ⁡ A ∈ C ℋ ↔ ⊥ ⁡ A ∈ S ℋ ∧ ∀ f ∈ Cauchy f : ℕ ⟶ ⊥ ⁡ A → ∃ x ∈ ⊥ ⁡ A f ⇝v x
31 1 29 30 sylanbrc ⊢ A ⊆ ℋ → ⊥ ⁡ A ∈ C ℋ