Metamath Proof Explorer


Theorem nnfoctb

Description: There exists a mapping from NN onto any (nonempty) countable set. (Contributed by Glauco Siliprandi, 17-Aug-2020)

Ref Expression
Assertion nnfoctb ⊢ A ≼ ω ∧ A ≠ ∅ → ∃ f f : ℕ ⟶ onto A

Proof

Step Hyp Ref Expression
1 simpr ⊢ A ≼ ω ∧ A ≠ ∅ → A ≠ ∅
2 reldom ⊢ Rel ⁡ ≼
3 2 a1i ⊢ A ≼ ω → Rel ⁡ ≼
4 brrelex1 ⊢ Rel ⁡ ≼ ∧ A ≼ ω → A ∈ V
5 3 4 mpancom ⊢ A ≼ ω → A ∈ V
6 0sdomg ⊢ A ∈ V → ∅ ≺ A ↔ A ≠ ∅
7 5 6 syl ⊢ A ≼ ω → ∅ ≺ A ↔ A ≠ ∅
8 7 adantr ⊢ A ≼ ω ∧ A ≠ ∅ → ∅ ≺ A ↔ A ≠ ∅
9 1 8 mpbird ⊢ A ≼ ω ∧ A ≠ ∅ → ∅ ≺ A
10 nnenom ⊢ ℕ ≈ ω
11 10 ensymi ⊢ ω ≈ ℕ
12 11 a1i ⊢ A ≼ ω → ω ≈ ℕ
13 domentr ⊢ A ≼ ω ∧ ω ≈ ℕ → A ≼ ℕ
14 12 13 mpdan ⊢ A ≼ ω → A ≼ ℕ
15 14 adantr ⊢ A ≼ ω ∧ A ≠ ∅ → A ≼ ℕ
16 fodomr ⊢ ∅ ≺ A ∧ A ≼ ℕ → ∃ f f : ℕ ⟶ onto A
17 9 15 16 syl2anc ⊢ A ≼ ω ∧ A ≠ ∅ → ∃ f f : ℕ ⟶ onto A