Metamath Proof Explorer


Theorem infcntss

Description: Every infinite set has a denumerable subset. Similar to Exercise 8 of TakeutiZaring p. 91. (However, we need neither AC nor the Axiom of Infinity because of the way we express "infinite" in the antecedent.) (Contributed by NM, 23-Oct-2004)

Ref Expression
Hypothesis infcntss.1 ⊢ A ∈ V
Assertion infcntss ⊢ ω ≼ A → ∃ x x ⊆ A ∧ x ≈ ω

Proof

Step Hyp Ref Expression
1 infcntss.1 ⊢ A ∈ V
2 1 domen ⊢ ω ≼ A ↔ ∃ x ω ≈ x ∧ x ⊆ A
3 ensym ⊢ ω ≈ x → x ≈ ω
4 3 anim1ci ⊢ ω ≈ x ∧ x ⊆ A → x ⊆ A ∧ x ≈ ω
5 4 eximi ⊢ ∃ x ω ≈ x ∧ x ⊆ A → ∃ x x ⊆ A ∧ x ≈ ω
6 2 5 sylbi ⊢ ω ≼ A → ∃ x x ⊆ A ∧ x ≈ ω