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 ( 𝑥𝐴𝑥 ≈ ω ) )

Proof

Step Hyp Ref Expression
1 infcntss.1 𝐴 ∈ V
2 1 domen ( ω ≼ 𝐴 ↔ ∃ 𝑥 ( ω ≈ 𝑥𝑥𝐴 ) )
3 ensym ( ω ≈ 𝑥𝑥 ≈ ω )
4 3 anim1ci ( ( ω ≈ 𝑥𝑥𝐴 ) → ( 𝑥𝐴𝑥 ≈ ω ) )
5 4 eximi ( ∃ 𝑥 ( ω ≈ 𝑥𝑥𝐴 ) → ∃ 𝑥 ( 𝑥𝐴𝑥 ≈ ω ) )
6 2 5 sylbi ( ω ≼ 𝐴 → ∃ 𝑥 ( 𝑥𝐴𝑥 ≈ ω ) )