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 AV
Assertion infcntss ωAxxAxω

Proof

Step Hyp Ref Expression
1 infcntss.1 AV
2 1 domen ωAxωxxA
3 ensym ωxxω
4 3 anim1ci ωxxAxAxω
5 4 eximi xωxxAxxAxω
6 2 5 sylbi ωAxxAxω