Metamath Proof Explorer


Theorem uffinfix

Description: An ultrafilter containing a finite element is fixed. (Contributed by Jeff Hankins, 5-Dec-2009) (Revised by Stefan O'Rear, 2-Aug-2015)

Ref Expression
Assertion uffinfix ⊢ F ∈ UFil ⁡ X ∧ S ∈ F ∧ S ∈ Fin → ∃ x ∈ X F = y ∈ 𝒫 X | x ∈ y

Proof

Step Hyp Ref Expression
1 ufilfil ⊢ F ∈ UFil ⁡ X → F ∈ Fil ⁡ X
2 filfinnfr ⊢ F ∈ Fil ⁡ X ∧ S ∈ F ∧ S ∈ Fin → ⋂ F ≠ ∅
3 1 2 syl3an1 ⊢ F ∈ UFil ⁡ X ∧ S ∈ F ∧ S ∈ Fin → ⋂ F ≠ ∅
4 uffix2 ⊢ F ∈ UFil ⁡ X → ⋂ F ≠ ∅ ↔ ∃ x ∈ X F = y ∈ 𝒫 X | x ∈ y
5 4 3ad2ant1 ⊢ F ∈ UFil ⁡ X ∧ S ∈ F ∧ S ∈ Fin → ⋂ F ≠ ∅ ↔ ∃ x ∈ X F = y ∈ 𝒫 X | x ∈ y
6 3 5 mpbid ⊢ F ∈ UFil ⁡ X ∧ S ∈ F ∧ S ∈ Fin → ∃ x ∈ X F = y ∈ 𝒫 X | x ∈ y