Theorem clatlubcl2

Description: Any subset of the base set has an LUB in a complete lattice. (Contributed by NM, 13-Sep-2018)

	Ref	Expression
Hypotheses	clatlubcl.b	⊢ 𝐵 = ( Base ‘ 𝐾 )
	clatlubcl.u	⊢ 𝑈 = ( lub ‘ 𝐾 )
Assertion	clatlubcl2	⊢ ( ( 𝐾 ∈ CLat ∧ 𝑆 ⊆ 𝐵 ) → 𝑆 ∈ dom 𝑈 )

Proof

Step	Hyp	Ref	Expression
1		clatlubcl.b	⊢ 𝐵 = ( Base ‘ 𝐾 )
2		clatlubcl.u	⊢ 𝑈 = ( lub ‘ 𝐾 )
3	1	fvexi	⊢ 𝐵 ∈ V
4	3	elpw2	⊢ ( 𝑆 ∈ 𝒫 𝐵 ↔ 𝑆 ⊆ 𝐵 )
5	4	biimpri	⊢ ( 𝑆 ⊆ 𝐵 → 𝑆 ∈ 𝒫 𝐵 )
6	5	adantl	⊢ ( ( 𝐾 ∈ CLat ∧ 𝑆 ⊆ 𝐵 ) → 𝑆 ∈ 𝒫 𝐵 )
7		eqid	⊢ ( glb ‘ 𝐾 ) = ( glb ‘ 𝐾 )
8	1 2 7	isclat	⊢ ( 𝐾 ∈ CLat ↔ ( 𝐾 ∈ Poset ∧ ( dom 𝑈 = 𝒫 𝐵 ∧ dom ( glb ‘ 𝐾 ) = 𝒫 𝐵 ) ) )
9		simprl	⊢ ( ( 𝐾 ∈ Poset ∧ ( dom 𝑈 = 𝒫 𝐵 ∧ dom ( glb ‘ 𝐾 ) = 𝒫 𝐵 ) ) → dom 𝑈 = 𝒫 𝐵 )
10	8 9	sylbi	⊢ ( 𝐾 ∈ CLat → dom 𝑈 = 𝒫 𝐵 )
11	10	adantr	⊢ ( ( 𝐾 ∈ CLat ∧ 𝑆 ⊆ 𝐵 ) → dom 𝑈 = 𝒫 𝐵 )
12	6 11	eleqtrrd	⊢ ( ( 𝐾 ∈ CLat ∧ 𝑆 ⊆ 𝐵 ) → 𝑆 ∈ dom 𝑈 )