Metamath Proof Explorer


Theorem clatlubcl

Description: Any subset of the base set has an LUB in a complete lattice. (Contributed by NM, 14-Sep-2011)

Ref Expression
Hypotheses clatlubcl.b 𝐵 = ( Base ‘ 𝐾 )
clatlubcl.u 𝑈 = ( lub ‘ 𝐾 )
Assertion clatlubcl ( ( 𝐾 ∈ CLat ∧ 𝑆𝐵 ) → ( 𝑈𝑆 ) ∈ 𝐵 )

Proof

Step Hyp Ref Expression
1 clatlubcl.b 𝐵 = ( Base ‘ 𝐾 )
2 clatlubcl.u 𝑈 = ( lub ‘ 𝐾 )
3 eqid ( glb ‘ 𝐾 ) = ( glb ‘ 𝐾 )
4 1 2 3 clatlem ( ( 𝐾 ∈ CLat ∧ 𝑆𝐵 ) → ( ( 𝑈𝑆 ) ∈ 𝐵 ∧ ( ( glb ‘ 𝐾 ) ‘ 𝑆 ) ∈ 𝐵 ) )
5 4 simpld ( ( 𝐾 ∈ CLat ∧ 𝑆𝐵 ) → ( 𝑈𝑆 ) ∈ 𝐵 )