Metamath Proof Explorer


Theorem tsettps

Description: If the topology component is already correctly truncated, then it forms a topological space (with the topology extractor function coming out the same as the component). (Contributed by Mario Carneiro, 13-Aug-2015)

Ref Expression
Hypotheses tsettps.a 𝐴 = ( Base ‘ 𝐾 )
tsettps.j 𝐽 = ( TopSet ‘ 𝐾 )
Assertion tsettps ( 𝐽 ∈ ( TopOn ‘ 𝐴 ) → 𝐾 ∈ TopSp )

Proof

Step Hyp Ref Expression
1 tsettps.a 𝐴 = ( Base ‘ 𝐾 )
2 tsettps.j 𝐽 = ( TopSet ‘ 𝐾 )
3 1 2 topontopn ( 𝐽 ∈ ( TopOn ‘ 𝐴 ) → 𝐽 = ( TopOpen ‘ 𝐾 ) )
4 id ( 𝐽 ∈ ( TopOn ‘ 𝐴 ) → 𝐽 ∈ ( TopOn ‘ 𝐴 ) )
5 3 4 eqeltrrd ( 𝐽 ∈ ( TopOn ‘ 𝐴 ) → ( TopOpen ‘ 𝐾 ) ∈ ( TopOn ‘ 𝐴 ) )
6 eqid ( TopOpen ‘ 𝐾 ) = ( TopOpen ‘ 𝐾 )
7 1 6 istps ( 𝐾 ∈ TopSp ↔ ( TopOpen ‘ 𝐾 ) ∈ ( TopOn ‘ 𝐴 ) )
8 5 7 sylibr ( 𝐽 ∈ ( TopOn ‘ 𝐴 ) → 𝐾 ∈ TopSp )