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 e. ( TopOn ` A ) -> K e. TopSp )

Proof

Step Hyp Ref Expression
1 tsettps.a
 |-  A = ( Base ` K )
2 tsettps.j
 |-  J = ( TopSet ` K )
3 1 2 topontopn
 |-  ( J e. ( TopOn ` A ) -> J = ( TopOpen ` K ) )
4 id
 |-  ( J e. ( TopOn ` A ) -> J e. ( TopOn ` A ) )
5 3 4 eqeltrrd
 |-  ( J e. ( TopOn ` A ) -> ( TopOpen ` K ) e. ( TopOn ` A ) )
6 eqid
 |-  ( TopOpen ` K ) = ( TopOpen ` K )
7 1 6 istps
 |-  ( K e. TopSp <-> ( TopOpen ` K ) e. ( TopOn ` A ) )
8 5 7 sylibr
 |-  ( J e. ( TopOn ` A ) -> K e. TopSp )