Metamath Proof Explorer


Theorem oppgtopn

Description: Topology of an opposite group. (Contributed by Mario Carneiro, 17-Sep-2015)

Ref Expression
Hypotheses oppgbas.1 𝑂 = ( oppg𝑅 )
oppgtopn.2 𝐽 = ( TopOpen ‘ 𝑅 )
Assertion oppgtopn 𝐽 = ( TopOpen ‘ 𝑂 )

Proof

Step Hyp Ref Expression
1 oppgbas.1 𝑂 = ( oppg𝑅 )
2 oppgtopn.2 𝐽 = ( TopOpen ‘ 𝑅 )
3 eqid ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
4 eqid ( TopSet ‘ 𝑅 ) = ( TopSet ‘ 𝑅 )
5 3 4 topnval ( ( TopSet ‘ 𝑅 ) ↾t ( Base ‘ 𝑅 ) ) = ( TopOpen ‘ 𝑅 )
6 1 3 oppgbas ( Base ‘ 𝑅 ) = ( Base ‘ 𝑂 )
7 1 4 oppgtset ( TopSet ‘ 𝑅 ) = ( TopSet ‘ 𝑂 )
8 6 7 topnval ( ( TopSet ‘ 𝑅 ) ↾t ( Base ‘ 𝑅 ) ) = ( TopOpen ‘ 𝑂 )
9 2 5 8 3eqtr2i 𝐽 = ( TopOpen ‘ 𝑂 )