Metamath Proof Explorer


Theorem isopn2

Description: A subset of the underlying set of a topology is open iff its complement is closed. (Contributed by NM, 4-Oct-2006)

Ref Expression
Hypothesis iscld.1
|- X = U. J
Assertion isopn2
|- ( ( J e. Top /\ S C_ X ) -> ( S e. J <-> ( X \ S ) e. ( Clsd ` J ) ) )

Proof

Step Hyp Ref Expression
1 iscld.1
 |-  X = U. J
2 difss
 |-  ( X \ S ) C_ X
3 1 iscld2
 |-  ( ( J e. Top /\ ( X \ S ) C_ X ) -> ( ( X \ S ) e. ( Clsd ` J ) <-> ( X \ ( X \ S ) ) e. J ) )
4 2 3 mpan2
 |-  ( J e. Top -> ( ( X \ S ) e. ( Clsd ` J ) <-> ( X \ ( X \ S ) ) e. J ) )
5 dfss4
 |-  ( S C_ X <-> ( X \ ( X \ S ) ) = S )
6 5 biimpi
 |-  ( S C_ X -> ( X \ ( X \ S ) ) = S )
7 6 eleq1d
 |-  ( S C_ X -> ( ( X \ ( X \ S ) ) e. J <-> S e. J ) )
8 4 7 sylan9bb
 |-  ( ( J e. Top /\ S C_ X ) -> ( ( X \ S ) e. ( Clsd ` J ) <-> S e. J ) )
9 8 bicomd
 |-  ( ( J e. Top /\ S C_ X ) -> ( S e. J <-> ( X \ S ) e. ( Clsd ` J ) ) )