Metamath Proof Explorer


Theorem isconn

Description: The predicate J is a connected topology . (Contributed by FL, 17-Nov-2008)

Ref Expression
Hypothesis isconn.1 𝑋 = 𝐽
Assertion isconn ( 𝐽 ∈ Conn ↔ ( 𝐽 ∈ Top ∧ ( 𝐽 ∩ ( Clsd ‘ 𝐽 ) ) = { ∅ , 𝑋 } ) )

Proof

Step Hyp Ref Expression
1 isconn.1 𝑋 = 𝐽
2 id ( 𝑗 = 𝐽𝑗 = 𝐽 )
3 fveq2 ( 𝑗 = 𝐽 → ( Clsd ‘ 𝑗 ) = ( Clsd ‘ 𝐽 ) )
4 2 3 ineq12d ( 𝑗 = 𝐽 → ( 𝑗 ∩ ( Clsd ‘ 𝑗 ) ) = ( 𝐽 ∩ ( Clsd ‘ 𝐽 ) ) )
5 unieq ( 𝑗 = 𝐽 𝑗 = 𝐽 )
6 5 1 eqtr4di ( 𝑗 = 𝐽 𝑗 = 𝑋 )
7 6 preq2d ( 𝑗 = 𝐽 → { ∅ , 𝑗 } = { ∅ , 𝑋 } )
8 4 7 eqeq12d ( 𝑗 = 𝐽 → ( ( 𝑗 ∩ ( Clsd ‘ 𝑗 ) ) = { ∅ , 𝑗 } ↔ ( 𝐽 ∩ ( Clsd ‘ 𝐽 ) ) = { ∅ , 𝑋 } ) )
9 df-conn Conn = { 𝑗 ∈ Top ∣ ( 𝑗 ∩ ( Clsd ‘ 𝑗 ) ) = { ∅ , 𝑗 } }
10 8 9 elrab2 ( 𝐽 ∈ Conn ↔ ( 𝐽 ∈ Top ∧ ( 𝐽 ∩ ( Clsd ‘ 𝐽 ) ) = { ∅ , 𝑋 } ) )