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 = J → j = 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