reserve x,X,X2,Y,Y2 for set;
reserve GX for non empty TopSpace;
reserve A2,B2 for Subset of GX;
reserve B for Subset of GX;

theorem
  for A,B being Subset of GX st A is a_component & B is connected
  & B<>{} & A misses B holds A misses Component_of B
proof
  let A,B be Subset of GX;
  assume that
A1: A is a_component and
A2: B is connected & B<>{} and
A3: A /\ B ={};
A4: A is connected by A1;
  assume A /\ Component_of B <>{};
  then consider x being Point of GX such that
A5: x in A /\ Component_of B by SUBSET_1:4;
  x in A by A5,XBOOLE_0:def 4;
  then
A6: Component_of x=Component_of A by A4,Th15;
A7: x in Component_of B by A5,XBOOLE_0:def 4;
  Component_of A=A & Component_of B=Component_of Component_of B by A1,A2,Th7
,Th11;
  then (Component_of B) /\ B={} by A2,A3,A7,A6,Th5,Th15;
  hence contradiction by A2,Th1,XBOOLE_1:28;
end;
