reserve FT for non empty RelStr,
  A,B,C for Subset of FT;

theorem
  FT is connected implies for A, B being Subset of FT st [#]FT = A \/ B
  & A <> {}FT & B <> {}FT & A is closed & B is closed holds A meets B
proof
  assume
A1: [#]FT is connected;
  given A, B being Subset of FT such that
A2: [#]FT = A \/ B & A <> {}FT & B <> {}FT & A is closed & B is closed &
  A misses B;
  thus contradiction by A1,A2;
end;
