reserve X for non empty TopSpace;
reserve x for Point of X;
reserve U1 for Subset of X;

theorem Th18:
  X is locally_connected iff for A being non empty Subset of X, B
  being Subset of X st A is open & B is_a_component_of A holds B is open
proof
  thus X is locally_connected implies for A being non empty Subset of X, B
  being Subset of X st A is open & B is_a_component_of A holds B is open
  proof
    assume
A1: X is locally_connected;
    let A be non empty Subset of X, B be Subset of X such that
A2: A is open and
A3: B is_a_component_of A;
    consider B1 being Subset of X|A such that
A4: B1=B and
A5: B1 is a_component by A3,CONNSP_1:def 6;
    reconsider B1 as Subset of X|A;
    A is locally_connected by A1,A2,Th17;
    then X|A is locally_connected;
    then B1 is open by A5,Th15;
    then B1 in the topology of X|A by PRE_TOPC:def 2;
    then consider B2 being Subset of X such that
A6: B2 in the topology of X and
A7: B1 = B2 /\ [#](X|A) by PRE_TOPC:def 4;
A8: B = B2 /\ A by A4,A7,PRE_TOPC:def 5;
    reconsider B2 as Subset of X;
    B2 is open by A6,PRE_TOPC:def 2;
    hence thesis by A2,A8;
  end;
  assume
A9: for A being non empty Subset of X, B being Subset of X st A is open
  & B is_a_component_of A holds B is open;
  let x;
  for U1 being non empty Subset of X st U1 is open & x in U1 ex x1 being
  Point of X|U1 st x1=x & x in Int(Component_of x1)
  proof
    let U1 be non empty Subset of X such that
A10: U1 is open and
A11: x in U1;
    x in [#](X|U1) by A11,PRE_TOPC:def 5;
    then reconsider x1=x as Point of X|U1;
    set S1=Component_of x1;
    reconsider S=S1 as Subset of X by PRE_TOPC:11;
    S1 is a_component by CONNSP_1:40;
    then S is_a_component_of U1 by CONNSP_1:def 6;
    then
A12: S is open by A9,A10;
    take x1;
    x in S by CONNSP_1:38;
    then S1 is a_neighborhood of x1 by A12,Th3,Th10;
    hence thesis by Def1;
  end;
  hence thesis by Th14;
end;
