The Mizar article:

Locally Connected Spaces

by
Beata Padlewska

Received September 5, 1990

Copyright (c) 1990 Association of Mizar Users

MML identifier: CONNSP_2
[ MML identifier index ]


environ

 vocabulary PRE_TOPC, TOPS_1, SUBSET_1, BOOLE, RELAT_1, CONNSP_1, RELAT_2,
      SETFAM_1, TARSKI, COMPTS_1, CONNSP_2;
 notation TARSKI, XBOOLE_0, SUBSET_1, STRUCT_0, PRE_TOPC, TOPS_1, CONNSP_1,
      COMPTS_1;
 constructors TOPS_1, CONNSP_1, COMPTS_1, MEMBERED;
 clusters PRE_TOPC, STRUCT_0, SUBSET_1, MEMBERED, ZFMISC_1;
 requirements SUBSET, BOOLE;
 definitions COMPTS_1;
 theorems TARSKI, ZFMISC_1, SETFAM_1, PRE_TOPC, TOPS_1, CONNSP_1, COMPTS_1,
      XBOOLE_0, XBOOLE_1;
 schemes PRE_TOPC;

begin
::
::            A NEIGHBORHOOD OF A POINT
::
definition let X be non empty TopSpace,x be Point of X;
  mode a_neighborhood of x -> Subset of X means
:Def1: x in Int(it);
existence
proof
  take Z=[#] X;
    Int([#] X) = [#] X by TOPS_1:43;
 hence x in Int(Z) by PRE_TOPC:13;
end;
end;

::
::               A NEIGHBORHOOD OF A SET
::

definition let X be non empty TopSpace,A be Subset of X;
  mode a_neighborhood of A -> Subset of X means
:Def2: A c= Int(it);
existence
proof
  take Z=[#] X;
    Int([#] X) = [#] X by TOPS_1:43;
 hence A c= Int(Z) by PRE_TOPC:14;
end;
end;

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

canceled 2;

theorem
  for A,B being Subset of X holds
A is a_neighborhood of x & B is a_neighborhood of x implies
A \/ B is a_neighborhood of x
proof
 let A,B be Subset of X;
 reconsider A1 = A, B1 = B as Subset of X;
   A1 is a_neighborhood of x & B1 is a_neighborhood of x implies
  A1 \/ B1 is a_neighborhood of x
 proof
   assume x in Int A1 & x in Int B1;
  then A1:  x in Int A1 \/ Int B1 by XBOOLE_0:def 2;
          Int A1 \/ Int B1 c= Int (A1 \/ B1) by TOPS_1:49;
    hence thesis by A1,Def1;
  end;
  hence thesis;
end;

theorem
  for A,B being Subset of X holds
A is a_neighborhood of x & B is a_neighborhood of x iff
A /\ B is a_neighborhood of x
proof
 let A,B be Subset of X;
  A is a_neighborhood of x & B is a_neighborhood of x iff
  x in Int A & x in Int B by Def1;
  then A is a_neighborhood of x & B is a_neighborhood of x iff
  x in Int A /\ Int B by XBOOLE_0:def 3;
  then A is a_neighborhood of x & B is a_neighborhood of x iff
  x in Int (A /\ B) by TOPS_1:46;
 hence thesis by Def1;
end;

theorem Th5:
 for U1 being Subset of X, x being Point of X st U1 is open & x in U1 holds
 U1 is a_neighborhood of x
proof
 let U1 be Subset of X, x be Point of X;
  assume U1 is open & x in U1;
   then x in Int U1 by TOPS_1:55;
 hence thesis by Def1;
end;

theorem Th6:
 for U1 being Subset of X, x being Point of X
 st U1 is a_neighborhood of x
 holds x in U1
proof
 let U1 be Subset of X, x be Point of X;
 assume U1 is a_neighborhood of x;
  then A1: x in Int (U1) by Def1;
        Int(U1) c= U1 by TOPS_1:44;
 hence thesis by A1;
end;

theorem Th7:
 U1 is a_neighborhood of x implies
 ex V being non empty Subset of X st
  V is a_neighborhood of x & V is open & V c= U1
proof
  assume U1 is a_neighborhood of x;
  then x in Int(U1) by Def1;
  then consider V being Subset of X such that
  A1:V is open and
  A2:V c= U1 and
  A3:x in V by TOPS_1:54;
  reconsider V as non empty Subset of X by A3;
  take V;
 thus thesis by A1,A2,A3,Th5;
end;

theorem Th8:
 U1 is a_neighborhood of x iff
  ex V being Subset of X st V is open & V c= U1 & x in V
proof
 A1:now
   given V being Subset of X such that
    A2:V is open & V c= U1 & x in V;
      x in Int U1 by A2,TOPS_1:54;
   hence U1 is a_neighborhood of x by Def1;
  end;
    now
   assume U1 is a_neighborhood of x;
    then consider V being non empty Subset of X such that
    A3:V is a_neighborhood of x and
    A4:V is open and
    A5:V c= U1 by Th7;
    take V;
      x in V by A3,Th6;
   hence ex V being Subset of X st V is open & V c= U1 & x in V by A4,A5;
  end;
 hence thesis by A1;
end;

theorem
   for U1 being Subset of X holds
  U1 is open iff for x st x in U1 ex A being Subset of X st
 A is a_neighborhood of x & A c= U1
proof
 let U1 be Subset of X;
 A1:now
   assume A2:U1 is open;
       let x; assume
      x in U1;
    then U1 is a_neighborhood of x by A2,Th5;
   hence ex A being Subset of X
   st A is a_neighborhood of x & A c= U1;
  end;
    now assume
   A3:for x st x in U1 ex A being Subset of X st
      A is a_neighborhood of x & A c= U1;
     for x being set holds x in U1 iff ex V being Subset of X st
                     V is open & V c= U1 & x in V
   proof
    let x be set;
    thus x in U1 implies ex V being Subset of X st V is open & V c= U1 & x in
V
     proof
      assume A4:x in U1;
       then reconsider x as Point of X;
       consider S being Subset of X such that
       A5: S is a_neighborhood of x & S c= U1 by A3,A4;
       consider V being Subset of X such that
       A6:V is open & V c= S & x in V by A5,Th8;
       take V;
      thus thesis by A5,A6,XBOOLE_1:1;
     end;
       given V being Subset of X such that
       A7:V is open & V c= U1 & x in V;
       thus x in U1 by A7;
     end;
   hence U1 is open by TOPS_1:57;
  end;
 hence thesis by A1;
end;

theorem
  for V being Subset of X holds
V is a_neighborhood of {x} iff V is a_neighborhood of x
 proof
let V be Subset of X;
  thus V is a_neighborhood of {x} implies V is a_neighborhood of x
  proof
    assume
       V is a_neighborhood of {x};
     then A1: {x} c= Int(V) by Def2;
       x in {x} by TARSKI:def 1;
     hence V is a_neighborhood of x by A1,Def1;
  end;
  assume V is a_neighborhood of x;
   then x in Int (V) by Def1;
   then for p being set holds p in {x} implies p in Int V by TARSKI:def 1;
   then {x} c= Int V by TARSKI:def 3;
 hence V is a_neighborhood of {x} by Def2;
end;

theorem Th11:
for B being non empty Subset of X, x being Point of X|B,
    A being Subset of X|B,
    A1 being Subset of X, x1 being Point of X
 st B is open & A is a_neighborhood of x & A = A1 & x = x1 holds
 A1 is a_neighborhood of x1
proof
 let B be non empty Subset of X, x be Point of X|B,
 A be Subset of X|B,
  A1 be Subset of X, x1 be Point of X such that
   A1: B is open and
   A2: A is a_neighborhood of x and
   A3: A=A1 & x=x1;
      x in Int A by A2,Def1;
    then consider Q1 being Subset of X|B such that
    A4: Q1 is open and
    A5: Q1 c= A and
    A6: x in Q1 by TOPS_1:54;
      Q1 in the topology of X|B by A4,PRE_TOPC:def 5;
    then consider Q being Subset of X such that
    A7: Q in the topology of X and
    A8: Q1 = Q /\ [#](X|B) by PRE_TOPC:def 9;
    A9: Q1 = Q /\ B by A8,PRE_TOPC:def 10;
    reconsider Q2 = Q as Subset of X;
      Q2 is open by A7,PRE_TOPC:def 5;
     then Q /\ B is open by A1,TOPS_1:38;
   then x1 in Int A1 by A3,A5,A6,A9,TOPS_1:54;
  hence thesis by Def1;
end;

Lm1:
 for X1 being SubSpace of X, A being Subset of X,
 A1 being Subset of X1 st
 A = A1 holds Int (A) /\ [#](X1) c= Int A1
proof
 let X1 be SubSpace of X, A be Subset of X,
 A1 be Subset of X1;
  assume A1: A = A1;
    for p being set holds p in Int (A) /\ [#](X1) implies p in Int A1
  proof
  let p be set;
   assume p in Int (A) /\ [#](X1);
        then A2:p in Int (A) & p in [#](X1) by XBOOLE_0:def 3;
        then consider Q being Subset of X such that
        A3:Q is open and
        A4:Q c= A and
        A5:p in Q by TOPS_1:54;
          ex Q1 being Subset of X1 st Q1 is open & Q1 c= A1 & p in Q1
        proof
            Q /\ [#] X1 c= [#] X1 by XBOOLE_1:17;
          then reconsider Q1=Q /\ [#] X1 as Subset of X1 by PRE_TOPC:16;
          take Q1;
A6:          Q1 c= Q by XBOOLE_1:17;
            Q in the topology of X by A3,PRE_TOPC:def 5;
          then Q1 in the topology of X1 by PRE_TOPC:def 9;
          hence thesis by A1,A2,A4,A5,A6,PRE_TOPC:def 5,XBOOLE_0:def 3,XBOOLE_1
:1;
        end;
      hence p in Int A1 by TOPS_1:54;
     end;
  hence thesis by TARSKI:def 3;
end;

theorem Th12:
for B being non empty Subset of X, x being Point of X|B,
    A being Subset of X|B,
    A1 being Subset of X, x1 being Point of X
 st A1 is a_neighborhood of x1 & A=A1 & x=x1 holds
 A is a_neighborhood of x
proof
 let B be non empty Subset of X, x be Point of X|B,
 A be Subset of X|B,
  A1 be Subset of X, x1 be Point of X such that
   A1: A1 is a_neighborhood of x1 and
   A2: A=A1 & x=x1;
   A3:Int (A1) /\ [#](X|B) c= Int A by A2,Lm1;
   A4:x1 in [#](X|B) by A2,PRE_TOPC:13;
     x1 in Int A1 by A1,Def1;
   then x1 in Int (A1) /\ [#](X|B) by A4,XBOOLE_0:def 3;
   hence thesis by A2,A3,Def1;
end;

theorem Th13:
for A being Subset of X, B being Subset of X st
 A is_a_component_of X & A c= B holds A is_a_component_of B
proof
 let A be Subset of X,B be Subset of X such that
  A1:A is_a_component_of X and
  A2:A c= B;
  A3:A is connected by A1,CONNSP_1:def 5;
       ex A1 being Subset of X|B st A=A1 &
     A1 is_a_component_of X|B
     proof
         B = [#](X|B) by PRE_TOPC:def 10;
       then reconsider C = A as Subset of X|B by A2,PRE_TOPC:16;
       take A1=C;
       A4:A1 is connected by A3,CONNSP_1:24;
         for D being Subset of X|B st D is connected holds A1 c= D implies A1=D
       proof
        let D be Subset of X|B such that
         A5:D is connected;
         assume A6:A1 c= D;
         reconsider D1=D as Subset of X by PRE_TOPC:39;
           D1 is connected by A5,CONNSP_1:24;
        hence thesis by A1,A6,CONNSP_1:def 5;
       end;
      hence thesis by A4,CONNSP_1:def 5;
     end;
    hence thesis by CONNSP_1:def 6;
end;

theorem
   for X1 being non empty SubSpace of X, x being Point of X,
     x1 being Point of X1 st
 x = x1 holds skl x1 c= skl x
proof
 let X1 be non empty SubSpace of X, x be Point of X, x1 be Point of X1;
  assume A1:x = x1;
     consider F being Subset-Family of X such that
A2: (for A being Subset of X holds A in F iff A is connected
    & x in A) and
A3: union F = skl x by CONNSP_1:def 7;
    reconsider Z = skl x1 as Subset of X by PRE_TOPC:39;
A4: skl x1 is connected by CONNSP_1:41;
A5: x1 in Z by CONNSP_1:40;
      Z is connected by A4,CONNSP_1:24;
    then Z in F by A1,A2,A5;
  hence skl x1 c= skl x by A3,ZFMISC_1:92;
end;

::
::            LOCALLY CONNECTED TOPOLOGICAL SPACES
::

definition let X be non empty TopSpace, x be Point of X;
   pred X is_locally_connected_in x means :Def3:
         for U1 being Subset of X st
           U1 is a_neighborhood of x ex V being Subset of X st
           V is a_neighborhood of x & V is connected & V c= U1;
end;

definition let X be non empty TopSpace;
  attr X is locally_connected means
   :Def4:  for x being Point of X holds X is_locally_connected_in x;
end;

definition
 let X be non empty TopSpace,
     A be Subset of X, x be Point of X;
 pred A is_locally_connected_in x means
  :Def5: for B being non empty Subset of X st A = B
   ex x1 being Point of X|B st x1=x & X|B is_locally_connected_in x1;
end;

definition
 let X be non empty TopSpace, A be non empty Subset of X;
 attr A is locally_connected means :Def6: X|A is locally_connected;
end;

canceled 4;

theorem Th19:
 for x holds X is_locally_connected_in x iff
 for V,S being Subset of X st V is a_neighborhood of x &
 S is_a_component_of V & x in S
 holds S is a_neighborhood of x
proof
 let x;
 thus X is_locally_connected_in x implies
      for V,S being Subset of X st V is a_neighborhood of x &
      S is_a_component_of V & x in S
      holds S is a_neighborhood of x
  proof
   assume A1:X is_locally_connected_in x;
   let V,S be Subset of X such that
       A2: V is a_neighborhood of x and
       A3: S is_a_component_of V and
       A4: x in S;
           consider V1 being Subset of X such that
       A5: V1 is a_neighborhood of x and
       A6: V1 is connected and
       A7: V1 c= V by A1,A2,Def3;
           consider S1 being Subset of X|V such that
       A8: S1=S & S1 is_a_component_of X|V by A3,CONNSP_1:def 6;
           reconsider V' = V as non empty Subset of X by A2,Th6;
             V1 c= [#](X|V) by A7,PRE_TOPC:def 10;
           then reconsider V2=V1 as Subset of X|V by PRE_TOPC:16;
             V2 is connected by A6,CONNSP_1:24;
           then V2 misses S1 or V2 c= S1 by A8,CONNSP_1:38;
      then A9: V2 /\ S1 = {}(X|V') or V2 c= S1 by XBOOLE_0:def 7;
             x in V2 & x in S1 by A4,A5,A8,Th6;
      then A10: Int V1 c= Int S by A8,A9,TOPS_1:48,XBOOLE_0:def 3;
             x in Int V1 by A5,Def1;
           hence thesis by A10,Def1;
  end;
  assume
    A11:for V,S being Subset of X st V is a_neighborhood of x &
       S is_a_component_of V & x in S holds
       S is a_neighborhood of x;
        for U1 being Subset of X st
      U1 is a_neighborhood of x ex V being Subset of X st
      V is a_neighborhood of x & V is connected & V c= U1
      proof
      let U1 be Subset of X; assume
        A12: U1 is a_neighborhood of x;
        then A13: x in U1 by Th6;
        A14: [#](X|U1) = U1 by PRE_TOPC:def 10;
        reconsider U1 as non empty Subset of X by A12,Th6;
              x in [#](X|U1) by A13,PRE_TOPC:def 10;
            then reconsider x1=x as Point of X|U1;
            set S1 = skl x1;
        A15: S1 is_a_component_of X|U1 by CONNSP_1:43;
            reconsider S=S1 as Subset of X by PRE_TOPC:39;
            take S;
        A16: S is_a_component_of U1 by A15,CONNSP_1:def 6;
        A17: x in S by CONNSP_1:40;
              S <> {} X & S1 is connected by CONNSP_1:40,41;
       hence thesis by A11,A12,A14,A16,A17,CONNSP_1:24,PRE_TOPC:14;
    end;
  hence X is_locally_connected_in x by Def3;
end;

theorem Th20:
 for x holds X is_locally_connected_in x iff
 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(skl x1)
proof
 let x;
  A1: X is_locally_connected_in x implies
   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(skl x1)
   proof
    assume A2: X is_locally_connected_in x;
     let U1 be non empty Subset of X such that
      A3: U1 is open and
      A4: x in U1;
            x in [#](X|U1) by A4,PRE_TOPC:def 10;
          then reconsider x1=x as Point of X|U1;
          take x1;
          reconsider S1=skl x1 as Subset of X|U1;
      A5: S1 is_a_component_of X|U1 by CONNSP_1:43;
          reconsider S=S1 as Subset of X by PRE_TOPC:39;
      A6: S is_a_component_of U1 by A5,CONNSP_1:def 6;
      A7: U1 is a_neighborhood of x by A3,A4,Th5;
            x in S by CONNSP_1:40;
          then S is a_neighborhood of x by A2,A6,A7,Th19;
          then S1 is a_neighborhood of x1 by Th12;
        hence thesis by Def1;
     end;
     (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(skl x1))
    implies X is_locally_connected_in x
   proof
    assume
     A8: 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(skl x1);
           for U1 being Subset of X st U1 is a_neighborhood of x
         ex V1 being Subset of X st
         V1 is a_neighborhood of x & V1 is connected & V1 c= U1
         proof
          let U1 be Subset of X; assume
              U1 is a_neighborhood of x;
               then consider V being non empty Subset of X such that
           A9: V is a_neighborhood of x and
           A10: V is open and
           A11: V c= U1 by Th7;
              x in V by A9,Th6;
               then consider x1 being Point of X|V such that
           A12: x1=x and
           A13: x in Int(skl x1) by A8,A10;
               set S1=skl x1;
           A14: S1 is a_neighborhood of x1 by A12,A13,Def1;
               reconsider S=S1 as Subset of X by PRE_TOPC:39;
               take S;
               A15: S <> {} X & S1 is connected by CONNSP_1:40,41;
                 S1 c= [#](X|V) by PRE_TOPC:14;
               then S c= V by PRE_TOPC:def 10;
           hence thesis by A10,A11,A12,A14,A15,Th11,CONNSP_1:24,XBOOLE_1:1;
        end;
     hence X is_locally_connected_in x by Def3;
    end;
   hence thesis by A1;
end;

theorem Th21:
  X is locally_connected implies
  for S being Subset of X st S is_a_component_of X holds
  S is open
proof
 assume A1:X is locally_connected;
 let S be Subset of X such that
 A2: S is_a_component_of X;
 A3: Int(S) c= S by TOPS_1:44;
      now
      let p be set;
       assume
       A4: p in S;
           then reconsider x=p as Point of X;
       A5: X is_locally_connected_in x by A1,Def4;
             S c= [#] X by PRE_TOPC:14;
       then A6: S is_a_component_of [#] X by A2,Th13;
             x in [#] X & [#] X is open by PRE_TOPC:13,TOPS_1:53;
           then [#] X is a_neighborhood of x by Th5;
           then S is a_neighborhood of x by A4,A5,A6,Th19;
       hence p in Int S by Def1;
     end;
    then S c= Int S by TARSKI:def 3;
    then Int S = S by A3,XBOOLE_0:def 10;
 hence S is open by TOPS_1:55;
end;

theorem Th22:
 X is_locally_connected_in x implies
 for A being non empty Subset of X st A is open & x in A holds
  A is_locally_connected_in x
proof
 assume A1: X is_locally_connected_in x;
 let A be non empty Subset of X such that
 A2: A is open and
 A3: x in A;
  reconsider B = A as non empty Subset of X;
A4:  [#](X|A) = A by PRE_TOPC:def 10;
    for C being non empty Subset of X st B = C
 ex x1 being Point of X|C st x1=x & X|C is_locally_connected_in x1
 proof let C be non empty Subset of X; assume
A5: B = C;
   then reconsider y=x as Point of X|C by A3,A4;
  take x1=y;
     for U1 being Subset of X|B st
   U1 is a_neighborhood of x1 ex V being Subset of X|B st
   V is a_neighborhood of x1 & V is connected & V c= U1
   proof
    let U1 be Subset of X|B such that
     A6: U1 is a_neighborhood of x1;
         reconsider U2=U1 as Subset of X by PRE_TOPC:39;
           U2 is a_neighborhood of x by A2,A5,A6,Th11;
         then consider V being Subset of X such that
     A7: V is a_neighborhood of x and
     A8: V is connected and
     A9: V c= U2 by A1,Def3;
           V is Subset of X|B by A9,XBOOLE_1:1;
         then reconsider V1= V as Subset of X|B;
         take V1;
    thus thesis by A5,A7,A8,A9,Th12,CONNSP_1:24;
   end;
  hence thesis by A5,Def3;
 end;
 hence A is_locally_connected_in x by Def5;
end;

theorem Th23:
 X is locally_connected implies
 for A being non empty Subset of X st A is open holds
  A is locally_connected
proof
  assume A1: X is locally_connected;
       let A be non empty Subset of X such that
   A2: A is open;
         for x being Point of X|A holds X|A is_locally_connected_in x
       proof
        let x be Point of X|A;
             x in [#](X|A) by PRE_TOPC:13;
       then A3: x in A by PRE_TOPC:def 10;
           then reconsider x1=x as Point of X;
             X is_locally_connected_in x1 by A1,Def4;
           then A is_locally_connected_in x1 by A2,A3,Th22;
            then ex x2 being Point of X|A st
 x2=x1 & X|A is_locally_connected_in x2 by Def5;
         hence X|A is_locally_connected_in x;
       end;
       then X|A is locally_connected by Def4;
  hence thesis by Def6;
end;

theorem Th24:
 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_of X|A by A3,CONNSP_1:def 6;
       reconsider B1 as Subset of X|A;
          A is locally_connected by A1,A2,Th23;
        then X|A is locally_connected by Def6;
        then B1 is open by A5,Th21;
     then B1 in the topology of X|A by PRE_TOPC:def 5;
        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 9;
    A8: B = B2 /\ A by A4,A7,PRE_TOPC:def 10;
       reconsider B2 as Subset of X;
          B2 is open by A6,PRE_TOPC:def 5;
   hence thesis by A2,A8,TOPS_1:38;
  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(skl 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 10;
          then reconsider x1=x as Point of X|U1;
          take x1;
          set S1=skl x1;
      A12: S1 is_a_component_of X|U1 by CONNSP_1:43;
          reconsider S=S1 as Subset of X by PRE_TOPC:39;
            S is_a_component_of U1 by A12,CONNSP_1:def 6;
      then A13: S is open by A9,A10;
            x in S by CONNSP_1:40;
          then S is a_neighborhood of x by A13,Th5;
          then S1 is a_neighborhood of x1 by Th12;
      hence thesis by Def1;
     end;
   hence thesis by Th20;
end;

theorem
   X is locally_connected implies
 for E being non empty Subset of X,
 C being non empty Subset of X|E st
 C is connected & C is open ex H being Subset of X st
 H is open & H is connected & C = E /\ H
proof
 assume A1: X is locally_connected;
  let E be non empty Subset of X,
  C be non empty Subset of X|E such that
  A2: C is connected and
  A3: C is open;
  A4: E <> {} X & C <> {} (X|E);
        C in the topology of X|E by A3,PRE_TOPC:def 5;
      then consider G being Subset of X such that
  A5: G in the topology of X and
  A6: C = G /\ [#](X|E) by PRE_TOPC:def 9;
  A7: C = G /\ E by A6,PRE_TOPC:def 10;
      reconsider G as non empty Subset of X by A6;
      consider x being Point of X|E such that
  A8: x in C by A4,PRE_TOPC:41;
        x in G by A6,A8,XBOOLE_0:def 3;
      then x in [#](X|G) by PRE_TOPC:def 10;
      then reconsider y=x as Point of X|G;
      set H1 = skl y;
  A9: H1 is_a_component_of X|G by CONNSP_1:43;
      reconsider H=H1 as Subset of X by PRE_TOPC:39;
 A10: H is_a_component_of G by A9,CONNSP_1:def 6;
      take H;
        G is open by A5,PRE_TOPC:def 5;
 hence H is open by A1,A10,Th24;
        H1 is connected by A9,CONNSP_1:def 5;
 hence H is connected by CONNSP_1:24;
      reconsider C1=C as Subset of X by PRE_TOPC:39;
 A11: C1 is connected by A2,CONNSP_1:24;
        C c= G by A6,XBOOLE_1:17;
      then C c= [#](X|G) by PRE_TOPC:def 10;
      then reconsider C2=C1 as Subset of X|G by PRE_TOPC:16;
        C2 is connected by A11,CONNSP_1:24;
      then C2 misses H or C2 c= H by A9,CONNSP_1:38;
 then A12: C2 /\ H = {}(X|G) or C2 c= H by XBOOLE_0:def 7;
A13:      x in H1 by CONNSP_1:40;
        C c= E by A7,XBOOLE_1:17;
 then A14: C c= E /\ H by A8,A12,A13,XBOOLE_0:def 3,XBOOLE_1:19;
        H /\ (G /\ E) c= C by A7,A8,A12,A13,XBOOLE_0:def 3,XBOOLE_1:28;
 then A15: (H /\ G) /\ E c= C by XBOOLE_1:16;
        H1 c= [#](X|G) by PRE_TOPC:14;
      then H c= G by PRE_TOPC:def 10;
      then E /\ H c= C by A15,XBOOLE_1:28;
 hence C = E /\ H by A14,XBOOLE_0:def 10;
end;

theorem Th26:
 X is_T4 iff for A,C being Subset of X
 st A <> {} & C <> [#] X & A c= C & A is closed & C is open
    ex G being Subset of X st G is open & A c= G & Cl G c= C
proof
 thus X is_T4 implies for A,C being Subset of X
   st A <> {} & C <> [#] X &
       A c= C & A is closed & C is open
       ex G being Subset of X st G is open & A c= G & Cl G c= C
  proof
   assume A1:for A,B being Subset of X st A <> {} & B <> {} &
             A is closed & B is closed & A misses B
             ex G,H being Subset of X st G is open & H is open &
             A c= G & B c= H & G misses H;
    let A,C be Subset of X such that
    A2: A <> {} & C <> [#] X and
    A3: A c= C and
    A4: A is closed and
    A5: C is open;
        set B=[#](X) \ C;
    A6: B <> {} by A2,PRE_TOPC:23;
          C` is closed by A5,TOPS_1:30;
    then A7: B is closed by PRE_TOPC:17;
          B c= [#](X) \ A by A3,XBOOLE_1:34;
        then B c= A` by PRE_TOPC:17;
        then A misses B by PRE_TOPC:21;
        then consider G,H being Subset of X such that
    A8: G is open & H is open and
    A9: A c= G & B c= H and
    A10: G misses H by A1,A2,A4,A6,A7;
        take G;
          for p being set holds p in Cl G implies p in C
        proof
          let p be set;
           assume A11:p in Cl G;
             then reconsider y=p as Point of X;
             A12: H` is closed by A8,TOPS_1:30;
                   G c= H` by A10,PRE_TOPC:21;
             then A13: y in H` by A11,A12,PRE_TOPC:45;
                   H` c= B` by A9,PRE_TOPC:19;
             then A14: y in B` by A13;
                   B`= [#] X \ ([#] X \ C) by PRE_TOPC:17;
          hence p in C by A14,PRE_TOPC:22;
        end;
    hence thesis by A8,A9,TARSKI:def 3;
  end;
  assume
    A15: for A,C being Subset of X st A <> {} & C <> [#] X &
        A c= C & A is closed & C is open
        ex G being Subset of X
        st G is open & A c= G & Cl G c= C;

          for A,B being Subset of X st A <> {} & B <> {} &
        A is closed & B is closed & A misses B
        ex G,H being Subset of X st G is open & H is open &
        A c= G & B c= H & G misses H
        proof
          let A,B be Subset of X such that
          A16: A <> {} & B <> {} and
          A17: A is closed & B is closed and
          A18: A misses B;
              set C = [#] X \ B;
                [#] X \ C = B by PRE_TOPC:22;
          then A19: C <> [#] X by A16,PRE_TOPC:23;
                A c= B` by A18,PRE_TOPC:21;
          then A20: A c= C by PRE_TOPC:17;
                C is open by A17,PRE_TOPC:def 6;
              then consider G being Subset of X such that
          A21: G is open & A c= G and
          A22: Cl G c= C by A15,A16,A17,A19,A20;
              take G;
              take H = [#] X \ Cl G;
                Cl(Cl G) = Cl G by TOPS_1:26;
              then Cl G is closed by PRE_TOPC:52;
         hence G is open & H is open by A21,PRE_TOPC:def 6;
                 C` c= (Cl G)` by A22,PRE_TOPC:19;
               then [#] X \([#] X \ B) c= (Cl G)` by PRE_TOPC:17;
               then [#] X \([#] X \ B) c= [#] X \ Cl G by PRE_TOPC:17;
         hence A c= G & B c= H by A21,PRE_TOPC:22;
                 G c= Cl G by PRE_TOPC:48;
               then [#] X \ Cl G c= [#] X \ G by XBOOLE_1:34;
               then H c= G` by PRE_TOPC:17;
         hence G misses H by PRE_TOPC:21;
        end;
   hence X is_T4 by COMPTS_1:def 6;
end;

theorem
   X is locally_connected & X is_T4 implies
 for A,B being Subset of X
 st A <> {} & B <> {} & A is closed & B is closed & A misses B holds
 (A is connected & B is connected implies
 ex R,S being Subset of X st R is connected & S is connected &
 R is open & S is open & A c= R & B c= S & R misses S)
proof
  assume A1: X is locally_connected & X is_T4;
   let A,B be Subset of X such that
 A2: A <> {} & B <> {} and
 A3: A is closed & B is closed and
 A4: A misses B;
     assume A5: A is connected & B is connected;
   A6: A c= B` by A4,PRE_TOPC:21;
   A7: B` is open by A3,TOPS_1:29;
         B = [#] X \ ([#] X \ B) by PRE_TOPC:22;
       then [#] X \ B <> [#] X by A2,PRE_TOPC:23;
       then B` <> [#] X by PRE_TOPC:17;
       then consider G being Subset of X such that
   A8: G is open & A c= G and
   A9: Cl G c= B` by A1,A2,A3,A6,A7,Th26;
   A10: Cl G misses B by A9,PRE_TOPC:21;
     A <> {} X by A2;
       then consider x being Point of X such that
  A11: x in A by PRE_TOPC:41;
  A12: x in G by A8,A11;
       reconsider G as non empty Subset of X by A2,A8,XBOOLE_1:3
;
         x in [#](X|G) by A12,PRE_TOPC:def 10;
       then reconsider y=x as Point of X|G;
       set H=skl y;
       reconsider H1=H as Subset of X by PRE_TOPC:39;
       take R=H1;
  A13: H is_a_component_of X|G by CONNSP_1:43;
A14:   H is connected by CONNSP_1:41;
A15:   H1 is_a_component_of G by A13,CONNSP_1:def 6;
         A c= [#](X|G) by A8,PRE_TOPC:def 10;
       then reconsider A1=A as Subset of X|G by PRE_TOPC:16;
         A1 is connected by A5,CONNSP_1:24;
       then A1 misses H or A1 c= H by A13,CONNSP_1:38;
  then A16: A1 /\ H = {} or A1 c= H by XBOOLE_0:def 7;
A17:   y in H by CONNSP_1:40;
         B <> {} X by A2;
       then consider z being Point of X such that
  A18: z in B by PRE_TOPC:41;
       A19: B c= (Cl G)` by A10,PRE_TOPC:21;
  then A20: z in (Cl G)` by A18;
       reconsider C = (Cl G)` as non empty Subset of X
       by A18,A19;
         z in [#](X|C) by A20,PRE_TOPC:def 10;
       then reconsider z1=z as Point of X|C;
       set V=skl z1;
       reconsider V1=V as Subset of X by PRE_TOPC:39;
       take S=V1;
  A21: V is_a_component_of X|C by CONNSP_1:43;
         Cl G = Cl(Cl G) by TOPS_1:26;
       then Cl G is closed by PRE_TOPC:52;
  then A22: (Cl G)` is open by TOPS_1:29;
A23:   V1 is_a_component_of (Cl G)` by A21,CONNSP_1:def 6;
A24:   V is connected by CONNSP_1:41;
         B c= [#](X|(Cl G)`) by A19,PRE_TOPC:def 10;
       then reconsider B1=B as Subset of X|(Cl G)` by PRE_TOPC:16;
         B1 is connected by A5,CONNSP_1:24;
       then B1 misses V or B1 c= V by A21,CONNSP_1:38;
  then A25: B1 /\ V = {} or B1 c= V by XBOOLE_0:def 7;
A26:   z1 in V by CONNSP_1:40;
         H c= [#](X|G) by PRE_TOPC:14;
  then A27: R c= G by PRE_TOPC:def 10;
         G c= Cl G by PRE_TOPC:48;
  then A28: R c= Cl G by A27,XBOOLE_1:1;
         V c= [#](X|(Cl G)`) by PRE_TOPC:14;
       then S c= (Cl G)` by PRE_TOPC:def 10;
then A29:     R /\ S c= Cl G /\ (Cl G)` by A28,XBOOLE_1:27;
         Cl G misses (Cl G)` by PRE_TOPC:26;
       then R /\ S c= {} X by A29,XBOOLE_0:def 7;
       then R /\ S = {} by XBOOLE_1:3;
   hence thesis by A1,A8,A11,A14,A15,A16,A17,A18,A22,A23,A24,A25,A26,Th24,
CONNSP_1:24,XBOOLE_0:def 3,def 7;
end;

theorem Th28:
  for x being Point of X, F being Subset-Family of X
  st for A being Subset of X holds A in F iff A is open closed
  & x in A holds F <> {}
proof
  let x be Point of X, F be Subset-Family of X such that
A1: for A being Subset of X holds A in F iff A is open closed
    & x in A;
      [#] X is open closed & x in [#] X by PRE_TOPC:13,TOPS_1:53;
  hence F <> {} by A1;
end;

::
::             A QUASICOMPONENT OF A POINT
::

definition let X be non empty TopSpace, x be Point of X;
  func qskl x -> Subset of X means
   :Def7:ex F being Subset-Family of X st
      (for A being Subset of X holds A in F iff
      A is open closed & x in A) & meet F = it;
   existence
   proof
     defpred P[set] means
       ex A1 being Subset of X st A1 = $1 & A1 is open closed & x in $1;
        consider F being Subset-Family of X such that
   A1:  for A being Subset of X holds A in F iff P[A]
          from SubFamExS;
        reconsider S = meet F as Subset of X;
      take S, F;
      thus for A being Subset of X holds A in F iff A is open closed & x in A
      proof
       let A be Subset of X;
       thus A in F implies A is open closed & x in A
       proof
         assume A in F;
         then ex A1 being Subset of X st A1 = A & A1 is open closed & x in A
by A1;
         hence thesis;
       end;
       thus thesis by A1;
      end;
      thus meet F = S;
   end;
   uniqueness
   proof let S,S' be Subset of X; assume
   A2:  (ex F being Subset-Family of X st
        (for A being Subset of X holds A in F
        iff A is open closed &
        x in A) & meet F = S) & ex F' being Subset-Family of X st
        (for A being Subset of X holds A in F'
        iff A is open closed & x in A) & meet F' = S';
        then consider F being Subset-Family of X such that
   A3:  (for A being Subset of X holds A in F
   iff A is open closed
        & x in A) & meet F = S;
        consider F' being Subset-Family of X such that
   A4: (for A being Subset of X holds A in F'
   iff A is open closed
        & x in A) & meet F' = S' by A2;
   A5: F <> {} & F' <> {} by A3,A4,Th28;
         now let y be set;
       A6:  now assume
A7:            y in S;
                 for B being set st B in F' holds y in B
               proof
                let B be set; assume
               A8: B in F';
                   then reconsider B1=B as Subset of X;
                     B1 is open closed & x in B1 by A4,A8;
                   then B1 in F by A3;
                hence y in B by A3,A7,SETFAM_1:def 1;
               end;
             hence y in S' by A4,A5,SETFAM_1:def 1;
           end;
             now assume
A9:            y in S';
                 for B being set st B in F holds y in B
               proof
                let B be set; assume
                A10: B in F;
                   then reconsider B1=B as Subset of X;
                      B1 is open closed & x in B1 by A3,A10;
                    then B1 in F' by A4;
                hence y in B by A4,A9,SETFAM_1:def 1;
               end;
             hence y in S by A3,A5,SETFAM_1:def 1;
           end;
         hence y in S iff y in S' by A6;
       end;
     hence S = S' by TARSKI:2;
   end;
end;

canceled;

theorem
    x in qskl x
proof
    consider F being Subset-Family of X such that
A1: (for A being Subset of X holds A in F iff A is open closed
    & x in A) and
A2: qskl x = meet F by Def7;
A3: F <> {} by A1,Th28;
      for A being set holds A in F implies x in A by A1;
  hence x in qskl x by A2,A3,SETFAM_1:def 1;
end;

theorem
   for A being Subset of X st A is open closed & x in A holds
   A c= qskl x implies A = qskl x
proof
 let A be Subset of X; assume
A1: A is open closed & x in A; assume
A2: A c= qskl x;
    consider F being Subset-Family of X such that
A3: for A being Subset of X holds
    (A in F iff A is open closed & x in A) and
A4: qskl x = meet F by Def7;
      A in F by A1,A3;
    then qskl x c= A by A4,SETFAM_1:4;
  hence thesis by A2,XBOOLE_0:def 10;
end;

theorem
   qskl x is closed
proof
    consider F being Subset-Family of X such that
A1: for A being Subset of X holds
    (A in F iff A is open closed & x in A) and
A2: qskl x = meet F by Def7;
      for A being Subset of X st A in F holds A is closed by A1;
 hence qskl x is closed by A2,PRE_TOPC:44;
end;

theorem
   skl x c= qskl x
proof
    consider F' being Subset-Family of X such that
A1: for A being Subset of X holds
    (A in F' iff A is open closed & x in A) and
A2: qskl x = meet F' by Def7;
A3:F' <> {} by A1,Th28;
      for B1 being set st B1 in F' holds skl x c= B1
    proof
     let B1 be set; assume
     A4: B1 in F';
         then reconsider B=B1 as Subset of X;
     A5: B is open closed & x in B by A1,A4;
         set S=skl x;
     A6: [#] X = B \/ B` by PRE_TOPC:18;
     A7: S = S /\ [#] X by PRE_TOPC:15
          .= (S /\ B) \/ (S /\ B`) by A6,XBOOLE_1:23;
           Cl B = B & B` is closed by A5,PRE_TOPC:52,TOPS_1:30;
         then Cl B = B & Cl B` = B` by PRE_TOPC:52;
         then Cl B misses B` & B misses Cl B` by PRE_TOPC:26;
     then A8: B,B` are_separated by CONNSP_1:def 1;
     A9: x in S by CONNSP_1:40;
           S /\ B c= B & S /\ B` c= B` by XBOOLE_1:17;
    then A10: S /\ B,S /\ B` are_separated by A8,CONNSP_1:8;
           S is connected by CONNSP_1:41;
then S /\ B = {}X or S /\ B` = {}X by A7,A10,CONNSP_1:16;
         then S misses B` by A5,A9,XBOOLE_0:def 3,def 7;
         then S c= B`` by PRE_TOPC:21;
      hence thesis;
    end;
  hence skl x c= qskl x by A2,A3,SETFAM_1:6;
end;

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