The Mizar article:

Remarks on Special Subsets of Topological Spaces

by
Zbigniew Karno

Received April 6, 1993

Copyright (c) 1993 Association of Mizar Users

MML identifier: TOPS_3
[ MML identifier index ]


environ

 vocabulary PRE_TOPC, BOOLE, SUBSET_1, TOPS_1, TOPS_3;
 notation TARSKI, XBOOLE_0, SUBSET_1, STRUCT_0, PRE_TOPC, TOPS_1, BORSUK_1,
      TSEP_1;
 constructors TOPS_1, BORSUK_1, TSEP_1, MEMBERED;
 clusters BORSUK_1, TSEP_1, PRE_TOPC, TOPS_1, MEMBERED, ZFMISC_1;
 requirements BOOLE, SUBSET;
 definitions PRE_TOPC, XBOOLE_0;
 theorems PRE_TOPC, TOPS_1, TOPS_2, TSEP_1, TDLAT_3, SUBSET_1, TARSKI,
      XBOOLE_0, XBOOLE_1;

begin
:: 1. Selected Properties of Subsets of a Topological Space.
reserve X for TopStruct, A for Subset of X;

theorem Th1:
 (A = {}X iff  A` = [#]X) &
   (A = {} iff  A` = the carrier of X)
 proof
  thus A = {}X iff  A` = [#]X
   proof thus A = {}X implies  A` = [#]X by PRE_TOPC:27;
    assume  A` = [#]X;
     then A`` = {}X by TOPS_1:8;
    hence thesis;
   end;
  hence A = {} iff  A` = the carrier of X by PRE_TOPC:12;
 end;

theorem Th2:
 (A = [#]X iff  A` = {}X) &
   (A = the carrier of X iff  A` = {})
 proof
  thus A = [#]X iff  A` = {}X
   proof thus A = [#]X implies  A` = {}X by TOPS_1:8;
    assume  A` = {}X;
     then A`` = [#]X by PRE_TOPC:27;
    hence thesis;
   end;
  hence A = the carrier of X iff  A` = {} by PRE_TOPC:12;
 end;

theorem Th3:
 for X being TopSpace, A,B being Subset of X holds
  Int A /\ Cl B c= Cl(A /\ B)
 proof
   let X be TopSpace, A,B be Subset of X;
      Int A c= A by TOPS_1:44;
   then (Int A) /\ B c= A /\ B by XBOOLE_1:26;
  then A1: Cl((Int A) /\ B) c= Cl(A /\ B) by PRE_TOPC:49;
  Int A /\ Cl B c= Cl((Int A) /\ B) by TOPS_1:40;
  hence thesis by A1,XBOOLE_1:1;
 end;

reserve X for TopSpace, A,B for Subset of X;

theorem Th4:
 Int(A \/ B) c= (Cl A) \/ Int B
 proof
      (Int  A`) /\ Cl  B` c= Cl(( A`) /\  B`) by Th3;
   then (Cl(( A`) /\  B`))` c= ((Int  A`) /\ Cl  B`)` by PRE_TOPC:19;
   then Int((( A`) /\  B`)`) c= (((Int  A`) /\ Cl  B`))` by TDLAT_3:4;
   then Int((( A`) /\  B`)`) c= (((Int  A`) /\ Cl  B`))`;
   then Int(( A``) \/ ( B``)) c= (((Int  A`) /\ Cl  B`))` & ( A``) = A
         by SUBSET_1:30;
   then Int(A \/ B) c= ((Int  A`) /\ Cl  B`)`;
   then Int(A \/ B) c= ((Int  A`) /\ Cl  B`)`;
   then Int(A \/ B) c= (Int  A`)` \/ (Cl  B`)` by SUBSET_1:30;
   then Int(A \/ B) c= Cl A \/ (Cl  B`)` by TDLAT_3:2;
  hence thesis by TOPS_1:def 1;
 end;

theorem Th5:
 for A being Subset of X st A is closed holds Int(A \/ B) c= A \/ Int B
 proof
  let A be Subset of X;
  assume A is closed;
   then Cl A = A & Int(A \/ B) c= (Cl A) \/ Int B by Th4,PRE_TOPC:52;
  hence thesis;
 end;

theorem Th6:
 for A being Subset of X st A is closed holds Int(A \/ B) = Int(A \/ Int B)
 proof
  let A be Subset of X;
  assume A is closed;
   then Int(A \/ B) c= A \/ Int B by Th5;
   then Int Int(A \/ B) c= Int(A \/ Int B) by TOPS_1:48;
  then A1: Int(A \/ B) c= Int(A \/ Int B) by TOPS_1:45;
      Int B c= B by TOPS_1:44;
   then A \/ Int B c= A \/ B by XBOOLE_1:9;
   then Int(A \/ Int B) c= Int(A \/ B) by TOPS_1:48;
  hence thesis by A1,XBOOLE_0:def 10;
 end;

theorem Th7:
 A misses Int Cl A implies Int Cl A = {}
 proof
  reconsider A' = A as Subset of X;
  assume A /\ Int Cl A = {};
   then A' misses Int Cl A' by XBOOLE_0:def 7;
   then (Cl A) misses (Int Cl A) by TSEP_1:40;
   then (Cl A) /\ (Int Cl A) = {} & Int Cl A c= Cl A
     by TOPS_1:44,XBOOLE_0:def 7;
  hence thesis by XBOOLE_1:28;
 end;

theorem
   A \/ Cl Int A = the carrier of X implies Cl Int A = the carrier of X
 proof
  reconsider A' = A as Subset of X;
  assume A \/ Cl Int A = the carrier of X;
then A' \/ Cl Int A' = the carrier of X;
   then (Int A) \/ (Cl Int A) = the carrier of X &
              Int A c= Cl Int A by PRE_TOPC:48,TDLAT_3:6;
  hence thesis by XBOOLE_1:12;
 end;

begin
:: 2. Special Subsets of a Topological Space.

definition let X be TopStruct, A be Subset of X;
 redefine attr A is boundary means
:Def1: Int A = {};
 compatibility by TOPS_1:84;
end;

theorem Th9:
 {}X is boundary
 proof
     Int({}X) = {} by TOPS_1:47;
  hence thesis by TOPS_1:84;
 end;

reserve X for non empty TopSpace, A for Subset of X;

theorem Th10:
 A is boundary implies A <> the carrier of X
 proof
  assume A1: Int A = {};
  assume A = the carrier of X;
   then A = [#]X by PRE_TOPC:12;
  hence contradiction by A1,TOPS_1:43;
 end;

reserve X for TopSpace, A,B for Subset of X;

theorem Th11:
 B is boundary & A c= B implies A is boundary
 proof
  assume B is boundary; then A1: Int B = {} by TOPS_1:84;
  assume A c= B; then Int A c= Int B by TOPS_1:48;
                 then Int A = {} by A1,XBOOLE_1:3;
  hence thesis by TOPS_1:84;
 end;

theorem
   A is boundary iff
  for C being Subset of X st  A` c= C & C is closed
   holds C = the carrier of X
 proof
  thus A is boundary implies
         for C being Subset of X st  A` c= C & C is closed
           holds C = the carrier of X
   proof assume A1: A is boundary;
     let C be Subset of X;
    assume  A` c= C;
     then  C` c=  A`` by PRE_TOPC:19;
     then A2:  C` c= A;
    assume C is closed;
     then  C` is open by TOPS_1:29;
     then  C` = {}X by A1,A2,TOPS_1:86;
    hence thesis by Th2;
   end;
  assume
A3: for C being Subset of X st  A` c= C & C is closed holds
               C = the carrier of X;
     now let G be Subset of X;
    assume G c= A & G is open;
     then  A` c=  G` &  G` is closed by PRE_TOPC:19,TOPS_1:30;
     then  G` = the carrier of X by A3;
    hence G = {} by Th1;
   end;
  hence thesis by TOPS_1:86;
 end;

theorem
    A is boundary iff
   for G being Subset of X st G <> {} & G is open holds ( A`) meets G
 proof
  thus A is boundary implies
        for G being Subset of X st G <> {} & G is open
         holds ( A`) meets G
   proof assume A1: A is boundary;
     let G be Subset of X;
    assume A2: G <> {};
    assume A3: G is open;
    assume ( A`) misses G;
     then G c= A by TOPS_1:20;
    hence contradiction by A1,A2,A3,TOPS_1:86;
   end;
  assume A4: for G being Subset of X st G <> {} & G is open
   holds ( A`) meets G;
  assume A5: Int A <> {};
      Int A is open & Int A c= A by TOPS_1:44;
   then ( A`) meets Int A & (Int A) misses  A` &
         ( A`) /\ Int A = (Int A) /\  A` by A4,A5,TOPS_1:20;
  hence contradiction;
 end;

theorem
    A is boundary iff
   for F being Subset of X holds F is closed
    implies Int F = Int(F \/ A)
 proof
  thus A is boundary implies for F being Subset of X holds
                              F is closed implies Int F = Int(F \/ A)
   proof assume A1: Int A = {};
     let F be Subset of X;
    assume F is closed;
     then Int(F \/ A) = Int(F \/ Int A) & F \/ {} = F by Th6;
    hence thesis by A1;
   end;
  assume for F being Subset of X holds
              F is closed implies Int F = Int(F \/ A);
   then Int {}X = Int({}X \/ A) & Int {}X = {}X by TOPS_1:47;
  hence thesis by Def1;
 end;

theorem
   A is boundary or B is boundary implies A /\ B is boundary
 proof
  assume A1: A is boundary or B is boundary;
     A /\ B c= A & A /\ B c= B by XBOOLE_1:17;
  hence thesis by A1,Th11;
 end;

definition let X be TopStruct, A be Subset of X;
 redefine attr A is dense means
:Def2: Cl A = the carrier of X;
 compatibility
  proof
   thus A is dense implies Cl A = the carrier of X
    proof assume A is dense;
      then Cl A = [#]X by TOPS_1:def 3;
     hence thesis by PRE_TOPC:12;
    end;
   assume Cl A = the carrier of X;
    then Cl A = [#]X by PRE_TOPC:12;
   hence thesis by TOPS_1:def 3;
  end;
end;

theorem
   [#]X is dense
 proof
     Cl([#]X) = [#]X by TOPS_1:27;
  hence thesis by TOPS_1:def 3;
 end;

reserve X for non empty TopSpace, A, B for Subset of X;

theorem Th17:
 A is dense implies A <> {}
 proof
  assume A is dense;
   then A1: Cl A = [#]X by TOPS_1:def 3;
  assume A = {};
   then Cl A = {}X by PRE_TOPC:52;
  hence contradiction by A1;
 end;

theorem Th18:
 A is dense iff  A` is boundary
 proof
  thus A is dense implies  A` is boundary
   proof
    assume A is dense;
     then ( A``) is dense;
    hence thesis by TOPS_1:def 4;
   end;
  assume  A` is boundary;
   then ( A``) is dense by TOPS_1:def 4;
  hence thesis;
 end;

theorem
   A is dense iff
  for C being Subset of X st A c= C & C is closed
   holds C = the carrier of X
 proof
  thus A is dense implies for C being Subset of X st
                           A c= C & C is closed holds C = the carrier of X
   proof assume A1: Cl A = the carrier of X;
     let C be Subset of X;
    assume A c= C & C is closed;
     then Cl A c= C & C c= the carrier of X by TOPS_1:31;
    hence thesis by A1,XBOOLE_0:def 10;
   end;
  assume A2: for C being Subset of X st
              A c= C & C is closed holds C = the carrier of X;
      A c= Cl A & Cl A is closed by PRE_TOPC:48;
   then Cl A = the carrier of X by A2;
  hence thesis by Def2;
 end;

theorem
   A is dense iff
  for G being Subset of X holds G is open
   implies Cl G = Cl(G /\ A)
 proof
  thus A is dense implies for G being Subset of X holds
                            G is open implies Cl G = Cl(G /\ A)
   proof assume A1: A is dense;
     let G be Subset of X;
    assume G is open;
     then Cl(G /\ Cl A) = Cl(G /\ A) & G /\ [#]X = G by PRE_TOPC:15,TOPS_1:41;
    hence thesis by A1,TOPS_1:def 3;
   end;
  assume
  for G being Subset of X holds G is open
      implies Cl G = Cl(G /\ A);
   then Cl [#]X = Cl([#]X /\ A);
   then [#]X = Cl([#]X /\ A) & [#]X /\ A = A by PRE_TOPC:15,TOPS_1:27;
   then Cl A = the carrier of X by PRE_TOPC:12;
  hence thesis by Def2;
 end;

theorem
   A is dense or B is dense implies A \/ B is dense
 proof
  assume A1: A is dense or B is dense;
      A c= A \/ B & B c= A \/ B by XBOOLE_1:7;
  hence thesis by A1,TOPS_1:79;
 end;

definition let X be TopStruct, A be Subset of X;
 redefine attr A is nowhere_dense means
:Def3: Int(Cl A) = {};
 compatibility
  proof
   thus A is nowhere_dense implies Int Cl A = {}
    proof assume A is nowhere_dense;
     then Cl A is boundary by TOPS_1:def 5;
     hence thesis by Def1;
    end;
   assume Int Cl A = {};
     then Cl A is boundary by Def1;
   hence thesis by TOPS_1:def 5;
  end;
end;

theorem Th22:
 {}X is nowhere_dense
 proof
     Cl({}X) = {}X & {}X is boundary by Th9,PRE_TOPC:52;
  hence thesis by TOPS_1:def 5;
 end;

theorem
   A is nowhere_dense implies A <> the carrier of X
 proof
  assume A is nowhere_dense;
   then A is boundary by TOPS_1:92;
  hence thesis by Th10;
 end;

theorem
   A is nowhere_dense implies Cl A is nowhere_dense
 proof
  assume A is nowhere_dense;
   then Cl A is boundary & Cl A is closed by TOPS_1:def 5;
  hence thesis by TOPS_1:93;
 end;

theorem
   A is nowhere_dense implies A is not dense
 proof
  assume A1: Int Cl A = {};
  assume A is dense;
   then Cl A = [#]X by TOPS_1:def 3;
  hence contradiction by A1,TOPS_1:43;
 end;

theorem Th26:
 B is nowhere_dense & A c= B implies A is nowhere_dense
 proof
  assume B is nowhere_dense;
   then A1: Cl B is boundary by TOPS_1:def 5;
  assume A c= B;
   then Cl A c= Cl B by PRE_TOPC:49;
   then Cl A is boundary by A1,Th11;
  hence thesis by TOPS_1:def 5;
 end;

theorem Th27:
 A is nowhere_dense iff ex C being Subset of X st
                          A c= C & C is closed & C is boundary
 proof
  thus A is nowhere_dense implies
     ex C being Subset of X
      st A c= C & C is closed & C is boundary
   proof
    assume A1: A is nowhere_dense;
    take Cl A;
    thus thesis by A1,PRE_TOPC:48,TOPS_1:def 5;
   end;
  given C being Subset of X such that
        A2: A c= C & C is closed and A3: C is boundary;
      Cl A c= C by A2,TOPS_1:31;
   then Cl A is boundary by A3,Th11;
  hence thesis by TOPS_1:def 5;
 end;

theorem Th28:
 A is nowhere_dense iff
  for G being Subset of X st G <> {} & G is open
   ex H being Subset of X
    st H c= G & H <> {} & H is open & A misses H
 proof
  thus A is nowhere_dense implies
        for G being Subset of X st G <> {} & G is open
         ex H being Subset of X
          st H c= G & H <> {} & H is open & A misses H
   proof assume A is nowhere_dense;
        then A1: Cl A is boundary by TOPS_1:def 5;
     let G be Subset of X;
    assume G <> {} & G is open;
     then consider H being Subset of X such that
            A2: H c= G & H <> {} & H is open and
             A3: Cl A misses H by A1,TOPS_1:87;
    take H;
        A c= Cl A by PRE_TOPC:48;
    hence thesis by A2,A3,XBOOLE_1:63;
   end;
  assume A4: for G being Subset of X st G <> {} & G is open
            ex H being Subset of X
              st H c= G & H <> {} & H is open & A misses H;
     for G being Subset of X st G <> {} & G is open
     ex H being Subset of X
      st H c= G & H <> {} & H is open & Cl A misses H
    proof let G be Subset of X;
     assume G <> {} & G is open;
      then consider H being Subset of X such that
             A5: H c= G & H <> {} & H is open and A6: A misses H by A4;
     take H;
     thus thesis by A5,A6,TSEP_1:40;
    end;
   then Cl A is boundary by TOPS_1:87;
  hence thesis by TOPS_1:def 5;
 end;

theorem
   A is nowhere_dense or B is nowhere_dense implies A /\ B is nowhere_dense
 proof
  assume A1: A is nowhere_dense or B is nowhere_dense;
     A /\ B c= A & A /\ B c= B by XBOOLE_1:17;
  hence thesis by A1,Th26;
 end;

theorem Th30:
 A is nowhere_dense & B is boundary implies A \/ B is boundary
 proof
  assume A is nowhere_dense;
   then A1: Cl A is boundary & Cl A is closed by TOPS_1:def 5;
      A c= Cl A by PRE_TOPC:48;
   then A2: B \/ A c= B \/ Cl A by XBOOLE_1:9;
  assume B is boundary;
   then B \/ Cl A is boundary by A1,TOPS_1:85;
  hence thesis by A2,Th11;
 end;

definition let X be TopStruct, A be Subset of X;
 attr A is everywhere_dense means
:Def4: Cl(Int A) = [#]X;
end;

definition let X be TopStruct, A be Subset of X;
 redefine attr A is everywhere_dense means
    Cl(Int A) = the carrier of X;
 compatibility
  proof
   thus A is everywhere_dense implies Cl Int A = the carrier of X
    proof assume A is everywhere_dense;
     then Cl Int A = [#]X by Def4;
     hence thesis by PRE_TOPC:12;
    end;
   assume Cl Int A = the carrier of X;
    then Cl Int A = [#]X by PRE_TOPC:12;
   hence thesis by Def4;
  end;
end;

theorem
   [#]X is everywhere_dense
 proof
      Int [#]X = [#]X by TOPS_1:43;
   then Cl Int [#]X = [#]X by TOPS_1:27;
  hence thesis by Def4;
 end;

theorem Th32:
 A is everywhere_dense implies Int A is everywhere_dense
 proof
  assume A is everywhere_dense;
   then Cl Int A = [#]X by Def4;
   then Cl Int Int A = [#]X by TOPS_1:45;
  hence Int A is everywhere_dense by Def4;
 end;

theorem Th33:
 A is everywhere_dense implies A is dense
 proof
  assume A is everywhere_dense; then A1: Cl Int A = [#] X by Def4;
      Int A c= A by TOPS_1:44;
   then [#]X c= Cl A & Cl A c= [#]X by A1,PRE_TOPC:14,49;
   then Cl A = [#]X by XBOOLE_0:def 10;
  hence thesis by TOPS_1:def 3;
 end;

theorem
   A is everywhere_dense implies A <> {}
 proof
  assume A is everywhere_dense;
   then A is dense by Th33;
  hence thesis by Th17;
 end;

theorem Th35:
 A is everywhere_dense iff Int A is dense
 proof
  thus A is everywhere_dense implies Int A is dense
   proof
    assume A is everywhere_dense;
     then Int A is everywhere_dense by Th32;
    hence Int A is dense by Th33;
   end;
  assume Int A is dense;
   then Cl Int A = [#]X by TOPS_1:def 3;
  hence A is everywhere_dense by Def4;
 end;

theorem Th36:
 A is open & A is dense implies A is everywhere_dense
 proof
  assume A is open;
   then A1: A = Int A by TOPS_1:55;
  assume A is dense;
  hence thesis by A1,Th35;
 end;

theorem
   A is everywhere_dense implies A is not boundary
 proof
  assume A is everywhere_dense;
   then A1: Cl Int A = [#]X by Def4;
  assume A is boundary;
   then Int A = {}X by TOPS_1:84;
  hence contradiction by A1,PRE_TOPC:52;
 end;

theorem Th38:
 A is everywhere_dense & A c= B implies B is everywhere_dense
 proof
  assume A is everywhere_dense; then A1: Cl Int A = [#]X by Def4;
  assume A c= B; then Int A c= Int B by TOPS_1:48;
    then Cl Int A c= Cl Int B & Cl Int B c= [#]X by PRE_TOPC:14,49;
    then [#]X = Cl Int B by A1,XBOOLE_0:def 10;
  hence thesis by Def4;
 end;

theorem Th39:
 A is everywhere_dense iff  A` is nowhere_dense
 proof
  thus A is everywhere_dense implies  A` is nowhere_dense
   proof assume A is everywhere_dense;
    then Cl Int A = [#]X by Def4;
    then (Cl Int A)` = {}X by Th2;
    then Int (Int A)` = {}X by TDLAT_3:4;
    then Int Cl  A` = {} by TDLAT_3:3;
    then Cl  A` is boundary by TOPS_1:84;
    hence thesis by TOPS_1:def 5;
   end;
  assume  A` is nowhere_dense;
   then Cl  A` is boundary by TOPS_1:def 5;
   then Int Cl  A` = {}X by TOPS_1:84;
   then Int (Int A)` = {}X by TDLAT_3:3;
   then (Cl Int A)` = {}X by TDLAT_3:4;
   then Cl Int A = [#]X by Th2;
  hence thesis by Def4;
 end;

theorem Th40:
 A is nowhere_dense iff  A` is everywhere_dense
 proof
  thus A is nowhere_dense implies  A` is everywhere_dense
   proof assume A is nowhere_dense;
    then Cl A is boundary by TOPS_1:def 5;
    then Int Cl A = {}X by TOPS_1:84;
    then Int (Int  A`)` = {}X by TDLAT_3:2;
    then (Cl Int  A`)` = {}X by TDLAT_3:4;
    then Cl Int  A` = [#]X by Th2;
    hence thesis by Def4;
   end;
  assume  A` is everywhere_dense;
   then Cl Int  A` = [#]X by Def4;
   then (Cl Int  A`)` = {}X by Th2;
   then Int (Int  A`)` = {}X by TDLAT_3:4;
   then Int Cl A = {} by TDLAT_3:2;
   then Cl A is boundary by TOPS_1:84;
  hence A is nowhere_dense by TOPS_1:def 5;
 end;

theorem Th41:
 A is everywhere_dense iff ex C being Subset of X st
                             C c= A & C is open & C is dense
 proof
  thus A is everywhere_dense implies ex C being Subset of X st
                                      C c= A & C is open & C is dense
   proof
    assume A1: A is everywhere_dense;
    take Int A;
    thus thesis by A1,Th35,TOPS_1:44;
   end;
  given C being Subset of X such that
        A2: C c= A & C is open and A3: C is dense;
      C c= Int A by A2,TOPS_1:56;
   then Int A is dense by A3,TOPS_1:79;
  hence thesis by Th35;
 end;

theorem
   A is everywhere_dense iff
  for F being Subset of X st F <> the carrier of X & F is closed
   ex H being Subset of X st
    F c= H & H <> the carrier of X & H is closed & A \/ H = the carrier of X
 proof
  thus A is everywhere_dense implies
        for F being Subset of X
           st F <> the carrier of X & F is closed
          ex H being Subset of X
          st F c= H & H <> the carrier of X &
                                    H is closed & A \/ H = the carrier of X
   proof assume A is everywhere_dense;
          then A1:  A` is nowhere_dense by Th39;
     let F be Subset of X;
    assume F <> the carrier of X;
      then F <> [#]X by PRE_TOPC:12;
      then [#]X \ F <> {} by PRE_TOPC:23;
     then A2:  F` <> {} by PRE_TOPC:17;
    assume F is closed;
      then  F` is open by TOPS_1:29;
     then consider G being Subset of X such that
           A3: G c=  F` and A4: G <> {} and A5: G is open and
           A6: ( A`) misses G by A1,A2,Th28;
    take H =  G`;
       F`` c= H by A3,PRE_TOPC:19;
    hence F c= H;
         H` <> {} by A4;
     then [#]X \ H <> {} by PRE_TOPC:17;
     then H <> [#]X by PRE_TOPC:23;
    hence H <> the carrier of X by PRE_TOPC:12;
    thus H is closed by A5,TOPS_1:30;
        ( A`) misses  H` by A6;
      then ( A`) /\  H` = {} by XBOOLE_0:def 7;
     then (A \/ H)` = {}X by SUBSET_1:29;
     then (A \/ H)` = {}X;
    hence A \/ H = the carrier of X by Th2;
   end;
  assume A7: for F being Subset of X
      st F <> the carrier of X & F is closed
               ex H being Subset of X
               st F c= H & H <> the carrier of X &
                               H is closed & A \/ H = the carrier of X;
      for G being Subset of X st G <> {} & G is open
     ex H being Subset of X
      st H c= G & H <> {} & H is open & ( A`) misses H
     proof let G be Subset of X;
      assume G <> {};
        then G`` <> {};
        then [#]X \  G` <> {} by PRE_TOPC:17;
        then  G` <> [#]X by PRE_TOPC:23;
       then A8:  G` <> the carrier of X by PRE_TOPC:12;
      assume G is open;
       then  G` is closed by TOPS_1:30;
      then consider F being Subset of X such that
             A9:  G` c= F and A10: F <> the carrier of X and
             A11: F is closed and A12: A \/ F = the carrier of X by A7,A8;
      take H =  F`;
         H c= G`` by A9,PRE_TOPC:19;
      hence H c= G;
           F <> [#]X by A10,PRE_TOPC:12;
        then [#]X \ F <> {} by PRE_TOPC:23;
      hence H <> {} by PRE_TOPC:17;
      thus H is open by A11,TOPS_1:29;
          (A \/ F)` = {}X & {}X = {} by A12,Th2;
       then (A \/ F)` = {};
      hence ( A`) /\ H = {} by SUBSET_1:29;
     end;
   then  A` is nowhere_dense by Th28;
  hence thesis by Th39;
 end;

theorem
   A is everywhere_dense or B is everywhere_dense implies
  A \/ B is everywhere_dense
 proof
  assume A1: A is everywhere_dense or B is everywhere_dense;
     A c= A \/ B & B c= A \/ B by XBOOLE_1:7;
  hence thesis by A1,Th38;
 end;

theorem Th44:
 A is everywhere_dense & B is everywhere_dense implies
  A /\ B is everywhere_dense
 proof
  assume A is everywhere_dense & B is everywhere_dense;
   then A1:   A` is nowhere_dense &  B` is nowhere_dense by Th39;
A2:   A` \/  B` = (A /\ B)` by SUBSET_1:30;
      A` \/  B` is nowhere_dense by A1,TOPS_1:90;
  hence thesis by A2,Th39;
 end;

theorem Th45:
 A is everywhere_dense & B is dense implies A /\ B is dense
 proof
  assume A is everywhere_dense;
   then A1:  A` is nowhere_dense by Th39;
  assume B is dense;
then A2:  B` is boundary by Th18;
A3:  A` \/  B` = (A /\ B)` by SUBSET_1:30;
      A` \/  B` is boundary by A1,A2,Th30;
  hence thesis by A3,Th18;
 end;

theorem
   A is dense & B is nowhere_dense implies A \ B is dense
 proof
  assume A1: A is dense;
  assume B is nowhere_dense;
   then A2:  B` is everywhere_dense by Th40;
      A \ B = B` /\ A by SUBSET_1:32;
    then A \ B = ( B`) /\ A;
  hence thesis by A1,A2,Th45;
 end;

theorem
   A is everywhere_dense & B is boundary implies A \ B is dense
 proof
  assume A1: A is everywhere_dense;
  assume B is boundary;
   then A2:  B` is dense by TOPS_1:def 4;
      A \ B = A /\ B` by SUBSET_1:32;
    then A \ B = A /\  B`;
  hence thesis by A1,A2,Th45;
 end;

theorem
   A is everywhere_dense & B is nowhere_dense implies A \ B is everywhere_dense
 proof
  assume A1: A is everywhere_dense;
  assume B is nowhere_dense;
   then A2:  B` is everywhere_dense by Th40;
      A \ B = A /\ B` by SUBSET_1:32;
    then A \ B = A /\  B`;
  hence thesis by A1,A2,Th44;
 end;

reserve D for Subset of X;

theorem
   D is everywhere_dense implies
   ex C,B being Subset of X st
     C is open & C is dense & B is nowhere_dense & C \/ B = D & C misses B
 proof
  assume D is everywhere_dense;
   then consider C being Subset of X such that
         A1: C c= D and A2: C is open and A3: C is dense by Th41;
  take C;
  take B = D \ C;
  thus C is open & C is dense by A2,A3;
        D c= [#]X by PRE_TOPC:14;
     then B c= [#]X \ C by XBOOLE_1:33;
    then A4: B c=  C` by PRE_TOPC:17;
       C is everywhere_dense by A2,A3,Th36;
    then  C` is nowhere_dense by Th39;
  hence B is nowhere_dense by A4,Th26;
  thus C \/ B = D & C misses B by A1,XBOOLE_1:45,79;
 end;

theorem
   D is everywhere_dense implies
  ex C,B being Subset of X
    st C is open & C is dense & B is closed &
      B is boundary & C \/ (D /\ B) = D &
      C misses B & C \/ B = the carrier of X
 proof
  assume D is everywhere_dense;
   then consider C being Subset of X such that
         A1: C c= D and A2: C is open and A3: C is dense by Th41;
  take C;
  take B =  C`;
  thus C is open & C is dense & B is closed & B is boundary
                                      by A2,A3,Th18,TOPS_1:30;
  thus C \/ (D /\ B) = (C \/ D) /\ (C \/  C`) by XBOOLE_1:24
                  .= (C \/ D) /\ [#]X by PRE_TOPC:18
                  .= C \/ D by PRE_TOPC:15
                  .= D by A1,XBOOLE_1:12;
     C misses B by PRE_TOPC:26;
  hence C /\ B = {} by XBOOLE_0:def 7;
     C \/ B = [#]X by PRE_TOPC:18;
  hence C \/ B = the carrier of X by PRE_TOPC:12;
 end;

theorem
   D is nowhere_dense implies
  ex C,B being Subset of X st C is closed & C is boundary &
      B is everywhere_dense & C /\ B = D & C \/ B = the carrier of X
 proof
  assume D is nowhere_dense;
   then consider C being Subset of X such that
         A1: D c= C and A2: C is closed and A3: C is boundary by Th27;
  take C;
  take B = D \/  C`;
  thus C is closed & C is boundary by A2,A3;
    A4:  C` c= B by XBOOLE_1:7;
        C is nowhere_dense by A2,A3,TOPS_1:93;
    then  C` is everywhere_dense by Th40;
  hence B is everywhere_dense by A4,Th38;
A5:C misses  C` by PRE_TOPC:26;
  thus C /\ B = (C /\ D) \/ (C /\  C`) by XBOOLE_1:23
            .= (C /\ D) \/ {}X by A5,XBOOLE_0:def 7
            .= D by A1,XBOOLE_1:28;
  thus C \/ B = D \/ (C \/  C`) by XBOOLE_1:4
            .= D \/ [#]X by PRE_TOPC:18
            .= [#]X by TOPS_1:2
            .= the carrier of X by PRE_TOPC:12;
 end;

theorem
   D is nowhere_dense implies
  ex C,B being Subset of X st C is closed & C is boundary & B is open &
      B is dense & C /\ (D \/ B) = D & C misses B & C \/ B = the carrier of X
 proof
  assume D is nowhere_dense;
   then consider C being Subset of X such that
         A1: D c= C and A2: C is closed and A3: C is boundary by Th27;
  take C;
  take B =  C`;
  thus C is closed & C is boundary & B is open & B is dense
                                     by A2,A3,TOPS_1:29,def 4;
A4:C misses  C` by PRE_TOPC:26;
  thus C /\ (D \/ B) = (C /\ D) \/ (C /\  C`) by XBOOLE_1:23
                  .= (C /\ D) \/ {}X by A4,XBOOLE_0:def 7
                  .= D by A1,XBOOLE_1:28;
     C misses B by PRE_TOPC:26;
  hence C /\ B = {} by XBOOLE_0:def 7;
     C \/ B = [#]X by PRE_TOPC:18;
  hence C \/ B = the carrier of X by PRE_TOPC:12;
 end;


begin
:: 3. Properties of Subsets in Subspaces.
reserve Y0 for SubSpace of X;

theorem Th53:
 for A being Subset of X, B being Subset of Y0
  st B c= A holds Cl B c= Cl A
 proof let A be Subset of X, B be Subset of Y0;
  assume A1: B c= A;
   then reconsider D = B as Subset of X by XBOOLE_1:1;
      Cl B = (Cl D) /\ [#]Y0 by PRE_TOPC:47;
   then Cl B c= Cl D & Cl D c= Cl A by A1,PRE_TOPC:49,XBOOLE_1:17;
  hence thesis by XBOOLE_1:1;
 end;

theorem Th54:
 for C, A being Subset of X,
     B being Subset of Y0 st
  C is closed & C c= the carrier of Y0 & A c= C & A = B holds Cl A = Cl B
 proof let C, A be Subset of X,
   B be Subset of Y0;
    assume A1: C is closed;
   assume A2: C c= the carrier of Y0;
   assume A3: A c= C;
   assume A4: A = B;
       Cl A c= Cl C by A3,PRE_TOPC:49;
    then Cl A c= C by A1,PRE_TOPC:52;
    then Cl A c= the carrier of Y0 by A2,XBOOLE_1:1;
    then Cl A c= [#]Y0 by PRE_TOPC:12;
    then Cl A = (Cl A) /\ [#]Y0 by XBOOLE_1:28;
  hence thesis by A4,PRE_TOPC:47;
 end;

theorem
   for Y0 being closed non empty SubSpace of X
  for A being Subset of X, B being Subset of Y0
   st A = B holds Cl A = Cl B
 proof let Y0 be closed non empty SubSpace of X;
    the carrier of Y0 is Subset of X by TSEP_1:1;
  then reconsider C = the carrier of Y0 as Subset of X;
   let A be Subset of X, B be Subset of Y0;
  assume A1: A = B;
    then A c= C & C is closed by TSEP_1:11;
  hence thesis by A1,Th54;
 end;

theorem Th56:
 for A being Subset of X, B being Subset of Y0
  st A c= B holds Int A c= Int B
 proof let A be Subset of X, B be Subset of Y0;
  assume A1: A c= B;
      Int A c= A by TOPS_1:44;
   then A2: Int A c= B by A1,XBOOLE_1:1;
   then Int A is Subset of Y0 by XBOOLE_1:1;
   then reconsider C = Int A as Subset of Y0;
   C is open by TOPS_2:33;
  hence thesis by A2,TOPS_1:56;
 end;

theorem Th57:
 for Y0 being non empty SubSpace of X,
     C, A being Subset of X,
     B being Subset of Y0 st
  C is open & C c= the carrier of Y0 & A c= C & A = B holds Int A = Int B
 proof let Y0 be non empty SubSpace of X,
   C, A be Subset of X,
  B be Subset of Y0;
    assume A1: C is open;
   assume A2: C c= the carrier of Y0;
   assume A3: A c= C;
   assume A4: A = B;
    then A5: B c= C & Int B c= B by A3,TOPS_1:44;
    then A6: Int B c= C & Int B is open by XBOOLE_1:1;
    then Int B is Subset of X by XBOOLE_1:1;
    then reconsider D = Int B as Subset of X;
          D is open by A1,A2,A6,TSEP_1:9;
    then D c= Int A & Int A c= Int B by A4,A5,Th56,TOPS_1:56;
  hence thesis by XBOOLE_0:def 10;
 end;

theorem
   for Y0 being open non empty SubSpace of X
  for A being Subset of X, B being Subset of Y0
   st A = B holds Int A = Int B
 proof let Y0 be open non empty SubSpace of X;
    the carrier of Y0 is Subset of X by TSEP_1:1;
  then reconsider C = the carrier of Y0 as Subset of X;
   let A be Subset of X, B be Subset of Y0;
  assume A1: A = B;
   then A c= C & C is open by TSEP_1:16;
  hence thesis by A1,Th57;
 end;

reserve X0 for SubSpace of X;

theorem Th59:
 for A being Subset of X, B being Subset of X0 st A c= B holds
  A is dense implies B is dense
 proof let A be Subset of X, B be Subset of X0;
  assume A1: A c= B;
   then A is Subset of X0 by XBOOLE_1:1;
   then reconsider C = A as Subset of X0;
  assume A is dense;
   then Cl A = [#]X & [#]X0 c= [#]X by PRE_TOPC:def 9,TOPS_1:def 3;
   then [#]X0 = (Cl A) /\ [#]X0 by XBOOLE_1:28;
   then Cl C = [#]X0 by PRE_TOPC:47;
   then C is dense by TOPS_1:def 3;
  hence thesis by A1,TOPS_1:79;
 end;

theorem Th60:
 for C, A being Subset of X, B being Subset of X0 st
    C c= the carrier of X0 & A c= C & A = B holds
  C is dense & B is dense iff A is dense
 proof let C, A be Subset of X,
  B be Subset of X0;
   assume A1: C c= the carrier of X0;
   assume A2: A c= C;
   assume A3: A = B;
  reconsider P = the carrier of X0
   as Subset of X by TSEP_1:1;
  thus C is dense & B is dense implies A is dense
   proof
    assume C is dense;
     then A4: Cl C = [#]X by TOPS_1:def 3;
    assume B is dense;
     then Cl B = [#]X0 by TOPS_1:def 3;
     then Cl B = P by PRE_TOPC:12;
     then P = (Cl A) /\ [#]X0 by A3,PRE_TOPC:47;
     then P c= Cl A by XBOOLE_1:17;
     then Cl P c= Cl Cl A by PRE_TOPC:49;
     then A5: Cl P c= Cl A by TOPS_1:26;
        [#]X c= Cl P by A1,A4,PRE_TOPC:49;
      then [#]X c= Cl A & Cl A c= [#]X by A5,PRE_TOPC:14,XBOOLE_1:1;
     then Cl A = [#]X by XBOOLE_0:def 10;
    hence thesis by TOPS_1:def 3;
   end;
  thus A is dense implies C is dense & B is dense by A2,A3,Th59,TOPS_1:79;
 end;

reserve X0 for non empty SubSpace of X;

theorem Th61:
 for A being Subset of X, B being Subset of X0 st A c= B holds
  A is everywhere_dense implies B is everywhere_dense
 proof let A be Subset of X, B be Subset of X0;
  assume A1: A c= B;
   then reconsider C = A as Subset of X0 by XBOOLE_1:1;
    A2: Int A c= Int C by Th56;
  assume A is everywhere_dense;
   then Int A is dense by Th35;
   then Int C is dense & Int C c= Int B by A1,A2,Th59,TOPS_1:48;
   then Int B is dense by TOPS_1:79;
  hence thesis by Th35;
 end;

theorem Th62:
 for C, A being Subset of X,
     B being Subset of X0 st
   C is open & C c= the carrier of X0 & A c= C & A = B holds
  C is dense & B is everywhere_dense iff A is everywhere_dense
 proof let C, A be Subset of X,
   B be Subset of X0;
   assume A1: C is open;
   assume C c= the carrier of X0;
    then reconsider E = C as Subset of X0;
      A2: E is open by A1,TOPS_2:33;
   assume A3: A c= C;
   assume A4: A = B;
     A5: Int B c= B by TOPS_1:44;
    then Int B is Subset of X by A4,XBOOLE_1:1;
    then reconsider D = Int B as Subset of X;
       Int B c= Int E by A3,A4,TOPS_1:48;
     then A6: Int B c= E & Int B is open by A2,TOPS_1:55;
     then D is open by A1,TSEP_1:9;
      then A7: D c= Int A by A4,A5,TOPS_1:56;
  thus C is dense & B is everywhere_dense
        implies A is everywhere_dense
   proof
    assume A8: C is dense;
    assume B is everywhere_dense;
     then Int B is dense by Th35;
     then D is dense by A6,A8,Th60;
     then Int A is dense by A7,TOPS_1:79;
    hence thesis by Th35;
   end;
  thus A is everywhere_dense implies
        C is dense & B is everywhere_dense
   proof
    assume A9: A is everywhere_dense;
    then C is everywhere_dense by A3,Th38;
    hence C is dense by Th33;
    thus thesis by A4,A9,Th61;
   end;
 end;

theorem
   for X0 being open non empty SubSpace of X
  for A,C being Subset of X,
      B being Subset of X0 st
    C = the carrier of X0 & A = B holds
   C is dense & B is everywhere_dense iff A is everywhere_dense
 proof let X0 be open non empty SubSpace of X;
        let A,C be Subset of X,
            B be Subset of X0;
  assume A1: C = the carrier of X0;
   then A2: C is open by TSEP_1:def 1;
  assume A = B;
  hence thesis by A1,A2,Th62;
 end;

theorem
   for C, A being Subset of X,
     B being Subset of X0 st
   C c= the carrier of X0 & A c= C & A = B holds
  C is everywhere_dense & B is everywhere_dense iff A is everywhere_dense
 proof let C, A be Subset of X,
   B be Subset of X0;
   assume A1: C c= the carrier of X0;
   assume A2: A c= C;
   assume A3: A = B;
  thus C is everywhere_dense & B is everywhere_dense
        implies A is everywhere_dense
   proof
    assume C is everywhere_dense;
     then A4: Int C is everywhere_dense by Th32;
     then A5: Int C is dense & Int C is open by Th33;
        Int C c= C by TOPS_1:44;
     then Int C is Subset of X0 by A1,XBOOLE_1:1;
     then reconsider D = Int C as Subset of X0;
    assume A6: B is everywhere_dense;
         D is everywhere_dense by A4,Th61;
     then A7: D /\ B is everywhere_dense by A6,Th44;
       A8: D /\ B c= Int C by XBOOLE_1:17;
      then D /\ B is Subset of X by XBOOLE_1:1;
      then reconsider E = D /\ B as Subset of X;
       A9: E is everywhere_dense by A5,A7,A8,Th62;
          E c= A by A3,XBOOLE_1:17;
    hence thesis by A9,Th38;
   end;
  thus A is everywhere_dense implies C is everywhere_dense &
                             B is everywhere_dense by A2,A3,Th38,Th61;
 end;

theorem Th65:
 for A being Subset of X,
     B being Subset of X0 st A c= B holds
  B is boundary implies A is boundary
 proof let A be Subset of X, B be Subset of X0;
  assume A1: A c= B;
   then reconsider C = A as Subset of X0 by XBOOLE_1:1;
     A2: Int A c= Int C by Th56;
     A3: Int C c= Int B by A1,TOPS_1:48;
  assume Int B = {};
   then Int C = {} by A3,XBOOLE_1:3;
   then Int A = {} by A2,XBOOLE_1:3;
  hence thesis by Def1;
 end;

theorem Th66:
 for C, A being Subset of X,
     B being Subset of X0 st
   C is open & C c= the carrier of X0 & A c= C & A = B holds
  A is boundary implies B is boundary
 proof let C, A be Subset of X,
    B be Subset of X0;
  assume C is open & C c= the carrier of X0 & A c= C & A = B;
    then A1: Int A = Int B by Th57;
  assume A is boundary;
   then Int A = {} by TOPS_1:84;
  hence thesis by A1,TOPS_1:84;
 end;

theorem
   for X0 being open non empty SubSpace of X
  for A being Subset of X,
      B being Subset of X0 st A = B holds
   A is boundary iff B is boundary
 proof let X0 be open non empty SubSpace of X;
   let A be Subset of X,
    B be Subset of X0;
  assume A1: A = B;
     the carrier of X0 is Subset of X by TSEP_1:1;
   then reconsider C = the carrier of X0 as Subset of X;
     A c= C & C is open by A1,TSEP_1:def 1;
  hence thesis by A1,Th65,Th66;
 end;

theorem Th68:
 for A being Subset of X,
     B being Subset of X0 st A c= B holds
   B is nowhere_dense implies A is nowhere_dense
 proof let A be Subset of X, B be Subset of X0;
  assume A1: A c= B;
   then A is Subset of X0 by XBOOLE_1:1;
   then reconsider C = A as Subset of X0;
        reconsider D = the carrier of X0 as Subset of X
         by TSEP_1:1;
       Int Cl A c= Cl A by TOPS_1:44;
    then (Int Cl A) /\ [#]X0 c= (Cl A) /\ [#]X0 by XBOOLE_1:26;
    then A2: (Int Cl A) /\ [#]X0 c= Cl C by PRE_TOPC:47;
    then (Int Cl A) /\ [#]X0 is Subset of X0 by XBOOLE_1:1;
    then reconsider G = (Int Cl A) /\ [#]X0 as Subset of X0;
      G is open by TOPS_2:32;
   then A3: G c= Int Cl C by A2,TOPS_1:56;
  assume B is nowhere_dense;
   then C is nowhere_dense by A1,Th26;
   then A4: Int Cl C = {} by Def3;
      now
     assume Int Cl A <> {};
      then A meets Int Cl A by Th7;
      then A5: A /\ Int Cl A <> {} by XBOOLE_0:def 7;
        C c= D;
      then A /\ Int Cl A c= D /\ Int Cl A by XBOOLE_1:26;
      then (Int Cl A) /\ D <> {} by A5,XBOOLE_1:3;
      then G <> {} by PRE_TOPC:12;
     hence contradiction by A3,A4,XBOOLE_1:3;
    end;
  hence thesis by Def3;
 end;

Lm1:
 for C, A being Subset of X,
     B being Subset of X0 st
   C is open & C = the carrier of X0 & A c= C & A = B holds
  A is nowhere_dense implies B is nowhere_dense
 proof let C, A be Subset of X,
    B be Subset of X0;
  assume A1: C is open;
  assume A2: C = the carrier of X0;
  assume A c= C & A = B;
   then A3: Cl B c= Cl A by Th53;
   then Cl B is Subset of X by XBOOLE_1:1;
   then reconsider D = Cl B as Subset of X;
    A4: Int D = Int Cl B by A1,A2,Th57;
  assume A is nowhere_dense;
   then Int Cl A = {} & Int D c= Int Cl A by A3,Def3,TOPS_1:48;
   then Int Cl B = {} by A4,XBOOLE_1:3;
  hence thesis by Def3;
 end;

theorem Th69:
 for C, A being Subset of X,
     B being Subset of X0 st
   C is open & C c= the carrier of X0 & A c= C & A = B holds
  A is nowhere_dense implies B is nowhere_dense
 proof let C, A be Subset of X,
    B be Subset of X0;
  assume A1: C is open;
  assume A2: C c= the carrier of X0;
  assume A3: A c= C & A = B;
  assume A4: A is nowhere_dense;
   A5:now assume C = {};
       then B = {}X0 by A3,XBOOLE_1:3;
       hence B is nowhere_dense by Th22;
      end;
     now assume C <> {};
    then consider X1 being strict non empty SubSpace of X such that
       A6: C = the carrier of X1 by TSEP_1:10;
      reconsider E = B as Subset of X1 by A3,A6;
     A7: E is nowhere_dense by A1,A3,A4,A6,Lm1;
        X1 is SubSpace of X0 by A2,A6,TSEP_1:4;
    hence B is nowhere_dense by A7,Th68;
   end;
  hence thesis by A5;
 end;

theorem
   for X0 being open non empty SubSpace of X
  for A being Subset of X,
      B being Subset of X0 st A = B holds
   A is nowhere_dense iff B is nowhere_dense
 proof let X0 be open non empty SubSpace of X;
  let A be Subset of X,
      B be Subset of X0;
  assume A1: A = B;
     the carrier of X0 is Subset of X by TSEP_1:1;
   then reconsider C = the carrier of X0 as Subset of X;
     A c= C & C is open by A1,TSEP_1:def 1;
  hence thesis by A1,Th68,Th69;
 end;


begin
:: 4. Subsets in Topological Spaces with the same Topological Structures.
theorem
   for X1, X2 being 1-sorted holds
 the carrier of X1 = the carrier of X2 implies
  for C1 being Subset of X1,
      C2 being Subset of X2 holds
    C1 = C2 iff  C1` =  C2`
 proof
  let X1, X2 be 1-sorted;
  assume the carrier of X1 = the carrier of X2;
     then [#]X1 = the carrier of X2 by PRE_TOPC:12;
    then A1: [#]X1 = [#]X2 by PRE_TOPC:12;
   let C1 be Subset of X1, C2 be Subset of X2;
  thus C1 = C2 implies  C1` =  C2`
   proof
    assume C1 = C2;
    hence  C1` = [#]X2 \ C2 by A1,PRE_TOPC:17
            .=  C2` by PRE_TOPC:17;
   end;
  thus  C1` =  C2` implies C1 = C2
   proof
    assume A2:  C1` =  C2`;
    thus C1 = [#]X1 \ ([#]X1 \ C1) by PRE_TOPC:22
           .= [#]X2 \  C2` by A1,A2,PRE_TOPC:17
           .= [#]X2 \ ([#]X2 \ C2) by PRE_TOPC:17
           .= C2 by PRE_TOPC:22;
   end;
 end;

reserve X1,X2 for TopStruct;

theorem Th72:
 the carrier of X1 = the carrier of X2 &
  (for C1 being Subset of X1,
       C2 being Subset of X2 st C1 = C2 holds
    (C1 is open iff C2 is open)) implies
  the TopStruct of X1 = the TopStruct of X2
 proof
  assume A1: the carrier of X1 = the carrier of X2;
  assume A2: for C1 being Subset of X1,
                 C2 being Subset of X2 st C1 = C2 holds
               (C1 is open iff C2 is open);
     the topology of X1 = the topology of X2
    proof
         now let D be set;
        assume A3: D in the topology of X1;
         then reconsider C1 = D as Subset of X1;
         reconsider C2 = C1 as Subset of X2 by A1;
            C1 is open by A3,PRE_TOPC:def 5;
         then C2 is open by A2;
        hence D in the topology of X2 by PRE_TOPC:def 5;
       end;
      then A4: the topology of X1 c= the topology of X2 by TARSKI:def 3;
         now let D be set;
        assume A5: D in the topology of X2;
         then reconsider C2 = D as Subset of X2;
         reconsider C1 = C2 as Subset of X1 by A1;
            C2 is open by A5,PRE_TOPC:def 5;
         then C1 is open by A2;
        hence D in the topology of X1 by PRE_TOPC:def 5;
       end;
      then the topology of X2 c= the topology of X1 by TARSKI:def 3;
     hence thesis by A4,XBOOLE_0:def 10;
    end;
  hence thesis by A1;
 end;

theorem Th73:
 the carrier of X1 = the carrier of X2 &
  (for C1 being Subset of X1,
       C2 being Subset of X2 st C1 = C2 holds
    (C1 is closed iff C2 is closed)) implies
  the TopStruct of X1 = the TopStruct of X2
 proof
  assume A1: the carrier of X1 = the carrier of X2;
  assume A2: for C1 being Subset of X1,
                 C2 being Subset of X2 st C1 = C2 holds
               (C1 is closed iff C2 is closed);
     now let C1 be Subset of X1,
           C2 be Subset of X2;
    assume C1 = C2;
then A3:   C1` =  C2` by A1;
    thus C1 is open implies C2 is open
     proof
      assume C1 is open;
       then  C1` is closed &  C1` =  C2` by A3,TOPS_1:30;
       then  C2` is closed by A2;
      hence thesis by TOPS_1:30;
     end;
    thus C2 is open implies C1 is open
     proof
      assume C2 is open;
       then  C2` is closed &  C1` =  C2` by A3,TOPS_1:30;
       then  C1` is closed by A2;
      hence thesis by TOPS_1:30;
     end;
   end;
  hence thesis by A1,Th72;
 end;

reserve X1,X2 for TopSpace;

theorem
   the carrier of X1 = the carrier of X2 &
  (for C1 being Subset of X1,
       C2 being Subset of X2 st C1 = C2 holds
    Int C1 = Int C2) implies
  the TopStruct of X1 = the TopStruct of X2
 proof
  assume A1: the carrier of X1 = the carrier of X2;
  assume A2: for C1 being Subset of X1,
          C2 being Subset of X2 st C1 = C2 holds
                Int C1 = Int C2;
     now let C1 be Subset of X1,
     C2 be Subset of X2;
    assume A3: C1 = C2;
    thus C1 is open implies C2 is open
     proof assume C1 is open;
      then C1 = Int C1 by TOPS_1:55;
      then C2 = Int C2 by A2,A3;
      hence thesis;
     end;
    thus C2 is open implies C1 is open
     proof assume C2 is open;
      then C2 = Int C2 by TOPS_1:55;
      then C1 = Int C1 by A2,A3;
      hence thesis;
     end;
   end;
  hence thesis by A1,Th72;
 end;

theorem
   the carrier of X1 = the carrier of X2 &
  (for C1 being Subset of X1,
       C2 being Subset of X2 st C1 = C2 holds
    Cl C1 = Cl C2) implies
  the TopStruct of X1 = the TopStruct of X2
 proof
  assume A1: the carrier of X1 = the carrier of X2;
  assume A2: for C1 being Subset of X1,
    C2 being Subset of X2 st C1 = C2 holds
                Cl C1 = Cl C2;
     now let C1 be Subset of X1,
     C2 be Subset of X2;
    assume A3: C1 = C2;
    thus C1 is closed implies C2 is closed
     proof assume C1 is closed;
      then C1 = Cl C1 by PRE_TOPC:52;
      then C2 = Cl C2 by A2,A3;
      hence thesis;
     end;
    thus C2 is closed implies C1 is closed
     proof assume C2 is closed;
      then C2 = Cl C2 by PRE_TOPC:52;
      then C1 = Cl C1 by A2,A3;
      hence thesis;
     end;
   end;
  hence thesis by A1,Th73;
 end;

reserve D1 for Subset of X1, D2 for Subset of X2;

theorem Th76:
 D1 = D2 & the TopStruct of X1 = the TopStruct of X2 implies
   (D1 is open implies D2 is open)
 proof
  assume A1: D1 = D2;
  assume A2: the TopStruct of X1 = the TopStruct of X2;
  assume D1 in the topology of X1;
  hence D2 is open by A1,A2,PRE_TOPC:def 5;
 end;

theorem Th77:
 D1 = D2 & the TopStruct of X1 = the TopStruct of X2 implies Int D1 = Int D2
 proof
  assume A1: D1 = D2;
  assume A2: the TopStruct of X1 = the TopStruct of X2;
    A3: Int D1 c= D1 by TOPS_1:44;
   then Int D1 is Subset of X2 by A1,XBOOLE_1:1;
   then reconsider C2 = Int D1 as Subset of X2;
    A4: Int D2 c= D2 by TOPS_1:44;
   then Int D2 is Subset of X1 by A1,XBOOLE_1:1;
   then reconsider C1 = Int D2 as Subset of X1;
      C2 is open & C1 is open by A2,Th76;
    then C1 c= Int D1 & C2 c= Int D2 by A1,A3,A4,TOPS_1:56;
  hence thesis by XBOOLE_0:def 10;
 end;

theorem Th78:
 D1 c= D2 & the TopStruct of X1 = the TopStruct of X2 implies Int D1 c= Int D2
 proof
  assume A1: D1 c= D2;
   then D1 is Subset of X2 by XBOOLE_1:1;
   then reconsider C2 = D1 as Subset of X2;
  assume the TopStruct of X1 = the TopStruct of X2;
   then Int D1 = Int C2 & Int C2 c= Int D2 by A1,Th77,TOPS_1:48;
  hence thesis;
 end;

theorem Th79:
 D1 = D2 & the TopStruct of X1 = the TopStruct of X2 implies
   (D1 is closed implies D2 is closed)
 proof
  assume A1: D1 = D2;
  assume A2: the TopStruct of X1 = the TopStruct of X2;
    then [#]X1 = the carrier of X2 by PRE_TOPC:12;
   then [#]X1 = [#]X2 by PRE_TOPC:12;
    then D1` = [#]X2 \ D2 by A1,PRE_TOPC:17;
   then A3: D1` = D2` by PRE_TOPC:17;
  assume D1 is closed;
   then D1` is open by TOPS_1:29;
   then D2` is open by A2,A3,Th76;
  hence D2 is closed by TOPS_1:29;
 end;

theorem Th80:
 D1 = D2 & the TopStruct of X1 = the TopStruct of X2 implies Cl D1 = Cl D2
 proof
  assume A1: D1 = D2;
  assume A2: the TopStruct of X1 = the TopStruct of X2;
    A3: D1 c= Cl D1 by PRE_TOPC:48;
   reconsider C2 = Cl D1 as Subset of X2 by A2;
    A4: D2 c= Cl D2 by PRE_TOPC:48;
   reconsider C1 = Cl D2 as Subset of X1 by A2;
      C2 is closed & C1 is closed by A2,Th79;
    then Cl D1 c= C1 & Cl D2 c= C2 by A1,A3,A4,TOPS_1:31;
  hence thesis by XBOOLE_0:def 10;
 end;

theorem Th81:
 D1 c= D2 & the TopStruct of X1 = the TopStruct of X2 implies Cl D1 c= Cl D2
 proof
  assume A1: D1 c= D2;
  assume A2: the TopStruct of X1 = the TopStruct of X2;
   then reconsider C2 = D1 as Subset of X2;
      Cl D1 = Cl C2 & Cl C2 c= Cl D2 by A1,A2,Th80,PRE_TOPC:49;
  hence thesis;
 end;

theorem Th82:
 D2 c= D1 & the TopStruct of X1 = the TopStruct of X2 implies
   (D1 is boundary implies D2 is boundary)
 proof
  assume A1: D2 c= D1;
   then D2 is Subset of X1 by XBOOLE_1:1;
   then reconsider C1 = D2 as Subset of X1;
  assume A2: the TopStruct of X1 = the TopStruct of X2;
  assume D1 is boundary;
   then C1 is boundary by A1,Th11;
   then Int C1 = {} & Int C1 = Int D2 by A2,Def1,Th77;
  hence thesis by Def1;
 end;

theorem Th83:
 D1 c= D2 & the TopStruct of X1 = the TopStruct of X2 implies
   (D1 is dense implies D2 is dense)
 proof
  assume A1: D1 c= D2;
   then D1 is Subset of X2 by XBOOLE_1:1;
   then reconsider C2 = D1 as Subset of X2;
  assume A2: the TopStruct of X1 = the TopStruct of X2;
  assume D1 is dense;
   then Cl D1 = the carrier of X1 & Cl D1 = Cl C2 by A2,Def2,Th80;
   then C2 is dense by A2,Def2;
  hence thesis by A1,TOPS_1:79;
 end;

theorem
   D2 c= D1 & the TopStruct of X1 = the TopStruct of X2 implies
   (D1 is nowhere_dense implies D2 is nowhere_dense)
 proof
  assume A1: D2 c= D1;
  assume A2: the TopStruct of X1 = the TopStruct of X2;
  assume D1 is nowhere_dense;
   then Cl D1 is boundary & Cl D2 c= Cl D1 by A1,A2,Th81,TOPS_1:def 5;
   then Cl D2 is boundary by A2,Th82;
  hence thesis by TOPS_1:def 5;
 end;

reserve X1,X2 for non empty TopSpace;
reserve D1 for Subset of X1, D2 for Subset of X2;

theorem
   D1 c= D2 & the TopStruct of X1 = the TopStruct of X2 implies
   (D1 is everywhere_dense implies D2 is everywhere_dense)
 proof
  assume A1: D1 c= D2;
  assume A2: the TopStruct of X1 = the TopStruct of X2;
  assume D1 is everywhere_dense;
   then Int D1 is dense & Int D1 c= Int D2 by A1,A2,Th35,Th78;
   then Int D2 is dense by A2,Th83;
  hence thesis by Th35;
 end;


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