Copyright (c) 1993 Association of Mizar Users
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;