### 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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