### Locally Connected Spaces

by

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;
```