let a, b be Real; ( a <> b iff Cl ].a,b.[ = [.a,b.] )
thus
( a <> b implies Cl ].a,b.[ = [.a,b.] )
( Cl ].a,b.[ = [.a,b.] implies a <> b )proof
assume A1:
a <> b
;
Cl ].a,b.[ = [.a,b.]
per cases
( a > b or a < b )
by A1, XXREAL_0:1;
suppose A3:
a < b
;
Cl ].a,b.[ = [.a,b.]then A4:
(a + b) / 2
< b
by XREAL_1:226;
thus
Cl ].a,b.[ c= [.a,b.]
by MEASURE6:57, XXREAL_1:25;
XBOOLE_0:def 10 [.a,b.] c= Cl ].a,b.[let p be
object ;
TARSKI:def 3 ( not p in [.a,b.] or p in Cl ].a,b.[ )A5:
].a,b.[ = { w where w is Real : ( a < w & w < b ) }
by RCOMP_1:def 2;
assume A6:
p in [.a,b.]
;
p in Cl ].a,b.[
[.a,b.] = { w where w is Real : ( a <= w & w <= b ) }
by RCOMP_1:def 1;
then consider r being
Real such that A7:
p = r
and A8:
a <= r
and A9:
r <= b
by A6;
a < (a + b) / 2
by A3, XREAL_1:226;
then A10:
(a + b) / 2
in ].a,b.[
by A5, A4;
now ( ( a < r & r < b & p in Cl ].a,b.[ ) or ( a = r & p in Cl ].a,b.[ ) or ( b = r & p in Cl ].a,b.[ ) )per cases
( ( a < r & r < b ) or a = r or b = r )
by A8, A9, XXREAL_0:1;
case A13:
a = r
;
p in Cl ].a,b.[
for
O being
open Subset of
REAL st
a in O holds
not
O /\ ].a,b.[ is
empty
proof
let O be
open Subset of
REAL;
( a in O implies not O /\ ].a,b.[ is empty )
assume
a in O
;
not O /\ ].a,b.[ is empty
then consider g being
Real such that A14:
0 < g
and A15:
].(a - g),(a + g).[ c= O
by RCOMP_1:19;
A16:
a - g < a - 0
by A14, XREAL_1:15;
A17:
].(a - g),(a + g).[ = { w where w is Real : ( a - g < w & w < a + g ) }
by RCOMP_1:def 2;
per cases
( a + g < b or a + g >= b )
;
suppose A18:
a + g < b
;
not O /\ ].a,b.[ is empty A19:
a + 0 < a + g
by A14, XREAL_1:6;
then A20:
a < (a + (a + g)) / 2
by XREAL_1:226;
A21:
(a + (a + g)) / 2
< a + g
by A19, XREAL_1:226;
then
(a + (a + g)) / 2
< b
by A18, XXREAL_0:2;
then A22:
(a + (a + g)) / 2
in ].a,b.[
by A5, A20;
a - g < (a + (a + g)) / 2
by A16, A20, XXREAL_0:2;
then
(a + (a + g)) / 2
in ].(a - g),(a + g).[
by A17, A21;
hence
not
O /\ ].a,b.[ is
empty
by A15, A22, XBOOLE_0:def 4;
verum end; end;
end; hence
p in Cl ].a,b.[
by A7, A13, MEASURE6:63;
verum end; case A25:
b = r
;
p in Cl ].a,b.[
for
O being
open Subset of
REAL st
b in O holds
not
O /\ ].a,b.[ is
empty
proof
let O be
open Subset of
REAL;
( b in O implies not O /\ ].a,b.[ is empty )
assume
b in O
;
not O /\ ].a,b.[ is empty
then consider g being
Real such that A26:
0 < g
and A27:
].(b - g),(b + g).[ c= O
by RCOMP_1:19;
A28:
b - g < b - 0
by A26, XREAL_1:15;
A29:
b + 0 < b + g
by A26, XREAL_1:6;
A30:
].(b - g),(b + g).[ = { w where w is Real : ( b - g < w & w < b + g ) }
by RCOMP_1:def 2;
per cases
( b - g > a or b - g <= a )
;
suppose A31:
b - g > a
;
not O /\ ].a,b.[ is empty A32:
(b + (b - g)) / 2
< b
by A28, XREAL_1:226;
A33:
b - g < (b + (b - g)) / 2
by A28, XREAL_1:226;
then
a < (b + (b - g)) / 2
by A31, XXREAL_0:2;
then A34:
(b + (b - g)) / 2
in ].a,b.[
by A5, A32;
(b + (b - g)) / 2
< b
by A28, XREAL_1:226;
then
(b + (b - g)) / 2
< b + g
by A29, XXREAL_0:2;
then
(b + (b - g)) / 2
in ].(b - g),(b + g).[
by A30, A33;
hence
not
O /\ ].a,b.[ is
empty
by A27, A34, XBOOLE_0:def 4;
verum end; end;
end; hence
p in Cl ].a,b.[
by A7, A25, MEASURE6:63;
verum end; end; end; hence
p in Cl ].a,b.[
;
verum end; end;
end;
assume that
A37:
Cl ].a,b.[ = [.a,b.]
and
A38:
a = b
; contradiction
[.a,b.] = {a}
by A38, XXREAL_1:17;
hence
contradiction
by A37, A38, MEASURE6:60, XXREAL_1:28; verum