let P be Subset of (TOP-REAL 2); ( |[(- 1),0 ]|,|[1,0 ]| realize-max-dist-in P implies P c= closed_inside_of_rectangle (- 1),1,(- 3),3 )
assume that
A1:
|[(- 1),0 ]| in P
and
A2:
|[1,0 ]| in P
and
A3:
for x, y being Point of (TOP-REAL 2) st x in P & y in P holds
dist |[(- 1),0 ]|,|[1,0 ]| >= dist x,y
; JORDAN24:def 1 P c= closed_inside_of_rectangle (- 1),1,(- 3),3
let p be set ; TARSKI:def 3 ( not p in P or p in closed_inside_of_rectangle (- 1),1,(- 3),3 )
assume A4:
p in P
; p in closed_inside_of_rectangle (- 1),1,(- 3),3
then reconsider p = p as Point of (TOP-REAL 2) ;
A5: dist |[(- 1),0 ]|,p =
sqrt ((((- 1) - (p `1 )) ^2 ) + ((0 - (p `2 )) ^2 ))
by Lm16, Lm18, TOPREAL6:101
.=
sqrt ((((- 1) - (p `1 )) ^2 ) + ((p `2 ) ^2 ))
;
A7:
now assume A8:
- 1
> p `1
;
contradictionthen
LSeg p,
|[1,0 ]| meets Vertical_Line (- 1)
by Lm17, Th8;
then consider x being
set such that A9:
x in LSeg p,
|[1,0 ]|
and A10:
x in Vertical_Line (- 1)
by XBOOLE_0:3;
reconsider x =
x as
Point of
(TOP-REAL 2) by A9;
A11:
x `1 = - 1
by A10, JORDAN6:34;
A12:
dist p,
|[1,0 ]| = (dist p,x) + (dist x,|[1,0 ]|)
by A9, JORDAN1K:29;
A13:
dist x,
|[1,0 ]| =
sqrt ((((x `1 ) - (|[1,0 ]| `1 )) ^2 ) + (((x `2 ) - (|[1,0 ]| `2 )) ^2 ))
by TOPREAL6:101
.=
sqrt (((- 2) ^2 ) + (((x `2 ) - 0 ) ^2 ))
by A11, Lm17, EUCLID:56
.=
sqrt (4 + ((x `2 ) ^2 ))
;
then
(dist p,|[1,0 ]|) + 0 > (dist |[(- 1),0 ]|,|[1,0 ]|) + 0
by A8, A11, A12, JORDAN1K:22, XREAL_1:10;
hence
contradiction
by A2, A3, A4;
verum end;
A14:
now assume A15:
p `1 > 1
;
contradictionthen
LSeg p,
|[(- 1),0 ]| meets Vertical_Line 1
by Lm16, Th8;
then consider x being
set such that A16:
x in LSeg p,
|[(- 1),0 ]|
and A17:
x in Vertical_Line 1
by XBOOLE_0:3;
reconsider x =
x as
Point of
(TOP-REAL 2) by A16;
A18:
x `1 = 1
by A17, JORDAN6:34;
A19:
dist p,
|[(- 1),0 ]| = (dist p,x) + (dist x,|[(- 1),0 ]|)
by A16, JORDAN1K:29;
A20:
dist x,
|[(- 1),0 ]| =
sqrt ((((x `1 ) - (|[(- 1),0 ]| `1 )) ^2 ) + (((x `2 ) - (|[(- 1),0 ]| `2 )) ^2 ))
by TOPREAL6:101
.=
sqrt (4 + ((x `2 ) ^2 ))
by A18, Lm16, Lm18
;
then
(dist p,|[(- 1),0 ]|) + 0 > (dist |[(- 1),0 ]|,|[1,0 ]|) + 0
by A15, A18, A19, JORDAN1K:22, XREAL_1:10;
hence
contradiction
by A1, A3, A4;
verum end;
3 >= p `2
by A6, Lm64, SQUARE_1:78;
hence
p in closed_inside_of_rectangle (- 1),1,(- 3),3
by A7, A14, A21; verum