let S be non empty TopSpace; for T being non empty Hausdorff TopSpace
for f, g being continuous Function of S,T holds
( ( for X being Subset of S st X = { p where p is Point of S : f . p <> g . p } holds
X is open ) & ( for X being Subset of S st X = { p where p is Point of S : f . p = g . p } holds
X is closed ) )
let T be non empty Hausdorff TopSpace; for f, g being continuous Function of S,T holds
( ( for X being Subset of S st X = { p where p is Point of S : f . p <> g . p } holds
X is open ) & ( for X being Subset of S st X = { p where p is Point of S : f . p = g . p } holds
X is closed ) )
let f, g be continuous Function of S,T; ( ( for X being Subset of S st X = { p where p is Point of S : f . p <> g . p } holds
X is open ) & ( for X being Subset of S st X = { p where p is Point of S : f . p = g . p } holds
X is closed ) )
{ p where p is Point of S : f . p <> g . p } c= the carrier of S
then reconsider A = { p where p is Point of S : f . p <> g . p } as Subset of S ;
A1:
[#] T <> {}
;
thus
for X being Subset of S st X = { p where p is Point of S : f . p <> g . p } holds
X is open
for X being Subset of S st X = { p where p is Point of S : f . p = g . p } holds
X is closed proof
let X be
Subset of
S;
( X = { p where p is Point of S : f . p <> g . p } implies X is open )
assume A2:
X = { p where p is Point of S : f . p <> g . p }
;
X is open
for
x being
set holds
(
x in X iff ex
Q being
Subset of
S st
(
Q is
open &
Q c= X &
x in Q ) )
proof
let x be
set ;
( x in X iff ex Q being Subset of S st
( Q is open & Q c= X & x in Q ) )
hereby ( ex Q being Subset of S st
( Q is open & Q c= X & x in Q ) implies x in X )
assume
x in X
;
ex Q being Element of bool the carrier of S st
( Q is open & Q c= X & x in Q )then consider p being
Point of
S such that A3:
x = p
and A4:
f . p <> g . p
by A2;
consider W,
V being
Subset of
T such that A5:
(
W is
open &
V is
open )
and A6:
f . p in W
and A7:
g . p in V
and A8:
W misses V
by A4, PRE_TOPC:def 10;
dom g = the
carrier of
S
by FUNCT_2:def 1;
then
[x,(g . p)] in g
by A3, FUNCT_1:def 2;
then A9:
x in g " V
by A7, RELAT_1:def 14;
take Q =
(f " W) /\ (g " V);
( Q is open & Q c= X & x in Q )
(
f " W is
open &
g " V is
open )
by A1, A5, TOPS_2:43;
hence
Q is
open
;
( Q c= X & x in Q )thus
Q c= X
x in Qproof
let q be
object ;
TARSKI:def 3 ( not q in Q or q in X )
assume A10:
q in Q
;
q in X
then
q in f " W
by XBOOLE_0:def 4;
then consider yf being
object such that A11:
[q,yf] in f
and A12:
yf in W
by RELAT_1:def 14;
q in g " V
by A10, XBOOLE_0:def 4;
then consider yg being
object such that A13:
(
[q,yg] in g &
yg in V )
by RELAT_1:def 14;
A14:
(
yg = g . q & not
yg in W )
by A8, A13, FUNCT_1:1, XBOOLE_0:3;
yf = f . q
by A11, FUNCT_1:1;
hence
q in X
by A2, A10, A12, A14;
verum
end;
dom f = the
carrier of
S
by FUNCT_2:def 1;
then
[x,(f . p)] in f
by A3, FUNCT_1:def 2;
then
x in f " W
by A6, RELAT_1:def 14;
hence
x in Q
by A9, XBOOLE_0:def 4;
verum
end;
given Q being
Subset of
S such that
Q is
open
and A15:
(
Q c= X &
x in Q )
;
x in X
thus
x in X
by A15;
verum
end;
hence
X is
open
by TOPS_1:25;
verum
end;
then A16:
A is open
;
let X be Subset of S; ( X = { p where p is Point of S : f . p = g . p } implies X is closed )
assume A17:
X = { p where p is Point of S : f . p = g . p }
; X is closed
X ` = A
hence
X is closed
by A16; verum