let X be set ; card { {X,[x,X]} where x is Element of X : x in X } = card X
set uG = X;
set A = { {X,[x,X]} where x is Element of X : x in X } ;
deffunc H1( set ) -> set = {X,[$1,X]};
consider f being Function such that
B:
dom f = X
and
D:
for x being set st x in X holds
f . x = H1(x)
from FUNCT_1:sch 3();
then X:
f is one-to-one
by FUNCT_1:def 4;
Y:
rng f = { {X,[x,X]} where x is Element of X : x in X }
proof
thus
rng f c= { {X,[x,X]} where x is Element of X : x in X }
XBOOLE_0:def 10 { {X,[x,X]} where x is Element of X : x in X } c= rng fproof
let y be
set ;
TARSKI:def 3 ( not y in rng f or y in { {X,[x,X]} where x is Element of X : x in X } )
assume
y in rng f
;
y in { {X,[x,X]} where x is Element of X : x in X }
then consider a being
set such that A1:
a in dom f
and B1:
f . a = y
by FUNCT_1:def 3;
y = {X,[a,X]}
by A1, B1, B, D;
hence
y in { {X,[x,X]} where x is Element of X : x in X }
by A1, B;
verum
end;
thus
{ {X,[x,X]} where x is Element of X : x in X } c= rng f
verumproof
let a be
set ;
TARSKI:def 3 ( not a in { {X,[x,X]} where x is Element of X : x in X } or a in rng f )
assume
a in { {X,[x,X]} where x is Element of X : x in X }
;
a in rng f
then consider x being
Element of
X such that A1:
a = {X,[x,X]}
and B1:
x in X
;
f . x = a
by A1, B1, D;
hence
a in rng f
by B, B1, FUNCT_1:def 3;
verum
end;
end;
{ {X,[x,X]} where x is Element of X : x in X } ,X are_equipotent
by B, X, Y, WELLORD2:def 4;
hence
card { {X,[x,X]} where x is Element of X : x in X } = card X
by CARD_1:5; verum