let L1 be lower-bounded continuous sup-Semilattice; for T being Scott TopAugmentation of L1
for B1 being Basis of T st not B1 is finite holds
not { (inf u) where u is Subset of T : u in B1 } is finite
let T be Scott TopAugmentation of L1; for B1 being Basis of T st not B1 is finite holds
not { (inf u) where u is Subset of T : u in B1 } is finite
let B1 be Basis of T; ( not B1 is finite implies not { (inf u) where u is Subset of T : u in B1 } is finite )
reconsider B2 = { (inf u) where u is Subset of T : u in B1 } as set ;
defpred S1[ set , set ] means ex y being Element of T st
( $1 = y & $2 = wayabove y );
reconsider B3 = { (wayabove (inf u)) where u is Subset of T : u in B1 } as Basis of T by Th27;
assume that
A1:
not B1 is finite
and
A2:
{ (inf u) where u is Subset of T : u in B1 } is finite
; contradiction
A3:
for x being set st x in B2 holds
ex y being set st
( y in B3 & S1[x,y] )
proof
let x be
set ;
( x in B2 implies ex y being set st
( y in B3 & S1[x,y] ) )
assume
x in B2
;
ex y being set st
( y in B3 & S1[x,y] )
then A4:
ex
u1 being
Subset of
T st
(
x = inf u1 &
u1 in B1 )
;
then reconsider z =
x as
Element of
T ;
take y =
wayabove z;
( y in B3 & S1[x,y] )
thus
y in B3
by A4;
S1[x,y]
take
z
;
( x = z & y = wayabove z )
thus
(
x = z &
y = wayabove z )
;
verum
end;
consider f being Function such that
A5:
dom f = B2
and
A6:
rng f c= B3
and
A7:
for x being set st x in B2 holds
S1[x,f . x]
from FUNCT_1:sch 5(A3);
B3 c= rng f
then
rng f = B3
by A6, XBOOLE_0:def 10;
then
B3 is finite
by A2, A5, FINSET_1:8;
then
T is finite
by YELLOW15:30;
hence
contradiction
by A1; verum