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Basicpropertiesandexamples
17
n−1
xn∈X\
U
BX(xk
,
8)
.As
d(xn
,
xm)>8
for
n/=m
,thesequence
k=1
(xn)n∈Ncannotadmitaconvergentsubsequence.
(b)⇒(c)
.Let
{Ui}i∈I
beanopencoveringof
X
;ofcourse,we
mayassumethatthesetofindices
I
isinfinite,forotherwisethere
isnothingtoprove.Moreover,let
F
bethefamilyofallthosenon-
emptysubsetsof
X
whichcannotbecoveredbyfinitelymanysets
Ui
,
andassumethatthefamily
F
isnon-empty.Forsure,
X∈F
,since
if
A⊆B
and
A∈F
,thenalso
B∈F
.Observealsothatif
A∈F
andif
{a1
,
...
,
ak}
isafinite
8
-netfor
X
,then
A∩BX(aj
,
8)∈F
for
some
j∈{
1,
...
,
k}
.Consequently,startingwith
A0:=X
,weareableto
constructasequenceofpoints
(xn)n∈N
andasequenceofsets
(An)n∈N
suchthatAn=An−1∩BX(xn,1
n)∈Fforn∈N.
Forevery
n∈N
,let
yn
beanarbitrarypointwhichbelongstothe
set
An
.Sincethesequence
(An)n∈N
isnon-increasing,foranypositive
integers
n
,
n0
suchthat
n>n0
wehave
yn∈An
0⊆BX(xn
0
,
n0)
1
.
Thus,
d(yn
,
ym)<d(yn
,
yn
0)+d(yn
0
,
ym)<4
n0
for
n
,
m>n0
.This
impliesthat
(yn)n∈N
isaCauchysequence,andtherefore,by(b),it
convergesinXtoapointy.
Since
{Ui}i∈I
isacoveringof
X
,thereisanindex
i0∈I
suchthat
y∈Ui
0
.Moreover,
Ui
0
isopen,sothereexistsan
8>
0suchthat
BX(y
,
8)⊆Ui
0
.Choosing
n∈N
sothat
n<1
1
48
and
d(yn
,
y)<
1
28
,weseethat
An⊆BX(xn
,
n)⊆BX(yn
1
,
n)⊆BX(y
2
,
8)⊆Ui
0
.
Hence,
An
iscoveredbythesingle-setfamily
Ui
0
,andthus
An/
∈F
–
acontradiction.Therefore,thefamily
F
isempty,meaningthat
X
can
becoveredbyfinitelymanysetsUi.
(c)⇒(a)
.Onthecontrary,letussupposethatthereisasequen-
ce
(xn)n∈N
whichdoesnothaveaconvergentsubsequence.Inparticu-
lar,theset
{xn|n∈N}
isinfinite.Then,foreachpoint
x∈X
thereis
aradius
rx>
0suchthatthesetofindices
Nx:={n∈N|d(xn
,
x)<
rx}
iseitheremptyorfinite.Thefamily
{BX(x
,
rx)}x∈X
isanopencov-
m
eringof
X
,andsofrom(c)wededucethat
X⊆
U
BX(yi
,
ry
i)
for
i=1
some
y1
,
...
,
ym∈X
.This,inturn,impliesthatatleastoneoftheballs
BX(yi
,
ry
i)
containsinfinitelymanytermsofthesequence
(xn)n∈N
.
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