Compactspacest
n analysis we learn that a b is compact and more generally
any closed bounded subset of IR is compact
Good properties of compact spaces
generalize the metric space notion of boundedness
Any continuous map f 9k IR achieves its maximum
compact
Def X a topological space A collection of open sets Ui ie
is an
opencover if Ui X
Def X is compact if every open cover contains a finite
subcollection that also covers X
i e F A E Ui finite sit Ut X
ies
f
subcollection
of opensets
Exi IR is not compact The covering IR Uz n ht2 has
no finite subcover
O l is not compact Yu I has no finite sub cover
Exe X o U Yul heh is compact
If X U Ui covers X there is some Uj containing O
Uj contains all pbut finitely many of the remaining points
so choose an open neighborhood Of each of the
remaining
points Th se and Ug form a finite subcover
Some basic results about compactspaces
This If A is compact and fA X continuous then
f A is compact
UUi f Ui
PI let be an open cover of f A Then
is an open coven of A which has a finite subcover
U f Uj
jeJefinite
Uj f f Uj so the sets Uj jet cover f A D
Them 0,1 is compact
PI let Ui be an open coven of Co D
Let A xc 0,1 I7 a finite subcover of Co x
Clearly OEA so At 4
We'll show A is open and closed Since 0,1 is connected this
will imply A 0,1
If a cover works for CO X then Xc Ui for some i so
Br x Elli some r O
So xeB x C A so A is open
To show A is closed suppose x is a limit point of A
XE Ui some i so x c Br E Ui some r
0 x 42 and.imits a finite cover so if we add Ui to the
finite cover we finite
get a cover of Co X D
You may recall from analysis that A C IR is compact A is closed and
bounded
In general how do closed sets relate to compact sets
Thin tf X is compact then
any closed A EX
is compact
PI let 1 be an open covering of A by sets open
in X Then AU X A is an open covering of X
Let B be a finite subcovering Then B covers A
after possibly removing X A so B is a finite subcover D
Are all compact setsclosed Sadly not ingeneral
E XE IR w the co finite topology is always compact
by a HW problem but not always closed
In Hausdorff spaces it is truethough
The tf X is a Hausdorff then compact set
space any
KEX is closed
PI We'll show Xlk is open
let x c Xlk For every yet there are disjoint neighborhoods
Uys y and Vy2x K is covered by Uy
Thus K has a finite subcover U U UzU VUK w
say
corresponding neighborhoods V Vz Vk of x w Ui disjointfrom
Vi Then V V h Nr is a finite intersection of open sets and
is thus an open neighborhood of
Moreover V is disjoint from U U Ulla so x c VE X K
Thus Xlk is open so K is closed D
It's much easier to check if mapsbetween compactHausdorff spaces
are homeomorphisms
Thmi A continuous bijection f X Y between compact Hausdorff
Spaces is a homeomorphism
PI We want to show that the images of open sets are open
so it suffices toshow the imagesof closed sets are closed
let A EX be closed Then A is compact Thus f A is compact
f A is closed since Y is Hausdorff D
Norte This onlyworks for compactHausdorff spaces
Consider the bijection
Hausdorff compact
notcompact Hausdorff
b b
f o 1 S defined x cos 2ITx Sin 21Tx
This is continuous but certainly not a homeomorphism
ProductsotcompactspacesI
This If X and Y are compact thenX Y is compact
PI let it be an open cover of X Y Eachelement of it
is the union of basis elements of the form U V where
UE X and EY are open
If those basiselements have a finite subcover then it does as well
by replacing U V w the open set in it in which it is contained
Thus we can assume it consists of basiselements of form Ui Vi
For x cX 3 4 is homeomorphic to 4 see HW so x Y is compact
So it has a finite subcover of the form i
Ui Vi
Where xc Ui Vi
ThenWIN Ui is a neighborhood of X and Ui Vi is a finite
cover of W Y
Thus for each xe X There's a finite subcover of Wx Y But the
Wx cover X so finitely many of them coven X
Thus F m xmeX st W U UW m
X and there is a finite
subcovering at it that covers Wx Y The union of these is
a finite subcovering of Xx Y D
Cori The product of finitely manycompact spaces is compact
In fact the product of infinitely many compact spaces given
the product topology is compact This is a deepresult called
Tychonoff's Theorem Proof requires Axiom of choice see section
37 of Munkresforproof
Uncountability of R
We can use properties of compact Hausdorff spacesto
give a slick
proof that IR is uncountable
Def X a topological
space x c X is an isolated point if x is open
in X
Isolated
points
theorem If X is a nonempty compact Hausdorff space
with no isolated points thin X is uncountable
0
PI Elgin If U C X open x cX F V open set V EU
and x T
I
I
1 x I
y
u
PImonfflamim choose ye U s t xt y This is possible since U
can't be a one point set
We can find W 2x andWy y disjoint neighborhoods Then
set Wy AU xy T since Wx is open and disjoint fromV
t y
o
V
I b lU
I it x i
s
et
k i l
Wii v
Wy
Now we use the claim to show uncountability
let f ft X be some function
By the claim set U X and find V EX sit T doesn't contain
f i
For n 1 apply the claim to f n and U Vn 1
Then I I 3 are nonemptyclosed sets w f n In
If Xlvi X any finite subcover would leaveout some
nonempty Ti Thus Xlvi f X so f Vi 4
Take Xe Afi Xt f n for any n so f I X is Ket
surjective so X is uncountable D
Cer Every closedinterval in IR is uncountable
