Appendix D
Metric Spaces
For those readers not already familiar with the elementary properties of metric spaces and the notion of compactness, this appendix presents a sufficiently
detailed treatment for a reasonable understanding of this subject matter. However, for those who have already had some exposure to elementary point set
topology (or even a solid introduction to real analysis), then the material in
this appendix should serve as a useful review of some basic concepts. Besides,
any mathematics or physics student should become thoroughly familiar with all
of this material.
D.l
Definitions and Open Sets
Let S be any set. Thena functiond: S x S IRis saidto be a metric on S
for all r, A,z-€ S:
if it hasthe followingproperties
(M1) d(r,s) > 0;
(M2) d(r,s): 0 if andonly1fr : s;
(M3) d(r,a) : d(a,r);
(Ma) d@,a)+ d(a,z) > d(r, z).
The real number d(r,g) is called the distance between x and y, and the set S
together with a metric d is called a metric space (S,d).
As a simple example, let S : lR and let d(r,g) : lt - gl for all r,g € IR.
From the properties of the absolute value, conditions (M1)-(M3) should be
obvious, and (M4) follows by simply noting that
lr - z1: lr - a * a - z l < l" - u l+ la- t l.
For our purposes,we point out that given any normedvector space(% ll ' ll)
we may treat V as a metric spaceby defining
d(n , s ): ll* -s ll
101
APPENDIX D, METHIC SPACES
t02
for every r,g € V. Using Theorem ??, the reader should have no
showing that this does indeed define a metric space (y'd). In fact, it is
see that lR.' forms a metric space relative to the standard inner product
associated norm.
Given a metric space (X,d) and any real number r ) 0, the open
radius r and center re is the set Ba(zo, r) c X defined by
trouble
easy to
and its
ball of
B a @ s , r ) : {r € X : d ( r , r o ) < r }.
Since the metric d is usually understood, we will generally leave off the subscript
Such a set is frequently referred to as an r-ball.
d and simply write B(rs,r).
We say that a subset (l of X is open if, given any point r € U, there exists
r > 0 and an open ball B(r,r) such that B(r,r) c U.
Probably the most common example of an open set is the open unit disk D1
l ht
r
r
1
'
Dv
tn K- oeflneo
D t - - {( r , g ) e R 2 , *" + 9 2 < l Y .
We see that given any point ro € D1, we can find an open bail B(rg,r) C D1 by
choosing r : 1,-d(re, 0) (see the picture in the proof of the following theorem).
The set
D 2 : {( r , y ) e R ' 1 2 + y ' < L }
is not open becausethere is no open ball centered on any of the boundary points
12 + A2 : 1 that is contained entirely within Dz.
The fundamental characterizations of open sets are contained in the following
three theorems.
Theorem
set.
D.L.
Let (X,d)
be a metric space. Then anu open ball is an open
Proof. Let B(rs,r) be an open bail in X and let u be any point in B(rs,r).
We musL find a B(r, r') contained in B(rs. r).
B(16,r)
103
D.1. DEFINITIO-IVSAAID OPEAI SETS
Since d(r, rs) 1r, we deflne rt :
have d(y,r) < r' and hence
d( a, r il
r - d(r,ts)'
Then for any A e B(t,r')
< d( a, x ) * d ( r , r g ) 1 r '
which shows that 9 € B(rg,r).
we
I d(r,16) : r
Therefore B(t,r')
C B(ts,r).
!
Theorem D.2. Let (X,d,) be a metric space. Then
(i) Both X and, o are open sets.
(ii) The intersection of a finite number of open sets is open.
(i,ii) The uni,on of an arbitrarg number of open sets is open.
Proof. (i) X is clearlyopensince for anyr € X andr ) 0 wehave B(t,r) c X.
The statement tha| a is open is also automatically satisfied since for any r € a
(there are none) and r > 0, we again have B(r,r) c a.
(ii) Let {(I},i e 1be a finite collection of open sets in X. Suppose {4}
is empty. Then nt/i : X because a point is in the intersection of a collection
of sets if it belongs to each set in the collection, so if there are no sets in the
y i"
collection, then every point of X satisfies this requirement. Hence nUi :
open by (i).
Now assume {U.} It not empty and let U : )U;.. lf U : O then it is open
(i),
so assume U + a. Suppose r eU sothat r € U; for every i e 1. Then
by
there exists B(r,r1) C [J.; for each i, and since there are only a finite number of
the ri we may let r: min{rl}. It follows that
B( r , r )
C B(r,r;)
cUi
for every i, and hence B(r,r) c aui : U. In other words, we have found an
open ball centered at each point of U and contained in U, thus proving that U
is open.
(iii) Let {U,} b" an arbitrary collection of open sets. If {U,} it empty, then
a is open by (i). Now suppose {t/,} is not empty and r € t-)Ui.
U : uUr:
Then z € Ut for some i, and hence there exists B(r,r1) C Ut C UU' so that uU,
!
is open.
Notice that part (ii) ofthis theorem requires that the collection be finite. To
see the necessity of this, consider the infinite collection of intervals in JRgiven
for 1/-n <. oo. The intersection of these sets is the point {0}
by (-lln,Iln)
which is not open in IR.
In an arbitrary metric space the structure of the open sets can be very
complicated. However, the most general description of an open set is contained
in the following.
Theorem D.3.,4 subset U of a rnetric space (X,d') is open if and only if it is
the un'ion of open balls.
APPEI,{DIX D. METRIC SPACES
104
Prool. Assume U is the union of open bails. By Theorem D.l each open ball
is an open set, and hence U is open by Theorem D.2(iii). Conversely, let U be
an open subset of X. For each r € [/ there exists at least one B(t,r) C U so
that U,EyB(r,r)
C U. On the other hand each r € [/ is contained in at least
tl
ThereforeU -l,EuB(r,r).
B( r , r ) s o t hat ( l c u, € u B ( r , r ) .
As a passing remark, note that a set is never open in and of itself. Rather,
a set is open only with respect to a specific metric space containing it. For
example, the set of nurnbers [0, 1) is not open when considered as a subset of
the real line because any open interval about the point 0 contains points not in
[0, 1). However, if [0, 1) is considered to be the entire space X, then it is open
by Theorem D.2(i).
If t/ is an open subset of a rnetric space (X,d), then its complemett U' :
X - U is said to be closed. In other words, a set is closed if and only if its
coniplement is open. For exarnple, a mornents thought should convince you that
the srrbset of lR.zdefined by {(r,y) e R' , 12 + a2 < 1} is a closed set. The
closed ball of radius r centered at 16 is the set B[rs,r) defined in the obvious
way by
Bfro,rl:
{z e X : d ( r s , " ) < r }.
We leave it to the reader (see Exercise D.3) to prove the closed set analogue
of Theorem D.2. The important difference to realize is that the intersection of
an arbitrary number of closed sets is closed, wliile only the union of a finite
number of closed sets is ciosed.
lf (X,d) is a metric space and Y C X, then Y may be considered a metric
space in its own right with the same metric d used on X. In other words, if we
let dlY denote the metric d restricted to points in Y, then the space (y,dly) is
said to be a subspace of the metric space (X,d).
Theorem D.4. Let (X,d) be a metric space and (Y,dlY) a metri'c subspaceof
X. Then a subsetW cY i,s open i.nY (i.e., open wi,th respectto the metri'c
dlY)rf and only i,f W --Y )U whereU i.s open in X.
Proof. LetW cY
such that the set
be open in l'and
supposer €W.
B a l y ( r , r ) : {a e Y
Then there exists r ) 0
: dlY(r,y) < r}
is a subset of 142.But this is clearlv the same as the open set
B a Q r , r ) : {A e X : d ( r , A ) < r }
restricted to only those points g that are in Y. Another way of saying this is
that
Balv(r,r) : Ba(r,r) nY.
105
D.2. CONTINUITY
SinceI'Z : J,ew Baly(r, r), it followsthat (seeExerciseA.1.1(ii)) W : tI o Y
whereU : o,ew Ba(r, r) is open in X (by TheoremD.2(iii)).
On the other hand, lel W : Y n U where U is open in X, and suppose
r €W. Then r € U so thereexistsr > 0 with
Ba @ ,r) : {A e X : d(n,g) < r} c U .
Bu t } , n Ba (r,r) i s j u s t Ba l y (z ,r ) : { A eY : dl Y (r,A ) < r} c I4l w hi chshow s
!
rhat W is open in Y.
Note that all of our discussionon metric spacesalsoappliesto normedvector
spaceswhered(r, g) : ll" - gll. Becauseof this, we canequaliywell discussopen
sets in any normedspaceV.
D.2
Continuity
(Y,d,y) be a mapping (where dx is the metric on X etc.).
Let /: (X,dx)
We say that / is continuous at zs € X if, given any real number e ) 0, there
exists a real number d > 0 such that dx(/(r), f(ro)) { e for every r € X with
dy(r,r6) { d. Equivalently, / is continuous at ,ro if for each ,8(/(16),e) there
exists 8(16,d) such lhar f (B(rs,6)) c B(f (z6),e). (Note these open balls are
defined with respect to two different metrics since they are in different spaces.
We do not want to clutter the notation by adding subscripts to B as in d1s and
dr,.) In words, "if you tell me how close you wish to get to the number /(26),
then I will tell you how close r rnust be to re in order that /(z) be that close."
If / is defined on a subset S c X, then / is said to be continuous on S if / is
continuous at every point of S.
For example, consider the rnapping / : (0, m) C R. + (0, oo) c lR defined
For any 16 € (0,oo) we have (using the absolute value as our
bV /(") :11r.
metric)
rko),_lj
tr(,)_
;l_+#
If r is such that lr - rol < d, then we see that
i:
1f@)-/ (ro )l< r1
l/.r0 |
d
.r10
< d means
In particular, if we choose6 < rof2, then r > rol2 (since lr-"ol
re - d ( r I rs*d), and hence d/(rre) < 26lrfi. Therefore, given any e ) 0,
if we pick 6 -_ :min{rof2,er2olZ} then we will have lf (") - /("0)l < e.
Fortunately one can usually tell by inspection (e.g., by drawing a picture if
necessary) whether or not a particular function is continuous without resorting
to clever calculations. The general definition is a powerful technique for proving theorems about classesof continuous functions satisfying given properties.
N{oreover, there is an intrinsic way to characterize continuous mappings that is
of tire utmost importance.
APPENDIX D. METHIC SPACES
106
(Y,dv). Then f is continuousif and'onla i'f
Theorem D.5. Let f t (X,dx)
U 'i,nY.
sets
all
open
in
X
is
open
for
f-'(U)
Prool. Suppose/ is continuousand U is an open subsetof Y' If r e f-t(U),
then /(r) € U so there exists B(f(t),e) C U (sinceU is open)' But the
continuity of / then impliesthere exists B(r,6) suchthat
f (B (r,5)) c B (J@ ),e) c U .
ThereforeB(r,5) is an r-ball (actually d-ball in this case)centeredon e and
-t(U) is open.
containedin /-1(U), and hencef
Conversely,assumef -t(U) is open wheneverU is, and let c € X be arbitrary. Then the open ball B(/(r),e) is an open set, so its inverseimage is an
open set containingr. Thereforethere existsan open ball B(r, d) containedin
this inverseimage,and it clearly has the property that f (B(r,6)) c B(f (r),e),
!
hence proving that / is continuous
Corollary. If f : (X,tty) -'-+(Y,d,y), then ! i,s continuous if and' only il f-t(F)
'i,sclosed,in X wheneaer F is closed i'n Y .
Proof. It was shown in Exercise A.2.1 that if A cY, then /-1(,4") : f-|(A)c'
for someopenset U C Y, andso
Ther ef or e, if F CY i s c l o s e d , t h e n F : U "
-1(U)c must be closedif and onlv if /
-t(U"):
:
D.5,
Theorem
by
f
f
/-1(I')
!
is continuous.
Note that if f : X + Y is continuous and [/ C Y is open, then /-1(U) is
open, but if A c X is open, then it is not necessarily true that /(A) is open' As
a simple example, consider the function / : IR + lR2 defined bv /(e) : (r,t2)'
It should be clear that the open ball t/ c lR2 shown below is an open set whose
inverse image is an open interval on lR U {A} (since some points of U are not
the image under / of any point in IR), but that the image under / of an open
intervalls part of the paiabola a : 12 which is not open as a subset of IR2'
f
- Or
107
D.3. COMPACTNESS
D.3
Compactness
Now supposethat (X, d) is a metric space,and let {U,} be a collectionof open
subsetsof X suchthat UU, : X. Sucha collectionof subsetsis calledan open
cover of X. A subcollection{7i} of the collection{U1} is said to be an open
subcover of X if avj : X. A space(X,d) is said to be compact if euery
open coverhas a finite subcover.Similarly,given a subsetA C X, a collection
be an
{U;} of open subsetsof X with the property that ,4 C Ut/, is said to
open cover of A. Equivalently,the collection{U,} of open subsetsof X is an
open coverof ,4 in X if the collection{Ui n A) is an open coverof the subset
A in the metric dl,4 (i.e., in the subspace,4).We then say that A is compact
if every open cover of ,4 has a finite subcover or, equivalently, A is compact
if the subspace,4 is compact. While this is not a particularly easyconceptto
thoroughly understandand appreciatewithout detailed study, its importance
to us is basedon the following two examples.
Example D.1. Consider the subset ,4 : (0' 1) of the real line R' We define
the collection{Ut,Uz,. . .} of open setsby
U n = (l f 2" + 1,1* l f 2" + \.
Thus {/r : (L14,314),fJz: (1/8, 718)etc.The collection{U.} clearlycovers/
since for any r € (0, 1) we can always find some (/, such that r € Ur. However,
A is not compact since given any finite number of the U, there exists e > 0 (so
that e € (0, 1)) which is not in any of the U,.
Example D.2. Let us show that the suhpaee {0' 1] of the real line is compact.
This is sometimescalled the Heine-Bnrel theorem, although we shall prove a
more general version below.
First note that the points 0 and 1 which are included in the subspace[0, 1]
are not in the set (0, 1) discussedin the previous example. However,if we have
positive real numbers o and b with a S b < 1, then the collection {U"} defined
abovetogether with the sets [0,o) and (b,1Jdoes indeed form an open cover for
[0, 1l (the sets [0,a) and (b, 1] are open by Theorem D'4)' It should be clear
ihut gi'u"tr the sets [0, a) and (b, 1] we can now choosea finite cover of [0' 1] by
including these sets along with a finite number af t'he Un. To prove that [0, 1]
is compact however, we must show that eny apen cover has a finite subcover.
Somewhat more generally let {Or} be any open cover of the interval [o, b]
in IR. Define
A : {a e [a, b] : [o, r] is coveredby a finite number of the O'].
We see that .4 f a sinceclearly a € A, and furthermore .4 is bounded above
by b. Therefore (by the Archimedeanaxiom) ,4 must have a least upper bound
APPEI{DIX D. METRIC SPACES
108
A < b. If ,4 k to be compact, then rve must have b e A. W will show
rn: s1np
that this is true by first proving that m e A, and ihen that m: b.
Since {O"} covers[o,b] andm e [o,b],_it follows-that ry Q O^ for some
O* € {O*}. Now, O- is an open subset of la, {, and hencethere are points in
O^ that are less thar. m, and points in O* that are greater than rn.
o^
a
n n 'r u
o
Becanse rn : sup.A, there is an z ( rn with r € o^such that the interval [4, r]
is covered by a finite number of the on, while [r, rzl] is covered try the single set
O-. Therefore fa,ml is covered by a fi.nite number of open sets so that' m € A.
Now suppose m I b. Then there is a point g with rn < A < b such that
ls covered
l*,g1 c O*. But we just showed that rn € /, so the interval fa,ml
by finitely many O, while [rn,gr] is covered by O*. Therefore g € '4 which
contradicts the definition of rn, and hence we must have m - b.
Arr important property of metric spaces is the following. Given two distinct
points r, A € X, there exist disjoint open sets U andV in X such that r €U
and g € V. That this does indeed hold for metric spaces is easy to prove by
considering open balls of radius d(r,y) l2 centered on each of the points r and
We sometimes refer to a
g. This property is called the Hausdorff property.
metric space as a "Hausdorff space" if we wish to emphasize this property.
The following theorems describe some of the most fundamental properties
of compact spaces.
Theorem
D.6. Any closeil subset of a cornpact space'is compact.
Proof . Let F c X be a closed subset of a compact space X. lf {Ui) is any open
cover of F, then (uU,) u F" is an open cover of X. Since X is compact, we
may select a finite subcover by choosing F" along with a finite number of the
t/i. But then lr is covered by this finite subcollection of theUi, and hence F is
I
compacl.
Theorem
D.7. Any compact subset of a metri'c space'is closed.
Proof. Let ,F be a compact subset of a metric space X. We will show that F"
is open. Fix any :r € Fc and suppose A € F. Since X is Hausdorff, there exist
a. As the point
open s et s Uo and V, s u c h t h a t r € U o , A € V o a n d U a a v a :
y varies over F, we see that {Vo : A € F} is an open cover for 'F. Since lr
is compact a finite number, say Vor,. . . ,Va-, will cover F. Corresponding to
109
D,3, COMPACTNESS
eachVaithere is aflon, arrdwe let U : );Uuu andV : U1Voo.By construction
x € (J, F c V and U n V : a. But then U is an open set containing u such
n
a, and henceF" is open.
that t/ )F:
Theorem D.8. Let (X,dx) be a compactspaceand let f be a continuous
functi,onfrom X onto a space(Y,dv). ThenY 'iscompact.
Proof. Let {U,} be any open coverof Y. Since/ is continuous,each/-1(t4) it
open in X, and hence{/-1(U,)} is an open coverfor X. But X is compactso a
finite numberof the /-1(Q), tuy {"f-t(I/n,),. . . , f -t(Ut )} coverX. Therefore
!
{Uor,...,Ur,} formsafinite subcoverfor Y, and henceY is compact.
Theorem D.9. Let {K1) be a collection of compactsubsetsof a metric space
X such that the 'i,ntersectionoJ eaeryfinite subcolleetionof {K1) is nonempty.
Then iK1 is nonempty.
Proof. Fix any K1 e {Kr] and assumeK1 ) (o,;s1K1): s. We will show this
leadsto a contradiction.First note that by our assumptionwe have(l;11K;) c
Kl, and hencefrom Example A.1 and Theorem A'1 we seethat
K1 C ( )i a1K i )' : U i + rK f .
T h u s { K;" } , i .f l ,i s a n o p e n c o verofK 1. B utK l i scompactsothatafi ni te
number of thesesets,say Kfr,.. ., Kf,, coverK1. Then
ol c (,x:rKi)
whichimplie
"
: Q!:'K1)'
K1n(.X=$L):
a.
However, this contradicts the hypothesis of the theorem.
n
Corollary. II {K^} i,s a sequenceof nonempty compact sets such that' Kn )
Kn+r, thennK. + a.
Proof. Th\s is an obvious special caseof Theorem D'9.
tr
As a particular application of this corollary we seethat if {-I'} is a nonempty
sequenceof intervals lon,bnl C IR such that 1' ) In+7t then t^t1rI a. Whlle
this result is basedon the fact that each1r is compact (ExampleD.2), we may
alsoprovethis directly as follows.If In: lon,bnl,we let S -- {o"}. Then S I a
and is boundedaboveby b1. By the Archimedeanaxiom, we let r : supS. For
any m,n € Z+ we have an I a*+n I b*+n -( b- so that z 4 b^ fot all m-
110
APPENDIX D. METRIC SPACES
Sincea- -{ r for ail m, we rnusthaver € la*,b^l : I* for eachm : 1,2,...
so that )1. + o. We now showthis result hoids in lR' as well.
Supposeo,b € lRnwhereat ( b' for eachi : I,...,fl. By an n-cell we mear
the set of ali points o € IRnsuch that ai < ri < bi for every i. In other words,
an n-cell is just an n-dimensionalrectangle.
of n-cells such that Iu ) Ix+t. Then
Theorem D.IO. Let {Ip} be a se(luence
nlp I o.
Proof. For eachk: 1,2,...the n-cell16 consistsof all pointsn € Rn with the
forevery l < i < n.sow el et 1i : [ni -.bi .l .N ow .
p ro p e rl y th a t a i < ri< bi
:
.
n
the
sequence
1,.. ,
for eachi
{1i} satisfiesthe hypothesesof the corollary
to TheoremD.9. Hencefor eachi : I,.. . , n thereexistsz' e fai, bi] for every
(r' ,...,2n) € R n, w e seethat z € 16 for every
If w e d efi ne
k :1 ,2 ,....
":
n
k :I,2 ,....
Theorem D.LL. Eueryn-cell is compact,
Proof. Lel 1 be an n-cell as defined above, and set d : [Dl:r(ao - or)tf'''.
Then if r,g € I we have li" - Stll< d (seeExampie ??). Let {U,} be any open
cover of 1 and assurne that it contains no finite subcover' We will show this
leads t o a t ' ont r adic l i o n .
Then we have 2" n-cells Qi
Let cj : (ai +U)12 for each j : I,...,n.
Since l has no
defined by the intervals [aj,C] and [c',ts] such that UQ;:1.
finite subcover, at least one of the Q;, which we call 11, can not be covered by
any finite number of the Ur. Next we subdivide /1 into another 2' n-cells and
continue in the same nlanner. We thus obtain a sequence {1"} of n-cells with
the following properl ies:
( a) 1' l I r ) I z
f .. . ;
(b) f. is not covered by any finite subcollection of the Un ;
k ) , , A € 1o im p l i e s l l " - g l l < 2 - o 6 .
BV (a) and Theorern D.10 there exists z € n/o, and since {U,} covers 1,
we must have z € Un for some k. Now, U;. is an open set in the metric space
IR', so there exists e ) 0 such that llz - yll < e implies a € Ur. If we choose
n sufficiently large that 2-od < e (that this can be done follows from Theorenr
!
A.3), then (c) inipiies Io Cflp which contradicts (b).
We are now in a position to prove the generalized Heine-Borel theorem.
Before doing so however, we first prove a simple result which is sometimes
taken a"sthe definition of a compact set. By way of terminology, any open set
[/ containing a point r is said to be a neighborhood of ;r, and the set U - {z}
is called a deleted neighborhood of z. We say that a point r e (X,d) is an
D.3. COMPACTT{ESS
111
accumulation point of A C X if every deletedneighborhoodof r intersects
A.
Theorem D.L2. Any infini'te subset A of a compact set K has a point of
i,n K.
accumulati,on
Prool. Supposeevery point e € K is not an accumulation point of ,4. Then
there exists a neighborhood u" of r such that u, contains at most a single
point of A, namelyr itself if r e A. Then clearlyno finite subcollectionof {u"}
I
covers,4 C K so thal K can not possibly be compact.
if and onlg if A
Theorem D. 13.,4 subsetA of a metric space(X, d) is closed,
contains all of its accumulationpoi'nts.
Proof. First supposethat,4 is closed.Let r € X be an accumulationpoint of
,4 and assumethat r ( A. Then r € A'which is an open set containingr that
does not intersect A, and hencecontradicts the fact that z is an accumulation
point of ,4. Thereforer must be an elementof A.
Conversely,suppose,4 containsall of its accumulationpoints. We show that
point of -4' then there
-4" is open. lf r € A and henceis not an accumulation
existsan opensel U conlainingr suchthal AaU : a. BuL then r €U C A'
tr
which implies A" is open.
we say that a subsetA c IR' is bounded if it can be enclosedin somen-cell.
The equivalenceof (i) and (ii) in the next theoremis calledthe (generalized)
of (i) and (iii) is a generalversion
Heine-Borel theorem, while the equivalence
of the Bolzano-Weierstrass theorem.
Theorem D.LA. Let A be a subsetof it**n.Then the followi'ng three properties
are equiualent:
(i) A is closedand bounded.
(ii) A is compact.
(lll) Eaery infi,ni,tesubsetof A has a poi'nt of accumulationin A.
ProoJ. (1) + (ii): If (i) holds, then ,4 can be enclosedby somen-cell which is
compactby TheoremD.11. But then ,4 is compactby TheoremD'6.
(ii) + (iii): This followsfrom TheoremD.12.
(iii) + (i): We assumethat every inflnite subsetof ,4 has an accumulation
point in A. Let us first show that ,4 must be bounded. If ,4 is not bounded, then
we can find an t7" € A suchthat ll"rll > k.
for eachpositiveintegerk:7,2,...
but containsno point of accumulationin
infinite
is
clearly
Then the set {16}
lR', so it certainly contains none in ,4. HenceA must be bounded.
172
APPENDIX D. \,TETRIC SPACES
We now show that ,4 must be closed. Again assume the contrary. Then
there exists 16 € IR' which is an accumulation point of ,4 but which does not
belong to ,4 (Theorem D.13). This means that for each k : I,2,... there exists
16 € Asuch that ll"r < 7lk. The set,9 : {r4} is then an infinite subset
"oll
of ,4 with r0 as an accumulation point. Since ro 4 A, we will be finished if
we can show that ,9 has no accutnulation point in ,4 (because the assumption
that ,4 is not closed then leads to a contradiction with the property described
in ( iii) ) .
First note that if a,b € IR', then Example ?? shows us that
lla+, b ll- llo- (-b )ll> ll" ll- lla .ll
Using this result, if g is any point of IR' other than 16 we have
rp - yl : l l x6 eo * ;ro - 9l l
2 llzo- s,ll- ll"r - "oll
> llro al - 1lk.
No matter how large (or srnall) ilro - gll is, we can alwaysfind a ks e Z+
suchthat llk < + ll"o - gll for everyfr > k6 (this is just TheoremA.3). Hence
o
l1' n- slltlllr
z
al1
for every k > ko. This shows that 3rcan not possibly be an accurnt-rlationpoint of
{"r} : S (because the open ball of radius } llro - gl] centered at 9 can contain
n
at most a finite number of elernents of S).
We remark that the irnplication "(i) implies (ii)" in this theorem is not true
in an arbitrary metric space (see Exercise D.5).
Let / be a rnapping from a set ,4 into lR'. Then / is said to be bounded
if there exists a real nurnber ,{f such that ll/(r)ll <i Il,[ for a]l r € A. If f is a
continuous mapping from a compact space X into IR', then /(X) is cornpact
(Theorem D.8) and hence closed and bounded (Theorem D.14). Thus we see
that any continuous function frorn a cornpact set into IR' is bounded. On the
1/r is not
R defined bV f @):
other hand, note that the function / : lR
bounded on the interval (0, 1). We also see that the function 9 : lR. lR defined
bv s@) - r for r € 10,1) never attains a maximum value, although it gets
arbitrarily close to 1. Note that both f and g are defined on non-compact sets.
We now show that a continuous function defined on a compact space takes
on its maximurn and minimum values at some point of the space.
Theorem D.LI. Let f be a conti,nuousreal-ualuedfuncti,on defined on a compact spaceX, and,letM:
supseX f @) andm:infsex
f @). Th,enthere eri,st
po'intsp,q € X such that f (p): M and f (q) : *.
D.3. COMPACTI]ESS
119
Proof. The abovediscussionshowedthat /(X) is a closedand boundedsubset
of lR..Henceby the Archimedeanaxiom, /(X) must havea sup and an inf. Let
M :s try f@ ). T h i s me a n sth a t gi vene) 0 thereexi stsz€ X suchthat
M-e1f(" )< M
(or elseM would not be the leastupper bound of /(X)). This just saysthat any
open ball centeredon /'1,1
intersects/(X), and henceM is an accumulationpoint
of /(X). But /(X) is closedso that TheoremD.13 tells us that M e f (X).
In other words, there exists p € X such that ttt : f(p). The proof for the
minimum is identical.
!
As an applicationof theseideas,we now prove the FundamentalTheorem
of Algebra.
Theorem D.16 (Fundamental Theorem of Algebra). The complernumber
fi,eldC, i,s algebrai,callyclosed.
Proof. Considerthe non-constant polvnomial
f (" ) : a o + a1z+ " ' + anz" e C [zl
where a. I 0. Recall that we view C as the set IR x IR: IR2,and let /i be any
(finite) real number. Then the absolutevalue function l/l : C + R. that takes
an y z € C to th e re a l n u mb e rl /( z)l i s cont,i nuous
on l he cl osedbal l B [0. R l of
radius ,R centeredat the origin. But B[0, R] is compact(TheoremD.14) so that
l/(z)l takes its minimum value at somepoint on the ball (TheoremD.15). On
t h e o th e r h a n d .i f w e w ri l e /(:) i n the form
f\z ) : o ,r" ( J :-
\ dr z "
1 -" -+ ...+
anz"-'
o,-'
anZ
+ r)
/
we see that l/(z)l becomes arbitrarily iarge as z becomes iarge. To be precise,
given any real C > 0 there exists -R ) 0 such that lzl > R implies lf(t)l> C.
We now combine these two facts as follows. Let z1 be arbitrary, and define
C:
t her e ex is t s Re ) 0 s u c h t h a t l / ( z ) l > l f ( " r ) l f o r a l l z e C
lf ( t ) l. Then
such that lz - zrl > R6 (i.e., for all z outside Blzl,Rsl). Since B[21,R6] is
compact, there exists a point zs € Blz1,R6] such that l/(zs)l < l/(z)l for all
z e Blz1, Rs].
Ro
ZO
Blz1, R6l
APPENDIX D. METHTC SPACES
tt4
In particular,l/(ro)l S l/(zr)l and hencewe seethat l/(zo)l < l/(z)l for all
z € C. In other words, z6 is an absoluteminimum of l/1. We claim that
f(zd : o.
To show f ("i :0, we assumethat /(ze) I 0 and arrive at a contradiction.
By a suitable choiceof constantsci we may write / in the form
... -t c,(z - ro)n.
f (r) : co+ q(z - ro)*
If f (zi l0 then cs: f (zs) 10. By assumptiondeg"f > 1, so we let rn be
the smallest integer greater than 0 such that c* * 0. Defining the new variable
w: z - 20, we may definethe polynomialfunction I by
I Q) : s(w) : cs I c*w* + w*+t h(w)
for some polynomial h.
Now let ri.r1be a complex number such that wT : -colc*
valuesof w : \wt for real ,\ with 0 < .\ < 1. Then
and consider all
c* ' tD * :c* \^w !:-co\*
and hence
f (") :9 (^?r1) -- co \* "o + \*+r wf+r h(\ul)
: coIl - A ^ + A * + tw i + l c;1h()Tr.,1)].
But l € [0,1] which is compact,and hence lwi+1c;rn(Xtr.'1)l is a continuous
function deflnedon a compact set. Then the imageof this function is a compact
subsetof IR (TheoremD.8) and so is closedand bounded(TheoremD.14). This
means there exists a number B > 0 such that
l uri + 1c;1n(.1.r)l 3 B
for all ) € [0,1], and therefore(since0 < ) < 1 implies0 < )- < 1)
+ \ +1w?+1cola1.ra;1;l
lo(r.t)l : l"olIr ^m
S l"ol{lt - }-l+ sm+r
l*7+r";1n(.1'u1)l}
< l"ol(t B).
+
^* ^*+L
Now recall that lcsl : l/(ro)l < l/(r)l for all z € C. If we can show that
o< 1-^* + ^* + 1
B<r
for sufficientlysmall ,\ with 0 < ) < 1, then we will haveshown that l/(z)l :
1g(,\tlr)l < lcol, u contradiction.But it is obviousthat ,\ can be chosenso that
+ ^n+ 1 B ( l i sthesameas
0 < 1 - S m1 \m+ rB . A ndtorequi rethat 1!
^m
for small enough ).
be
satisfied
certainly
requiring that lB < 1 which can
D,3. COMPACTNESS
1i 5
Exercises
1. Show the absolute value function is continuouson lR.
2. Show the norm on a vector spaceI/ definesa metric on I/.
3. Let (X, d) be a metric space.Prove:
(a) Both X and Z are closedsets.
(b) The intersection of an arbitrary number of closedsets is closed.
(c) The union of a finite number of closedsets is closed.
4. Let A be the subset of [0, 1] consisting of all z e [0, 1] whose decimal
expansioncontains only the digits 4 and 7. Explain whether or not .4 is
countable,densein [0,1] or compact.
5. Show {x : llxll, < 1} is closedand bounded but not compact in the space
12(seeExample ??).
6. A metric space is said to be separable if it contains a countable dense
subset. Prove that IR' is separable. fHi,nt: Consider the set of all points
in lR' with rational coordinates.]