Metric Spaces (Bath Lecture Notes)

8/6/2019 Metric Spaces (Bath Lecture Notes)

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Brief Notes on Metric Spaces, MATH0041

1. Metrics and Norms

1.1 Definition: A non-empty set X with a function d : X x X --+ l R is a metric space ifa) d(x,x) = 0;b) d(x,y) >Oifx~y;c) d(x, y) = d (y , x );d) d(x, y) + d(y, z ) ~ d(x, z) for all x, y, z E X.Condition (d) is called the triangle inequality.1.2 Examples: lR n with d(x, y) = [x - yl, the Euclidean metric; any nonempty set X with d(x, y) = 1 ifx ~ y, the discrete metric; Z with the metric d(n, m) = k!l if pk divides (n - m) and pkH does not (p aprime), the p-adic metric.1.3 A norm on a real vector space V (possibly infinite-dimensional) is a function II . II : V --+ lR such thata) 11011= 0;b) Ilvll > 0 if v ~ 0;c) II'xvll = 1,Xlllvllif ,XE lR ;d) Ilv + wll ~ Ilvll + Ilwll1.4 Examples: the Euclidean norm Ilxll = [x ] on lRn ; the max-norm Ilxll = max(lxil) on lRn ; the max- orsup-norm IIIII = max{l/(t)11 t E [0,1]} on e[O, 1] (the set of continuous functions I :[0,1]--+ l R ) .1.5 Proposition. If V is a vector space and 1111is a norm then d(x,y) = Ilx - yll is a metric on V.Proof: We need to check 1.1 (a-d). 1.1(a) follows immediately from 1.3(a) and 1.1(b) from 1.3(b). For 1.1(c),notice that d(x, y) = Ilx - yll

= II(-l)(y - x)11= I-lilly - xii= Ily-xll= d(y,x).

For 1.1(d)d(x, z) = Ilx - zll

= II(x - y) + (y - z)11~ Ilx - yll + Ily - zll= d(x, y) + d(y, z).

1.6 Lemma. If x , y , z E X then Id (x , y) - d(y, z)1 ~ d(x, z)Proof: Use the triangle inequality twice:

d(x, y ) ~ d(x, z) + d (z , y ) and d(y, z ) ~ d(x, y) + d(x, z)1


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sod(x ,y) - d (y ,z) ~ d (x ,z ) and d (y,z ) - d (x ,y ) ~ d (x ,z ).

1.7 Definition. If Y ~ X then the subspace metric on Y is just the restriction of d to Y x Y.1.8 Definition. If X, dx and Y, dv are metric spaces then the product metric on X x Y is drr given by

d rr((x ,y ), (X l,y l) ) = m ax{dx(x , X l), dy (y , y l)} .It is a metric on X x Y. To check the triangle inequality (the other three conditions are immediate)

drr((XI, yd , (X 2 ' Y 2 )) + drr((X 2, Y 2), (X 3, Y3 ))= max(dX(xI, X2), d y(y I' Y2 )) + max(dx (X 2 ' X 3), dY(Y2 ' Y 3 ))~ m ax(dX(xI ,X2) + dX(X 2 ' X 3 ), dy (y I' Y 2 ) + dY(Y2,Y3))~ m ax(dX(xI ,X3) ,dY(YI ,Y3))= di: ( (Xl, yd , (X 3 , Y 3 )).

We can also do this for any finite product X = Xl X ... X Xn of metric spaces, setting drr(x, y)maxi dx, ( X i , Y i) .1.9 The distance between two nonempty subsets A and B of a metric space X is defined to be d(A , B) =inf { d ( a , b) I a E A , b E B} . If A = { a } is a set with just one point we write d ( a , B ) (the distance from a toB) instead of d ( {a } ,B ) .The distance function d(A , B) on sets is not a metric itself: it can be zero even though A - I - B. In fact itcan be zero even when An B = 0, for instance A = (-1,0) and B = (0,1) in R1.10 A subset A ~ X is said to be bounded ifthere exists M E lRsuch that d(x , y ) < M for all X , Y EA . Thediameter of A is defined to be sup{d(x, y) I X , Y EA }. A bounded metric space is one for which X itself isbounded.Bounded intervals in lRare bounded sets. A discrete metric space is bounded (take M = 1).1.11 The (open) ball of radius r > 0 around, or centred at, c E X is

Br(c) = { x E X I d (c, x ) < r}.

1.12 Every ball is bounded: in fact the diameter of Br(c) is at most 2r. Small balls around c are containedin bigger ones, that is, Br(c) ~ BR (C ) if R > r.

2. Sequences and convergence.

2.1 A sequence in a metric space X is a function f : N -+ X. More usually one expresses this by writingf (n) = a n and calling the sequence ( an ) .

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2.2 A sequence ( a n ) in X ten d s or converges to a E X if' t iE> 0 3N 'tin > N d (a n , a ) < E.

In this case we write a n -+ a as n -+ 00, or lim a n = a . The point a is called the limit of the sequence: wen--+ooshall see shortly that a sequence has at most one limit. A sequence that has a limit is said to be convergent .2.3 A subsequence of a sequence f : N -+ X is fog : N -+ X, where 9 : N -+ N is a strictly increasingfunction. Again one usually expresses this differently: a subsequence of ( a n ) is ( a n J with ni+l > ii; for alliE N.2.4 Proposition. In a metric space X, a sequence ( a n ) tends to a if and only if d (a n , a ) -+ 0 in RProof: d (a n , a ) -+ 0 means 'tiE > 0 3N 'tin > N I d ( an , a )1 < E. But d (a n , a ) ~ 0 so we can drop the modulussigns, and then we see the definition of convergence in 2.2.2.5 A sequence ( a n ) is said to be bounded if and only if { a n } is a bounded subset of X, that is,

2.6 Theorem. Let X be a metric space and let (a n) , (bn ) be sequences in X.a) if a n -+ a and bn -+ b then d (a n , bn ) -+ d (a , b ) ;b) the sequence ( a n ) has at most one limit;c) if ( a n ) is convergent then it is bounded;d) if ( a n J is a subsequence of ( a n ) and a n -+ a then a n i -+ a .Proof: (a) We calculate directly, using 1.6:

Id (a n , bn ) - d (a , b )1 = Id (a n , bn ) - d (a n , b ) + d (a n , b) - d (a , b )1~ Id (a n , bn ) - d (a n , b )1 + Id (a n , b) - d (a , b )1~ d (b , bn ) + d (a n , a )-+ 0 as n -+ 00.

(b) If a n -+ a and a n -+ b then put bn = a n in part (a). Then 0 = d (a n , bn ) -+ d (a , b) so d (a , b) = 0 so a = b .(c) Choose E > 0: it doesn't matter what E is at all, so let us take E = 1. Then if a n -+ a we can find N E Nsuch that d (a n , a ) < 1 for n > N. Now take

M = max{max d (a i' a j) , 1 + max d (a i' a ), 2}.2,J~N 2~NI claim that for any m, n E N d (am , a n ) ~ M . Ifm and n are both less than or equal to N then d (am , a n ) ~max d (a i,a j) ~ M . Ifm ~ N < n then d (a m , a n ) ~ d (a m ,a )+d (a , a n ) ~ l+m a xd (a i,a ) ~ M . Ifm,n > N2,J~N 2~Nthen d (a m , a n ) ~ d (am , a ) + d (a , a n ) ~ 2 ~ M .(d) We must show that given E > 0 there is an N such that if j> N then d ( an j , a ) < E. For this E there isan N' such that if n > N' then d (a n , a ) < E: but then if j> N' we have n j > N' so d (a nj, a ) < E. S O all wehave to do is take N = N'.

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2.7 We shall show that a sequence in ]R k converges in the Euclidean metric if and only if each sequence ofcoefficients converges. Take a sequence (an) with an = (a~l), ... , a~k)). Then an -+ a = (a(l), ... ,a(k)) ifand only if

Iclaim that this is the same as saying that a~) -+ a(i) for each i= 1,2, ... , k. If lLJi(a~) - a(i))2 < 1 0 thenL:i(a~) - a(i))2 < 1 0 2 so for each iwe have (a~) - a(i))2 < 1 0 2 , and then la~) - a(i) I < E . So

Conversely suppose a~) -+ a(i) for each i. Then, given 1 0 > 0, we can choose N so large that la~) -a(i) I < E / V kfor all iif n > N. If we do that then as long as n > N we have

2.8 If d is the discrete metric on X then an -+ a if and only if there is an N such that an = a for n > N.This is because if 1 0 ~ 1 then d(an, a) < 1 0 implies an = a.2.9 Proposition: Ify ~X, a E Y and (an) is a sequence in Y, then an -+ a in Y with the subspace metricif and only if an -+ a in X. IfXI, ... , Xk are metric spaces and X = Xl X ... X Xk with the product metric,then the sequence (an) in X tends to a E X if and only if a~) -+ a(i) in Xi for each i.Proof: an -+ a in Y means dy(an,a) -+ 0 as n -+ 00. But dy(an,a) = d(an,a) so dy(an,a) -+ 0 ifand only if d(an, a) -+ O. Similarly, an -+ a means drr(an,a) -+ 0, that is maxidxi(a~),a(i)) -+ O. Butmaxi dx, (a~) , a(i)) -+ 0 if and only if dx, (a~) , a(i)) -+ 0 for each i.2.10 The max norm 111100on e[O, 1] is defined by 11/1100= max I/(t)l. Of course we can replace 0 and 1 by099any a, b ]R with a ~ b. More generally can define a norm II . Ilinlty, called the sup norm or the uniformnorm, on the set of bounded and continuous (or even just bounded), real-valued functions on any interval I,closed or not. It is defined by II Ilinlty = sup I/(t)l.tElWe need to check that 111100eally is a norm. This means checking 1.3(a)-(d). First (a): 1101100 max(O) = o .tElThen (b): if I is not the zero function then there is some TE l such that I (T ) - I - 0, and 11/1100= max I/(t)1 ~tElI/(T)I > O. For (c), note that IAf(t)1 = IAII/(t)1 so

IIAflloo = max IAf(t)1tEl= IAImax I/(t)1

tEl= IAlll/lioo.Finally, the triangle inequality, (d):

111+ glloo = max I/(t) + g(t)1tEl~ max(l/(t)1 + Ig(t)1)tEl~ max I/(t)1 + max Ig(t)1tEl tEl= 11/1100+ Ilglloo.

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I I 1 1 0 0 is called the uniform norm because In -+ I in this normed space if and only if In tends uniformly toI in the usual sense.2.11 Two metrics d l , d2 on the same set X are said to be equivalent if there are positive constants C l, C 2 > 0such that

The idea here is that if two metrics are equivalent they will give us the same notion of convergence: see 2.13.It's not quite true that if two metrics give the same notion of convergence they are equivalent: however,equivalence has the virtue of being easy to check. In any case examples where two inequivalent metrics givethe same notion of convergence are usually artificial.In the same way, one says that two norms I I . 1 1 1 and I I . 1 1 2 on a real vector space X are equivalent if thereare positive constants C l, C 2 > 0 such that for all x E X

To justify the use of the word "equivalent" we need to show that equivalence of metrics (resp. norms) isindeed an equivalence relation. We also need to show that the two uses of the word are compatible. Let usdo that first.Lemma. Two norms on a vector space X are equivalent if and only if the corresponding metrics areequivalent.Proof: If the metrics d l (x, y) = I l x - y l l l and d2 (x, y) = I l x - Y l 1 2 are equivalent then

and in particular if we take y = 0 we see that the norms are equivalent. Conversely, if I I . 1 1 1 and I I . 1 1 2 areequivalent we need only to write this down for the vector x - y to get the statement that the metrics areequivalent.In view of this we need only show that equivalence of metrics on a given set X is an equivalence relation:equivalence of norms is just a special case of that.Proposition. Equivalence of metrics on a given non-empty set X is an equivalence relation.Proof: Reflexivity: take C l = C 2 = l.Symmetry: if ds is equivalent to d2 then, for all X ,y E X, cld l(x ,y ) ~ d2 (x , y) so, since ci > 0, d l(x ,y ) ~~d2(X ,y) . Similarly d l(x ,y ) ~ ~ d2 (X ,y ) so d2 is equivalent to dl.Transitivity: if cld l(x ,y ) ~ d2 (x ,y ) ~ c2d2 (x , y) and C~d2 (X ,y ) ~ d3 (x ,y ) ~ C~d 2 (X ,y ) for all X ,y E X, sod s is equivalent to d a and d a is equivalent to d 3, then

and hence dl is equivalent to d3.

2.12 Proposition. The norms I I . I l k and I I 1 1 0 0 on ]R n are equivalent, for any k. In particular the differentI I . I l k are all equivalent to each other.

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Ilxll~ = (max IXil)k~LIXilk = Ilxll~~ nmaxlxilk= n(llxlloo)k

and, taking kth roots

Since 11111orresponds to the product metric and 11112orresponds to the Euclidean metric this, together with2.13 below, shows that convergence in the Euclidean metric is the same as convergence in each coordinate.We proved this by hand in 2.7.2.13 Theorem. If dl and d2 are equivalent metrics on a non-empty set X then an -+ a in (X, dd if andonly if an -+ a in (X, d2).Proof: By symmetry it is enough to show this in one direction only. Suppose an -+ a in (X, dd. We wantto show that an -+ a in (X,d2). Given E > 0, we choose N such that dl(an,a) < E/C2 if n > N. Thend2(an, a) ~ C2dl (an' a) < E, so we have what we want.2.14 We should check that what we are doing is not trivial: how do we know that there are any inequivalentpairs of metrics? In fact there are lots: here are two examples.X = J R , dl is the Euclidean metric, d2 is the discrete metric. We have seen that in the discrete metricthe only convergent sequences are the eventually constant sequences (see 2.8). On the other hand there areplenty of convergent sequences in the Euclidean metric that are not eventually constant, for instance an = ~.So dl and d2 cannot be equivalent, because if they were then according to 2.13 they ought to give the sameconvergent sequences.If X = C(O, 1) then 111100and 11111are inequivalent, where 11/111= J~ I/(t)ldt. (It is easy to check thatII . 111is a norm.) Consider the function

an (t ) = { 0 1 - nx if 0 < X ~ ~if~


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Suppose they were, so ' t/x,y E X cld l(x ,y ) ~ d 2( x ,y ) ~ c2d l(x , y ) . Choose x > 1/cl and y = x + 1. Thend 2( x , y ) = ~ > Cl = c ld l (x , y ) , a contradiction.Fortunately this doesn't matter much because it practically never happens. In particular (we aren't goingto prove this, though it's not especially hard) it doesn't happen for metrics given by norms.

3. Open and closed sets.

3.1 A subset U ~ X of a metric space ( X , d ) is said to be open in X if' t/x E U 3E > 0 B,(x ) ~ U.

Remember that B, (x ) = { y E X I d ( x , y ) < E}: it depends on X , and so does openness. It's not a propertyof U on its own.A subset Z ~ X is closed if X \ Z ~ X is open.Notice that closed does not mean "not open". It is quite possible for a set to be both open and closed in X,and usually most subsets of X are neither open nor closed.3.2 Examples. X ~ X is open: you may take Eto be anything. 0 ~ X is also open: this is true accordingto the definition above, because of a logical quibble about the meaning of't/x E 0 , but if you prefer you maymodify youir definition so that it says explicitly that 0 ~ X is open. As a result, 0 ~ X and X ~ X are alsoboth closed.Open balls are open, as you would hope. But this is not quite trivial to prove. Take U = Br (c) ~ X andsuppose x E U. Then d ( x , c) < r. Choose E= r - d ( x , c) >, which is positive. I claim that B ,(x ) ~ U.Suppose y E B, (x ) : then

d ( y , c) ~ d (y , x ) + d ( x , c) < E + d ( x , c) = rso Y E Br(c ) = U.If X is a discrete metric space then every subset of X is open, and therefore every subset of X is closed aswell. If U ~ X and X is discrete, suppose x E U: then B l ( X ) = { x } ~ U, so U is open.3.3 Theorem. Let X be a metric space.a) If A is nonempty and U ~ X is open for every 0: E A then U U is open in X.aEAnb) If Ul, ... ,Un are finitely many sets, each open in X, then n U, is open in X.i=lProof: (a) Suppose x E U Ua. Then x is in one of the Uas, say x E Uao. So for some E > 0 we haveaEAB ,(x ) ~ c.; ~ U o; so U u: is open.aEA aEAn(b) Suppose x E n Ui. Then x E U, for each i, so there is an e, > 0 such that B 'i (x ) CUi. Takei=l nE= min{El' ... ,En} . Then B ,(x ) ~ B 'i(X ) ~ U , for all i, so B,(x ) ~ n U, as required.i=lPart (b) breaks down if there are infinitely many of the U, because the minimum no longer exists (and theinfimum is no good as it might be zero). The intersection of all the open intervals (- ~, ~) C lRis {O } whichis not open in R

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From (a) and (b) it follows (by De Morgan's rules) thata') If A is nonempty and Za ~ X is closed for every 0: E A then n Za is closed in X.

aEAnb') If Zl, . . . ,Zn are finitely many sets, each closed in X , then U Zi is open in X .i=l

3.4 The collection of all the open subsets of a metric space X is called the topology of (X , d ) .Proposition. A nonempty set U ~ X is open if and only if U can be written as a union of open balls.Proof: If U can be written as a union of open balls then U is open by 3.3(a), since the open balls are openthemselves (3.2). Conversely, supose U is open. Then for each x E U there is a positive number Ex such thatB ,x (x ) ~ U. Consider W = U B ,x (x ). By definition W is a union of open balls: I claim W = U. IfwE W

xEUthen there is an x E U such that w E B,x (x ) ~ u, so W ~ U. If x E U then x E B ,Jx ) ~ w, so U ~ W.3.5 Theorem. If two metrics d , d ' on X are equivalent then the metric spaces (X , d ) and (X , d ') have thesame topology.Proof: Suppose U c ( X , d ) is open in the sense of d . Then there exists 10 > 0 such that B,(x ) ~ U . Weneed to show that U is also open in the sense of d'. Let us use B' rather than B to denote a ball in thesense of d': thus B~ (c ) = {y E X I d' (c , y) < r} . Then we must show that there exists 10 ' > 0 such thatB~ /(x ) ~ U. Since d and d ' are equivalent, there exist Cl,C2 > 0 such that c ld (x , y ) < d ' ( x , y ) < c2d ( x , y ) .Choose 10 ' = C l E. Then if y E B~,(x ) we have

d ( x , y ) < d ' (x , y ) /c l < E' lc l = 10so Y E B,(x ) ~ U , as required. Thus every open subset in the sense of d is also open in the sense of d ': theother way round follows by symmetry.If two metrics d and d ' on a set X yield the same topology we say they are t opo log ic a l ly e qu iv a l en t . We havejust proved that equivalent metrics are topologically equivalent. In order to prove this what we had to dowas check that every ball in the sense of d contained a ball in the sense of d ' (and vice versa).It is not true that if d and d ' are topologically equivalent then they are equivalent. The pathology in 2.14 isa counterexample: in that example the metrics are not equivalent but the topology is the same (every set isopen) in both cases.

3.6 If Y ~ X the interior of Y , written Int(Y) or yo, is defined byInt(Y) = U{U ~ X I U is open in X, U ~ Y}.

The c losure of Y , written Y, is defined byY = n {Z ~ X I Z is closed in X ,Z;2 Y } .

By 3.3, Int(Y) is open and Y is closed. You can think of Int(Y) as being the biggest open set containedin Y, and Y as being the smallest open set containing Y. Directly from the definitions it follows that ifU ~ Y is open in X (the qualification "in X") is essential then U ~ Int(Y), and similarly that if Z ;2 Y isclosed in X then Z ;2 Y.In X = JR , take Y = (0,1]. Then Int(Y) = (0,1) and Y = [0,1]. But Int(Y) can be very much smaller thanY: for instance, if X = J R 2 and Y is the x-axis then Int(Y) = 0 , because no ball is contained in Y.

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The boundary of a subset Y ~ X is denoted 8Y and is defined to be 8Y = Y \ lnt(Y).3.7 Theorem. Suppose X is a metric space and Y ~ X. Thena) lnt Y ~ Y ~ Y ;b) 8Y and Yare closed and lnt(Y) is open in X;c) lnt(Y) = Y if and only if Y is open, Y = Y if and only if Y is closed, and 8Y ~ Y if and only if Y isclosed;d) lnt(X \ Y) = X \ Y and X \ Y = X \ lnt(Y).Proof: (a) follows directly from the definitions of lnt(Y) and Y.(b) follows from the definitions and 3.3: notice that 8Y = Y n (X \ lnt(Y)) and is therefore the intersectionof two closed sets, so it is closed by 3.3(b').(c) If Y = lnt(Y) then certainly Y is open because lnt(Y) is open by part (b); if Y is open then Y E {U ~X I U is open in X, U ~ Y} so the union of all those sets contains Y and is therefore equal to Y. Theargument for Y is similar. If Y is closed then 8YY = Y: if Y is not closed then there is a point in Y \ Y(since Y - I - Y) and this point is also in 8Y \ Y.(d) X \ Y is open and contained in X \ Y, so X \ Y ~ lnt (X \ Y). Also, X \ lnt (X \ Y) is closed and containsY, so X \ lnt(X \ Y) ;2 Y, that is, lnt(X \ Y) ~ X \ Y. SOlnt(X \ Y) = X \ Y. Applying this to X \ Y weget X \ X \ Y = lnt X \ (X \ Y) = lnt Y so X \ Y = X \ lnt Y.3.8 Theorem. If Z ~ X then Z is closed in X if and only if the following holds: if an E Z for all n E Nand an -+ a in X, then a E Z.Proof: Suppose first that Z is closed, so U = X \ Z is open. If an -+ a but a ( j_ Z then a E U, so there existsI:> 0 such that B,(a) ~ U. So if d(an, a) < I: then an E U, and this is a contradiction. Therefore a E Z.Conversely, suppose Z is not closed. Then U is not open so 3a E U [oralle > OB,(a) ~ U. Choose suchan a E U and take I:= ~. Then B,(a) ~ U so B,(a) nZ - I - 0 . Choose an E B,(a) nZ. Then an -+ a, butan E Z and a ( j_ Z.This means that a set Z is closed in X if you don't fall out of it by taking limits. The limit of a sequence ofpoints of Z might not exist at all (in X), but if it does and Z is closed then it's in Z.

3.9 A subset Y ~ X is dense if Y = X. It is sparse of lnt(") = 0 . A metric space X is said to be separableifthere is a countable (or finite) dense subset Y ~ X. For instance lRwith the Euclidean metric is separablebecause < Q clRis a countable dense set.3.10 If x E X a subset Y ~ X is called a neighbourhood of x if x E lnt(Y). If Y is itself open it is called anopen neighbourhood of x.Proposition. A sequence an converges to a in X if and only if for every open neighbourhood U of x thereexists N such that an E U for n > N.Proof: If an -+ a and U '" a is open, choose I:> 0 such that B,(a) ~ U. Then choose N such that if n > Nthen d(an, a) < 1:. Then for n > N we have an E B, (a) ~ U, as required. For the converse, suppose thatfor every open neighbourhood U of x there exists N such that an E U for n > N. Then in particular this istrue for U = B,(a) for every I:> 0, which is precisely what the definition of convergence to a requires.3.11 IfX is a metric space and D ~ A ~ X, we say that D is relatively open in A if D is open in A with thesubspace metric dA.

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For example if D = (0,1) C A = J R cX = C (or A is the x-axis in J R 2 ) then D is relatively open in A ,though D is not open in X.Theorem. IfD ~ A ~ X then D is relatively open in A if and only if D = A n U for some open set U ~ X .Proof: The point is that a ball in (A , d A ) is a ball in X intersected with A . More precisely, if B:;! - ( x ) = { y EA I dA ( y , X ) < r} then B:;! - ( x ) = Br ( x ) n A where Br ( x ) = { y E X I d ( y , x ) < r} is the usual ball in X .If D = A n U then ' tIx E D 3Ex > 0 B ,Jx ) ~ U . So B,Jx) n A ~ D ; but n., ( x ) n A is the ball in (A , d A)with centre x and radius Ex, so we have shown that D is open in A.Conversely, if D is relatively open in A then there is a ball in the sense of (A , d A) centred at x whichis contained in D. So there is an epsilon.; such that B~ Jx ) ~ D , that is B,Jx) n A ~ D . Now takeU = U B,x (x): we havexED

A nU=A n U B,Jx) = U(AnB , Jx ) ) ~D.xED xED

For eample, [0, ~) is relatively open in [0,1) as it is [0,1) n (-1,1).3.12 If A ~ X and A is open in X then D ~ A is relatively open in A if and only if D is open in X .Proof: IfD is relatively open in A then D = A nU for some open U ~ X, and the intersection of two opensets is open so D is open in X . If D is open in X then we can write D = A nU simply by taking U = D .

4. Continuity.

4.1 Let (X , d ) and (X I , d l) be metric spaces and f : X -+ X a function. Then the following conditions areequivalent:i) ' tIx E X 'tiE > 038> 0 ( d ( x , y ) < 8) ===} (d l ( f ( x ) , f ( y ) ) < E) ;ii) for all x E X , if ( x n ) is a sequence in ( X , d ) and X n -+ x , then f( x n ) -+ f ( x ) in ( X I , d l) ;iii) if U ~ X I is open in X I then f - l (U ) ~ X is open in X .If one of these conditions holds then f is said to be continuous on X.Proof: (i) ===} (ii). Suppose X n -+ x and E > O . Then there exists 8 > 0 such that if d ( x n ' x ) < 8 thend ( f ( xn ) , f ( x ) ) < E. But 3N 'tin > N d (x n ' x ) < 8 so 3N 'tin > N d ( f ( xn ) , f ( x ) ) < E, i.e. f (x n ) -+ f ( x ) .(ii) ===} (iii). If U ~ X I is open then Z = X I \ U is closed and f- 1(Z) = X \ f- l (U ) = W. It is enough toshow that W is closed. By 3.8 it is enough to show that if Wn E Wand Wn -+ W E X then W E W. But ifWn -+ W then f (w n ) -+ f (w) E X I , and f( wn ) E Z and Z is closed, so by 3.8 f (w) E Z, i.e. w E W.(iii) ===} (i). Choose E > O . U = B, ( f ( x ) ) is an open subset of X I so f - l (U) is an open subset of X .Moreover x E t:' (U), so by the definition of an open set 38 > 0 B8 (x) ~ t:' (U). But that means preciselythat if d ( y , x ) < 8 (so Y E B8 (X ) ) then d( f ( y ) , f ( x ) ) < E (so f( y E U) ) .One sometimes says that f is continuous at a particular x E X if (i) or (ii) holds for that particular valueof x (strictly, we haven't checked that these two statements are equivalent, because we went via (iii) whichdoesn't mention a particular x). But continuity at just one point is almost never any use: what one needsis continuity in a neighbourhood of a point, and we say that f : X -+ X I is continuous near x E X is thereis a neighbourhood Y of x such that the restriction of f to Y is continuous.4.2 This agrees with all previous definitions of continuity of maps J R -+ J R , C -+ C, etc., because condition (i)is the usual definition in those cases.

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If t E [0,1] the evaluation map ev, : e[O, l ] -+ lR given by eVt(f) = f(t) is continuous when lR had theEuclidean metric and e [O , 1] has the uniform metric (or anything else sensible). Itis easiest to check this using(ii): if i-. -+ fin e [O , 1] with the uniform metric then fn -+ f pointwise, so fn(t) = eVt(fn) -+ f(t) = eVt(f),for any t E [0,1].The map I: e [O ,l] -+ e [O , 1] (with the uniform metric) given by 1(f)(t) = J~f(x ) d x is continuous. This ismost easily checked using (i): if Ilf - glloo < E then

1I1(f) - 1(g)11 = sup { I r f(x )d x - r g(X)dxl}099 10 10= sup I r ( f(x ) - g(X ))dX I099 10~ sup r I f(x ) - g(x )ld x09910~ 1 1 I f(x ) - g(x )ld x~ sup If(x ) - g(x )1

O:S;x:S; l= Ilf-gil

so we can simply take 8 = E.4.3 IfA ~ X then the inclusion map (A, dA) -+ (X, d) is continuous. We can check any ofthe three conditionseasily: for instance, condition (iii) is just 3.12 and (ii) is 2.9. If (Xl, dd and X2, d2) are metric spaces thenthe projection map 7rl : Xl X X2 -+ Xl, where Xl x X2 has the product metric drr, is continuous. Againthere is nothing new to check here: condition (ii) is part of 2.9.

n4.4 Either by induction or by a direct calculation, the projection maps tt; : n X i -+ X i are continuous (withi=lthe product metric) for any finite product. One way of thinking about the product metric is to say that it

is defined precisely so as to make this be the case. But we can't really deal with the product of infinitelymany X i. For one thing the set n Xa is problematic (it is an axiom rather than a theorem that it isn't

aEAempty: this is the Axiom of Choice); for another, the definition of product metric we had breaks downbecause the maximum may not exist.4.5 We could also use this to prove 2.9. If instead of appealing to 2.9 to prove that the projections arecontinuous we check condition (i) directly then we know condition (ii) must also be true, and that is 2.9(an converges if and only if a~) converges for each i). Let us check (i) directly. If drr(a, b) < E thenmaxdi(a(i) ,b(i)) < E so dl (7rl(a),7rl (b)) = dl (a(l ) ,b(l )) < E. S O we can simply take 8 = E.4.6 If X , Y and Z are metric spaces and f : X -+ Y and 9 : Y -+ Z are continuous then gf : X -+ Z iscontinuous. This is easy if you use condition (iii): suppose U ~ Z is open. Then g-l (U) ~ Y is open, sof l(g-l(U)) ~ X is open, but f l(g-l(U)) = (gf)-l(U).4.7 Definition. A continuous map f : (X, d) -+ (X I, dl) is called a homeomorphism (do not confuse with"homomorphism") if f is bijective (so it has a 2-sided inverse function f-l) and t:' :X I -+ X is alsocontinuous.The metric spaces X and X I are said to be homeomorphic if there is a homeomorphism f : X -+ X I. Youshould think of homeomorphism as being "isomorphism of topology" .

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4.8 Definition. A continuous map I :(X, d ) -+ (XI, d ') is called an isometry if' N we have

so ( a n ) is Cauchy.Suppose ( an ) is Cauchy. Choose N such that d (a n , am ) < 1 for n , m > N, and put

M = 1+ m ax{d (a n , am ) I n ,m ~ N + 1}.12


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Then, ifn,m ~ N we have d ( a n , am ) ~ m a x { d (a n , a m ) I n,m ~ N + 1} < M; ifn ~ N < m we haved (a n ,a m ) ~ d (a n ,aNH ) +d (a N+ l,a m )

< m ax { d (a n , a m ) I n,m ~ N + 1}+ 1 = M;and if n,m > N then d (a n , a m ) < 1~M . So in all cases d (a n , a m ) < M so the sequence is bounded.What we want to do now is to get some idea of how much difference there is between Cauchy and convergent.The first thing we have is a criterion which tells us, sometimes, that a Cauchy sequence is in fact convergent.5.3 Proposition. If a Cauchy sequence has a convergent subsequence then it is convergent itself.Proof : Suppose ( a n ) is a Cauchy sequence in some metric space ( X , d ) and a n j -+ a E X . Given 1 0 > 0 wechoose N such that d (a n , a m ) < 1 0 / 2 if n,m > Nand d (a n i, a ) < 1 0 / 2 if i> N. Now suppose n > Nandchoose i> n, so that ii; > n also. Then

so a n -+ a .5.4 A metric space (X , d ) is said to be complete if every Cauchy sequence in (X , d ) is convergent.This means, roughly, that everything that ought to have a limit, does. If we know that a metric space iscomplete then we can tell that a sequence is convergent without having to know first what the limit is goingto be. This means we can use convergence as a way of constructing or finding things we didn't previouslyknow about: completeness actually asserts that under certain circumstances some point of X exists, ratherthan just telling us more things about points we already knew existed.5.5 lR with the usual metric is complete. It is probably best to regard this as an axiom: in one ratherpopular approach this is pretty close to being the definition of R lRnis complete (with the Euclidean,or equivalently with the product, metric - if two metrics are equivalent then they give the same Cauchysequences). If (an) is a Cauchy sequence in lRnwith respect to the product metric drr then for m, n > N wehave maxi la ~ ) - a } 2 1 = d i: (an, am) < 1 0 ; but then for each iwe get I a ~ ) - a ~ I < 1 0 , so the sequence ( a ~ ) ) isa Cauchy sequence in lRand therefore converges. So an converges also.Q is not complete, because 3,3.1,3.14,3.141,3.1415 ... is obviously Cauchy but does not converge in Q since7f ( j_ Q. lR\ {O } (with the usual metric) is also not complete, because the Cauchy sequence l/n no longerconverges (somebody has taken the limit). The p-adic metrics on Z and Q are not complete metrics (thatis, they do not yield complete metric spaces) either: you can easily check that a n = pn is Cauchy but notconvergent.5.6 Proposition. If (X , d ) is a complete metric space and 0 - I - A cX then (A, d A ) is complete if and onlyif A is a closed subset of X.Proof : If A is complete, take a sequence ( a n ) in A that converges to a E X . We need to show that in facta E A. But since ( a n ) is convergent in X it is Cauchy in X and therefore it is Cauchy in A too, sinced (a m , a n ) < 1 0 if and only if d A (am ' a n ) < E. SO ( a n ) must converge in A, and the limit it has in A must bethe same as its limit in X, since a sequence has at most one limit. So a E A.Conversely, suppose A is closed and ( a n ) is a Cauchy sequence in A. Then ( a n ) is a Cauchy sequence in Xalso, so it has a limit a E X . Since A is closed, a E A, but now a n -+ a E A so ( a n ) is a convergent sequencein A and A is complete.5.7 Theorem. If a , bE lRand a < b then the space era , b] with the uniform metric is complete.For this we need the following lemma from real analysis.

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Lemma. If ( f n ) is a sequence of continuous functions on some interval [ a , b ] and i-. tends to f uniformly(that is, sup I f n ( t ) - f ( t ) 1 -+ 0 as n -+ 00), then f is also continuous on [ a , b ] .One expresses this informally by saying that "a uniform limit of continuous functions is continuous".P r o o f o f L e m m a : Choose N so large that ' t i n > N ' t i t E [ a , b ] l f n ( t ) - f ( t ) 1 < E/3 (by the uniform convergencethis is possible). If n > N we have

I f ( t ) - f ( s ) 1 = I f ( t ) - f n ( t ) + f n ( t ) - f n ( s ) + f n ( s ) - f ( s ) 1~ I f ( t ) - f n ( t ) 1 + I f n ( t ) - f n ( s ) 1 + I f n ( s ) - f ( s ) 1~ E/3 + E/3 + E/3 = E

for s E ( t - 8, t + 8), where 8 is such that 'tis E ( t - 8, t + 8) I f n ( t ) - f n ( s ) 1 < E/3. Such a 8 exists because i-.is continuous.This proves the lemma.P r o o f o f t h e t h e o r e m : Suppose that ( f n ) is a Cauchy sequence in ( C [ a , b ] , I I I I( x , ) , Choose N such that' t I n ,m > N I l f n - fml l < E. Then if n ,m > N, for any t E [ a , b ]

so the sequence (J n(t)) is a Cauchy sequence in lR, and this is true for any t E [a , b ]. So we can take thelimit. We define f ( t ) to be lim f n ( t ) .n--+ooSo far there is absolutely no reason why f should be continuous (and if it isn't it's no good to us). Whenwe defined f ( t ) we did so without paying any attention to f ( s ) for s close to t , which might have ended upbeing something totally different. But if we can prove that fn in fact converges uniformly to t,which wehaven't done yet, then by the lemma f will be continuous, and we'll have finished. So we need to know thatI l f n - fl l = sup I f n ( t ) - f ( t ) 1 tends to zero as n -+ 00.We know by the fact that ( f n ) is a Cauchy sequence in ( C [ a , b ] , 1 1 1 1 0 0 ) that

' t i E > 0 3N ' t i t E [ a , b ] ' t i n , m > N I f n ( t ) - fm (t ) 1 < E/2 .We fix n, t and E and simply take the limit of this as m -+ 00, a limit in lR:

' t i E > 0 3N ' t i t E [ a , b ] ' t i n > N I f n ( t ) - f ( t ) 1 ~ E/2 .But this is exactly what we want: ' t i E > 0 3N ' t i n > N sup I f n ( t ) - f ( t ) 1 ~ E/2 < E.5.8 Completion. (Not really part of the course, but ... ) If (X , d ) is any metric space there is a metric space( X , d ) , such that ( X , d ) is complete and there is a dense subset Y ~ X such that (Y, d y) is isometric to(X,d). Moreover, there is only one such ( X , d ) up to isometry (in other words, any other such space isisometric to that one).This means that if I've got an incomplete metric space there is a unique way to make it a little bit bigger sothat it becomes complete. ( X , d ) is called the completion of X with respect to the metric d. You only needto know one example: lR, with the usual metric, is (by definition if you like) the completion of < Q with theusual metric. If you do this with the p-adic metric on < Q instead you get a kind of alternative lReality called< Q p : this unlikely-looking object turns out to be very important in number theory.The way you make X is this: you define it to be the set of all Cauchy sequences in X, modulo an equivalencerelation that says that two Cauchy sequences are the same if they really ought to have the same limit. So

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(a n ) "" (bn ) (where ( an ) and (b n ) are Cauchy sequences in (X , d ) ) if and only if lim d (a n , bn ) = 0, andn--+ooelements of X are equivalence classes: X is inside X because you think of x E X as corresponding to theconstant sequence at x . To define d you simply say that d([ (a n )], [(bn )]) = lim d (a n , bn ). You then have ton--+oocheck that this all makes sense and really defines a metric space, etc. It isn't hard.5.9 Completeness, unlike most other things we've mentioned, is not a topological property. That is, in orderto tell whether a metric space is complete or not it's not enough to know what the open sets are: youreally have to know what the metric is. For instance, take N with the discrete metric. That is complete,because the only Cauchy sequences are the eventually constant sequences and they converge. Every subsetis open because in } = Bl(n) for any n E N and if A ~ N then A = UnEA { n } = U nE A B 1(n ), which isopen. But if we take the metric on N where the distance between successive points is d ( n , n + 1) = I/ n 2 ,so d ( n , m) = L:~=~l 1Ir2 , then the sequence ( an ) = n is Cauchy (because L:~=-nl 1Ir2 < L:~n 1I r2 < E if nis large enough). On the other hand every set is still open, because {n} = B1/2 n 2 (n). So this new metricspace has exactly the same open sets as the old one, and therefore exactly the same convergent sequences,but not the same Cauchy sequences: in particular, (n ) does not converge though it is Cauchy.

6. Contractions.

6.1 Suppose (X, d ) and (XI, d ') are metric spaces. We say that a map f : X -+ XI sa tisfie s a L ip sc hitzcondition if there is a a constant k, 0 ~ k < 00 (the Lipschitz constant) such that

\ fx,y E X d '(J(x ) ,f (y )) ~ kd (x ,y ) .For instance, if f is an isometry then f satisfies a Lipschitz condition with k = 1 (and moreover the inequalityis actually an equality).If f satisfies a Lipschitz condition then f is automatically continuous, so we don't need to specify this inthe definition. It is easy to see this: if k = 0 then f is constant and therefore continuous, and if k > 0 then,given E > 0, we take 8 = E lk . Then if d ( x , y ) < 8 we have d ' (J( x) , f ( y ) ) < k8 = E so f is continuous.6.2 A contraction or con tr action m a ppin g is a map f : X -+ XI which satisfies a Lipschitz condition withO~k


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(i) f has a unique fixed point ~ EX;(ii) if a o E X and a n = f ( an - d for n E N then a n -+ ~ as n -+ 00.Proof: Let k < 1 be the Lipschitz constant (we may as well assume k > 0 as the case k = 0 is the trivialcase of a constant map). We shall prove (ii) first, so we choose a point a o EX.Notice first of all that

This is easy to prove by induction: the case n = 0 is trivial andd (a n+ l' a n+ 2) = d (J(a n) , f (a n+d )

~ kd (a n , a n+d~ kn+ ld (a o, a d .

I claim that the sequence ( an ) is Cauchy. Ifm > n we haveml

d (a n , am ) ~ Ld (a r, a r+ !)r=nmlr=n

ml-n

r=O

which is less than E if n is large enough.So, since (X , d ) is complete, ( a n ) converges and we may write ~ = lim a n . We need to check that ~ is an--+oofixed point. But

d (~ , f (~ ) ) ~ d (~ ,a n ) + d (a n, f (~ ))= d (~ , a n) + d (J(a n -d , f (~ ) )~ d (~ ,a n) + kd(an- l ' ~ )< E if n is sufficiently large.

So d(~, f ( ~ ) ) = 0, so ~ = f ( ~ ) It remains to show that there is no other fixed point. If e is a fixed point then

d ( ~ , e ) = d (J(~ ) , f (e) ) ~ kd (~ , e )so either 1 ~ k, which is impossible, or d(~, e) = o .6.6 In the case of a function f : lR -+ lRwith 1 f '( t ) 1 ~ k < 1 this means that the equation f ( x ) = x alwayshas a unique solution. This can be proved directly using the mean value theorem, so what we have foundcan be thought of as an extension of the mean value theorem. (There are other ways of thinking of it whichmay be more illuminating.)6.7 Simple iteration. In the case above the contraction mapping theorem tell us how to solve f ( x ) = x : makea guess, ao, and repeatedly apply f so as to get a better guess. For example, suppose f ( x ) = ~ex on [0,1].Then f : [0,1] -+ [0,1] since ~ ~ ~ex ~ ~e < 1 if 0 ~ x ~ 1, and f ' ( x ) = f ( x ) so 1 f '( x )1 ~ ~e < 1. In other

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words f : [0,1] -+ [0,1] is a contraction mapping. Take ao = ~; then a l = f (~) = ~e~ ~ 0.5496, etc. Butit's rather slow: ten iterations gives alO ~ 0.6182 and the correct solution is 0.61906 ....6.8 The Newton-Raphson method. This is another way of solving equations iteratively and tends to be quitequick. We are now in a position to say when it works (and why). Suppose f : ]R -+ ]R is twice differentiable.Suppose that f ' ( x ) - I - 0 for all x E ]R and that there is a k with 0 ~ k < 1 such that

If(x )f" (x ) I < kf'(x )2 -

for all x E R Then f (x ) = 0 has a unique solution ~ E ]R , obtained by taking some starting value ao andputting an+l = a n - ff/(an)) (and then ~ = lim an ) .

an n--+oo

The point is that g;]R -+ ]R given by g(x) = x - f / ~ ~ ) is a contraction mapping because gl(X) = f ( ; : ( : ; ~ x ) ,and g(x) = x if and only if f (x ) = O .In actual practice one usually doesn't have such good behaviour for all x E ]R but only for x in some closedinterval X: then you have to check that g(x) is still in X.6.9 Theorem. (Pica rd 's Theor em ) If f : ]R 2 -+ ]R has continuous first derivative then given xo, to E ]R thedifferential equation with initial condition

x = f( t, x ); x(to) = X ohas a unique solution for to - 8 ~ t ~ to + 8 if 8 is small enough.By x I mean ~~. This theorem means that we can always find a solution locally to a simple differentialequation like this, but things may go wrong after a long time.Proof: We give ]R and ]R 2 the Euclidean metric and put

D = [to - 1, to + 1] x [xo - 1, Xo + 1] C ]R2;c = 1+ max {If(t, x)l, (t, x ) ED}c'=l+m ax{I~ ~ (t'X )I' ( t,X )ED}.

These maxima exist (because D is compact, see the next section): this is why we need to restrict attentionto D rather than working with the whole of ]R2. However there is nothing very special about the precisechoice ofD we have made. Choose 8 such that 0 < 8 < min(~, ~): note that 8 < 1. Now we look at a closedball in C[to - 8 , to + 8] , putting

Z = {x E C[to - 8,to +8]1 Ix (t) - xol ~ c8 if It - tol ~ 8}.Thus Z = Bcc5(XO) , the closed ball in C[to - 8, to + 8] of radius c8 with centre the constant function Xo.Because Z is closed, it is complete. We define a map T : Z -+ Z by

T(x)(t) = Xo + i t f(s, x (s)) d sto

for x EX, t E [ to - 8, to + 8]. Ifwe can prove that T has a fixed point in Z we shall have solved our differentialequation, because T(x) = x if and only if x = f( t, x ) and x (O) = X o. So we need to check that T : Z -+ Z isa contraction mapping.

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First of all we need to check that T is indeed a map from Z to Z. So we must check that T(x) E Z if x E Z.But

IT(x )( t) - x o l = 1 1 : I(s, x (s)) -~ 1 1 : II(s,x (s)) Ids I~ 1 1 : c d s l~ - s .

Then we need to check that T is a contraction mapping. If x , y E Z then for some t E [to - 8, to + 8]IIT(x ) - T(y ) lloo = IT(x )( t) - T(y )( t)1

= 1 1 : I(s,x (s)) - I (s,y (s)) -= l i t { l X (S ) 81(s,w) dW} d s l

to y(s) 8w~ l i t 1 1 X(S ) 81(s, w) dwl d s l

to y(s) 8w

~ 1 1 : I l ~ s \ S ) I 811~ w) I dwl d s l~ 1 1 : I l ~ s \ S ) c ' dwl d s l= 1 1 : c'lx (s) - y (s)1 -~ 1 1 : c ' l l x - Y ll o o -~ 8 c ' l l x - Y l l o o

and 8 c ' < 1 so T is a contraction.

7. Compactness.

7.1 The Bolzano-Weierstrass theorem in real analysis says that a bounded sequence in lRhas a convergentsubseqence.7.2 A metric space (X, d) is said to be sequen tia lly com pa ct if every sequence in X has a convergent subse-quence. IfA ~ X we say that A is compact if A is empty or if (A, d A ) is compact, where d A is the subspacemetric.7.3 Any bounded closed interval [a , b ] ~ lRis sequentially compact. This is one way of stating the Bolzano-Weierstrass theorem: if a ~ an ~ band (anj) is convergent then a ~ lim an j ~ b, so any sequence in [a , b ]J--+OOhas a subsequence which converges in [a,b]. But (0,1] is not compact because the sequence an = lin doesnot have a subsequence which converges in (0,1].7.4 Proposition. Suppose (X , d ) is sequentially compact. Then(i) X is complete;

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uniformly, without having to know where you are first (compare being able to chosse N first in the definitionof uniform convergence). You do know that once you've made a choice of 8, say 8a, at a point x E X it willwork in some small neighbourhood of x (usually a ball), and you take these neighbourhoods to be your Uas.(For this reason the index set A is most often simply X.) Then you want to take the inf (or sup) of the 8as:in general you can't (or you can but you might get zero). But if you know that in practice you only needfinitely many of the Uas you can take the sup or inf of those finitely many 8as, which is all right.7.12 Theorem A metric space (X , d ) is compact if and only if it is sequentially compact.We'll prove this later on (7.17).7.13 Proposition. If Z is a compact metric space and f : Z -+ lR is continuous then f is bounded andattains its bounds.That is, there exists ~ E Z such that f (~ ) = sup f (x ) .

xEZProof: If f is unbounded we can find a n E Z such that I f (an)1 -+ 00. But then we can find a convergentsubsequence ( anJ with limit a E Z, and f (anJ -+ f (a ) < 00, which is a contradiction.If no such ~ exists then g(x) = ( f(x ) - SUPt f ( t))- l is a continuous (negative) function so it is boundedbelow by -M say. But then f(x ) ~ SUPt f( t) - 11M for all x E Z and that contradicts the definition of thesupremum.7.14 A continuous image of a compact set is compact. That is, if f : X -+ Y is continuous and Z ~ X iscompact then f(Z) ~ Y is compact.This is easiest via sequential compactness. If ( an ) is a sequence in f (Z) then choose a n E Z such thatf( a n ) = b. Then ( an ) has a convergent subsequence a n i -+ a E Z and, applying t,we have bn i -+ f (a ) Ef (Z).7.15 The Heine-Borel theorem states (in the language we have now) that [0,1] is a compact subset of R7.16 A circle is compact; so is a sphere or a torus. A sphere with a point missing is not compact. A closeddisc in C is compact; an open disc is not.7.17 We shall now prove 7.12.(Compact ===} sequentially compact.) Suppose ( an ) is a sequence in a compact metric space X and ( an )has no convergent subsequence. Then for each x E X there is an rx > 0 such that {n E N I a n E BrJx)}is finite: if this were not true for some x we could choose a subsequence tending to x . The Brx (x ) between

Nthem cover X, so we choose a finite subcover, X =~ BrX i ( X i ) . Ifm > 1~~)max{ n E N I a n E n., ( x ) } ) ,

2-1 - -which exists since the sets are finite, then am ( j_ X, which is a contradiction.(Sequentially compact ===} compact.) Take any open covering {Ua 1 0 : E A} of X. For each x E X we define

r (x ) = sup{r > 0 I : 3 0 : such that Br(x ) ~ Ua }.This means that Br(x ) (x ) is (roughly speaking) the biggest ball around X that can be fitted into one of thepatches in the cover. A really big ball might spread outside all the patches that contain x. It's possible thatr (x ) = 00, that is, the set is unbounded, but that will make no difference.We then put s(x) = min(~r(x), 1). The 1 is put there to make sure that we are writing something finite:the ~ (which could as well be any constant less than 1) is there to make sure s(x) isn't as big as r (x ) . Bydefinition there are Ua s which contain B s( x) ( x) : choose one of them for each x E X and call it Ua(x ) .The idea is that we shall cover X with finitely many of the B s( x) ( x) s and therefore with finitely many of theUa(x )s . Ifwe can do this we'll have shown that {Ua } has a finite subcover.

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Suppose we can't do this. Then we can pick a sequence ( a n ) in X by starting with any a l and requiring thata n ( j_ U B s( a i) ( a i) . Such an a n must always exist, otherwise the first n -1 of the Bs(ai) (c,) have covered X,i< nwhich we supposed couldn't be done. Having got this sequence we can apply sequential compactness toextract a subsequence ( a n , ) tending to a E X.Ifm > n we have d (a m , a n ) ~ s(a n ) since if d (a m , a n ) < s( a n ) then am E Bs(an) ( a n ) which we forbade. So

and, taking the limit as j-+ 00, we get0= d (a , a ) ~ lim s(a n j) ~ 0J--+=

so s (an j ) -+ 0 as j-+ 00.On the other hand we can choose jEN such that d (a n j, a ) < ~s(a ) , and if we do that then Bs(a ) /2 (a n ,} ~Bs(a ) (a ) ~ U a ( a ) . (Check: if y E B s( a) /2 ( an ,) then d (y , a ) ~ d ( y , a n ,} + d (a n j, a ) < ~s ( a ) + ~s ( a ) = s (a ) , soy E B s( a ) ( a ). ) This means that r (a n ,} ~ ~ s(a ) , since r ( a n ,} is as big as a ball centred at a n j can be withoutneeding another patch to cover it, and the ~s(a)-ball, we have just checked, needs only one patch, namelyUa( a ) . So

s (an ,} = min ( r ( an ,} j 2 , 1) ~ min (s(a) j4, 1 ) ,but this is a positive constant so s( a n j) does not tend to zero.Now we have reached a contradiction, so our assumption that we couldn't cover X with the Bs(x ) (x)s musthave been false. This gives a finite sub cover of {Ua}, so X is compact.

8. Connectedness.

8.1 Theorem. The following conditions on a metric space (X,d) are equivalent:(i) X = X o UXl with X O , Xl open and nonempty and X o nXl = 0 .(ii) X = X o UXl with X o, Xl closed and nonempty and X o nXl = 0 .(iii) There exists a continuous function I :X -+ E, where E has the discrete metric, such that I is notconstant.Proof: (i) is equivalent to (ii) because if X o and Xl are both open (or both closed) then X o = X \ Xl andXl = X \ X o are both closed (or both open).(i) implies (iii) because I ( x ) = i if x E Xi is such a function; (iii) implies (i) because we can take Xi = 1-1(i)(calling two of the points of E by the names 0 and 1).8.2 A metric space X is said to be disconnected if one (and hence all) of the conditions 8.1(i)-(iii) holds. Ifthey do not hold (that is if no such decomposition or function exists) then X is said to be con n ected . X issaid to be l oc a ll y c on n ec te d if every point of X has a connected neighbourhood.8.3 If a discrete metric space has more than one point then it is disconnected, since we can take X o = { x } .lR with the usual metric is connected because of the intermediate value theorem: the function that takes thevalue 0 on X o and 1 on Xl is continuous so by the intermediate value theorem it takes some other values aswell, which is a contradiction. lRnis also connected, as we shall see in a moment (it follows at once from 8.4).8.4 Proposition. If X and Yare connected then X x Y is connected.

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Proof: If X x Y is disconnected let f :X x Y -+ E be nonconstant, E discrete. Choose y E Y and consider9y(X) = f (x , y ) . This is a continuous function 9y : X -+ E , so it is constant, say 9y(X) = h(y) . But nowh : Y -+ E is continuous, as h ( y ) = f ( x , y ) for any x E X, and so it is also constant. But that means f isconstant.8.5 If XI ~ X is closed, open and connected then XI is said to be a connected component of X. Any locallyconnected metric space X can be written uniquely as a disjoint union of connected components, namely asthe union of all the connected components of X. Any two different connected components XI, X" of Xmust be disjoint, as if XI nX" - I - 0, XI then XI is disconnected as XI = (XI nX") u (XI \ X"). By "locallyconnected" I mean that every point x E X has a connected neighbourhood.8.6 A metric space X is path-connected if given any two points xo, Xl E X there is a continuous map"( : [0,1]-+ X such that "((0) = Xo and "((1) = Xl.8.7 Proposition. If X is path-connected then it is connected.Proof: If X is disconnected then there is a nonconstant continuous function f : X -+ E. Put h ( x ) = 1 ifX > 1, if X < 0, x otherwise: this is a continuous function h : ]R -+ [0,1]. Then f 0 "( 0 h : ]R -+ E is anonconstant continuous function ]R -+ E, contradicting 8.3.8.8 The converse is false: the set

{ ( x , y ) E ]R 2 I x ~ 0, y = sin(l/x) if x - I - O ]is connected but not path-connected. But this doesn't often happen.

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Metric Spaces (Bath Lecture Notes)

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