ANDREAS LEOPOLD KNUTSEN
Abstract. These notes are written as supplementary notes for the course MAT211- Real Analysis, taught at the University of Bergen, Fall 2019. They are meant to supplement [Ru].
Contents
1. Functions, relations and equivalence relations 1
2. Subsets of metric spaces 3
3. Connected sets 4
4. Cantor’s construction ofRfrom Qby means of Cauchy sequences 5
5. Completion of a metric space 9
6. An alternative description of compact sets 13
7. Spaces of functions and the Arzelà-Ascoli theorem 17
References 20
1. Functions, relations and equivalence relations This section complements [Ru, 1.5 and 2.1-3].
We start with a couple of definitions (almost) left out in [Ru]. We also recall the following terminology not used in [Ru]: We denote afunction (or mapping) f (see [Ru, Def. 2.1]) from a set Ato a set B by f :A→B and say that
• f is injectiveor aninjection if it is one-to-one;
• f is surjectiveor asurjection if it is onto;
• f is bijectiveor a bijectionif it is both one-to-one and onto.
We also recall that if f is bijective, there is a well-defined inverse function (or inverse mapping)
g:B →Adefined asg(y) =the unique x∈A such thatf(x) =y.
Applying f first and then g simply bringsx back tox. The inverse mapping g is often denoted by f−1, which should not be confused with the notation of inverse images as in [Ru, Def. 2.2].
Definition 1.1. Let X and Y be sets. The Cartesian product of X and Y, denoted X×Y, is the set of ordered pairs (x, y) such thatx∈X andy ∈Y. In other words,
X×Y ={(x, y)|x∈X, y∈Y}.
Definition 1.2. LetX be a set. ArelationR onXis a subset of the Cartesian product X×X, that is,R⊂X×X.
One often denotes xRy, x ∼y, or x ≡y to mean (x, y)∈ R. We will in these notes stick to the notationx∼y.
Date: 12.11.2019.
1
Definition 1.3. Anequivalence relationon a setX is a relation satisfying the following properties:
(i) Reflexivity: x∼x for all x∈X.
(ii) Symmetry: ifx∼y, theny∼x, for allx, y∈X.
(iii) Transitivity: ifx∼y and y∼z, thenx∼z, for all x, y, z∈X.
Example 1.4. Theorder relation
x∼y⇐⇒x < y
on R defined in [Ru, Def. 1.5] is not an equivalence relation, as (i) fails (x < x fails to hold), and also (ii) fails (if x < y it is not true that also y < x). However, (iii) holds.
Example 1.5. The relation on the set of all sets in [Ru, Def. 2.3] defined by A∼B⇐⇒ ∃bijectionf :A→B
(meaning in the language of [Ru, Def. 2.3] that the sets A and B can be put in one- to-one correspondence) is an equivalence relation: Reflexivity holds because the identity mapping id:A→A mappingx to xis a bijection. Symmetry holds because a bijection f :A → B has an inverse mapping f−1 : B → A that is still a bijection. Transitivity holds because if f :A→B and g:B →C are bijections, then the composed map
g◦f :A→Cdefined by(g◦f)(x) =g(f(x)) is a bijection.
The next definition will play an important role in §§4-5.
Definition 1.6. Given an equivalence relation∼ on a setX and x∈X, we define the equivalence class [x]of xto be the set:
[x] :={y∈X|y∼x}.
Any element in the equivalence class is called arepresentativeof the equivalence class.
We often denote theset of equivalence classes byX/∼.
Using the properties of equivalence relations, it is not difficult to prove that anyx∈X lies in precisely one equivalence class. Indeed, by reflexivity, x ∼ x, so that x ∈ [x], whencex lies in at least one equivalence class. Ifx∈[y1]andx∈[y2], thenx∼y1 and x∼y2. By symmetry, we havey1 ∼x. But this, together with x∼y2 and transitivity, yields that y1 ∼y2, that is, [y1] = [y2]. Hence, X/∼is a collection of disjoint subsets1 ofX whose union is the whole ofX. (Such a collection is called apartition.)
We conclude this section with a couple of more examples of equivalence relations, to shed some more light on this notion, although the examples themselves will not play any role in the remainder of the course.
Example 1.7. Fix any m∈N. Let X=Z. The relation on Zdefined by x∼y⇐⇒mdividesx−y
is an equivalence relation.
The equivalence class of any n∈Z is
[n] ={k∈Z mdividesn−k}={n, n±m, n±2m, n±3m, . . .}.
Clearly [n] contains a unique number n0 satisfying 0 ≤ n0 ≤ m−1. Thus the set of equivalence classes Z/ ∼, which is often denoted by Z/mZ or Zm, is in one-to-one correspondence with the set
{0,1,2, . . . , m−1}.
1We recall that two setsA andB aredisjointifA∩B=∅, cf. [Ru, Def. 2.9], which means that A andB have no elements in common.
Example 1.8. The relation onC given by
z1 ∼z2⇐⇒ |z1|=|z2| is an equivalence relation.
Letz∈C. Then the equivalence class of z is
[z] ={c∈C| |c|=|z|},
which consists of all points lying on the circle of radius|z|about the origin in the complex plane. Thus the set of equivalence classes is in one-to-one correspondence with the set of circles about the origin in the complex plane (including the circle with radius zero, that is, the origin itself), which is again in one-to-one correspondence with the set of nonnegative real numbers.
Exercise 1.9. Prove that the relations in the last two examples are indeed equivalence relations.
2. Subsets of metric spaces This section complements [Ru, 2.29-30].
As mentioned in [Ru, 2.16], any subsetY of a metric spaceX(whose metric we denote by dX just to be precise) is automatically a metric space with metric function dY the restrictionof dX toY, that is:
dY(p, q) =dX(p, q) for all p, q∈Y.
It is immediate that dY satisfies the axioms of a metric, asdX does.
When the metric onY ⊂X is the restriction of the metric onX, we callY asubspace ofX.
We have to be a bit careful when considering subsetsE ⊂Y ⊂X. Certain properties ofEdepend on whether we considerE as a subset of the metric spaceY or of the metric spaceX. This is for instance the case of properties of being “open” and “closed”, as well as the concept of “closure” and “complement”, as the following example shows:
Example 2.1. (a) Consider E = (0,1) ⊂Y =R ⊂ X = R2 (where we think of R as thex-axis inR2). ThenE is open as a subset ofY =R, but not as a subset ofX=R2, because any neighborhood inX of a point in E, which is an open disc, will necessarily contain points outside thex-axis. The complement ofE is of course also totally different when considered inRor R2.
(b) ConsiderE = (0,1)⊂Y = (0,∞) =R+ ⊂X=R. Then the closure ofE inY is (0,1], wheras the closure of E inX is [0,1].
The complement ofE inY is[1,∞)and the complement ofEinXis(−∞,0]∪[1,∞).
To distinguish between the complement ofEinY and inXone can denote the former byY \E and the latter by X\E, as it causes no ambiguity. We recall that for any sets A andB, one has by definition
A\B ={x∈A|x6∈B}.
For any pointp∈E, we will in these notes denote a neighborhood of radius r of pin Y byNrY(p)and in X by NrX(p).
Similarly, we will denote the closure ofE as a subset of Y by EY and as a subset of X by EX.
It is easy to see that
NrY(p) = NrX(p)∩Y; (1)
EY = EX ∩Y.
(2)
Exercise 2.2. Verify (1) and (2).
Hence, for instance,E is open as subset ofY if and only if to eachp∈E, there is an r >0such thatNrY(p) =NrX(p)∩Y ⊂E, which means thatq ∈E wheneverd(p, q)< r andq ∈Y, which is the definitionin [Ru, 2.29] ofE beingopen relative to Y (one often also uses the expression “E isopen in Y"). Therefore, [Ru, Thm. 2.30] says that:
(3) E is an open subset of Y ⇐⇒E =Y ∩G for some open subsetG ofX.
Exercise 2.3. Prove that
(4) E is a closed subset of Y ⇐⇒E =Y ∩W for some closed subset W ofX.
Therefore, (3) and (4) give nice criteria for comparing the notions of “open” and
“closed” in the two spaces X and Y.
There are other properties of E that are independent of whether E is considered as a subset of Y or of X, that is, they are only dependent of E itself, which is of course also a metric space by restricting the metric to E. We call such properties intrinsic.
An example of such a property is the property of being compact, by [Ru, Thm. 2.33].
Another property that is intrinsic is the property of beingconnected, although it does not seem so when looking at the definition [Ru, 2.45]. We will consider the latter property in the next section.
3. Connected sets
This section complements [Ru, 2.45-47]. In particular, we will give in Definition 3.3 an equivalent definition of a metric space being connected.
We recall [Ru, Def. 2.45], which we reformulate as follows, using the notation AqB to mean A∪B withA∩B =∅ (this is called the disjoint union of the sets A and B, where we recall that two setsA andB aredisjointif A∩B =∅, cf. [Ru, Def. 2.9]):
Definition 3.1. LetX be a metric space and E⊂X.
• E is disconnected if one can write E = AqB, with A, B 6= ∅, A∩B =∅ and A∩B =∅.
• E isconnectedotherwise.
We remark that the closures above are taken in X, that is, A = AX and B = BX with the notation of the previous section. We will stick to this notation below.
We will need the following simple observation:
Lemma 3.2. Let A and B be disjoint nonempty sets in a metric space X. Let E =AqB⊂X. Then the following conditions are equivalent:
(i) A∩B =∅ andA∩B =∅;
(ii) A and B are both open relative to E;
(iii) A and B are both closed relative to E.
Proof. The equivalence of (ii) and (iii) is clear, sinceA=E\B is the complement ofB inE, and B =E\Ais the complement ofA inE, and A is open as subset ofE if and only if its complement inE is closed, by [Ru, Thm. 2.23].
To prove that (i) implies (iii), we use (2) and the distributive law of unions and intersections [Ru, (10), p. 28] and obtain
AE =A∩E =A∩(A∪B) = (A∩A)∪(A∩B).
NowA∩A =A, as A⊂A, and A∩B =∅ by assumption. Therefore, AE =A, which means thatAis closed as a subset ofE, by [Ru, Thm. 2.27]. The proof forB is identical.
To prove that (iii) implies (i), we note thatB =E∩B asB ⊂E and use again (2) to obtain
A∩B =A∩E∩B =AE ∩B.
By the assumption thatA is closed relative to E, we have AE =A, whence AE ∩B = A∩B=∅by assumption. This proves thatA∩B =∅, and the proof thatA∩B=∅is
identical.
By this lemma, the following is an equivalent definition of disconnectedness and con- nectedness, which is independent of the “big space” X in which E lives. This is often used as a definition in many books.
Definition 3.3. LetE be a metric space.
• E is disconnectedif one can write E =AqB, with A, B 6=∅ and A and B are open subsets ofE.
• E is connected otherwise; that is, E is connected if it cannot be written as a disjoint union of two nonempty open subsets.
Of course, by Lemma 3.2, we may substitute “open” by “closed” in the latter definition.
The following gives yet another property that is taken as definition in many books:
Exercise 3.4. Prove that a metric spaceE is connected if and only if the only subsets ofE that are both open and closed are E and ∅.
We conclude this section with a remark and an additional exercise.
Remark 3.5. By [Ru, Thm. 2.47] a subsetE ⊂Ris connected if and only if it has the property that whenever x, y ∈E and z ∈R satisfies x < z < y, then alsoz ∈E. Sets with this property are precisely the sets of the form (a, b),[a, b),(a, b]or [a, b], where
a=
(infE, if E is bounded below
−∞, otherwise and
b=
(supE, if E is bounded above
∞, otherwise.
Exercise 3.6. LetX be a metric space with the discretemetric, that is (cf. [Ru, Exc.
10, p. 44], for anyp, q∈X:
d(p, q) =
(0, if p=q, 1, if p6=q.
What are the connected subsets of X?
4. Cantor’s construction of R from Q by means of Cauchy sequences This section complements [Ru, 3.8-12 and 1.19].
In this section, we will go through the construction of the real numbers by using Cauchy sequences, due to the German mathematician Georg Cantor (1845–1918) [Ca].
This idea will be generalized when we construct completions of arbitrary metric spaces in the next section. The aim is to give a different, independent proof of [Ru, Thm.
1.19] (whose proof follows the different approach of the German mathematician Julius Dedekind (1831–1916) [De]):
Theorem 4.1. There exists an ordered field R with the least-upper-bound property and containing Q as an ordered subfield.
Before starting the proof, we remark that we do not need to know beforehand whatR is to talk about distances in Qand Cauchy sequences: Indeed, the number |p−q| ∈Q for anyp, q∈Qand satisfies the definition of a metric. Moreover, we say that a sequence {pn}inQis aCauchy sequenceif for anyrational >0, there is an integerN such that
|pm−pn| < whenever m, n ≥ N and that it converges to x ∈ Q if for any rational >0, there is an integerN such that|pn−x|< whenever n≥N.
We recall the following:
Proposition 4.2. There are Cauchy sequences inQthat do not converge to any element in Q.
Proof. This can be proved with the same technique used in [Ru, Ex. 1.1 and 1.9(a)]
to prove that Q neither satisfies the least-upper-bound nor the greatest-lower bound- property. The association in [Ru, (3)] of a rational number q to any rational number p defines a sequence{pn} recursively by
(5) pn+1 =pn− p2n−2
pn+ 2 = 2− 2 pn+ 2
which is easily checked (by induction) to be increasing ifp21 <2and decreasing ifp21 >2.
The sequence is a Cauchy sequence. If the limit exists, sayL= limpn, then taking limits in (5) shows thatL2 = 2, whence the limit cannot exist in Q, by [Ru, Ex. 1.1].
Exercise 4.3. Prove that {pn} in the proof of Proposition 4.2 is a Cauchy sequence (without using the knowledge that it converges inR).
Remark 4.4. There are other recursive relations yielding Cauchy sequences whose limits L, if they exist, must satisfy L2 = 2, for instancexn+1 = x2n +x1
n.
The moral of Cantor’s construction is the following: since Q lacks limits of Cauchy sequences inQ, we add those “limits” representing them simply by the Cauchy sequences themselves. The elements of Q can be viewed as constant Cauchy sequences in Q. In this language, the real number √
2 would be represented by the Cauchy sequence {pn} in the proof of Proposition 4.2. However, we have to take into account that different Cauchy sequences may yield the same limits, for instance different starting points p1 above yield different Cauchy sequences{pn}still representing√
2, and the same applies for the sequencesxn in Remark 4.4. It is to avoid this ambiguity that equivalences enter the picture:
Definition 4.5. Two Cauchy sequences{pn} and{qn} inQare said to beequivalentif the sequence{|pn−qn|}inQconverges to 0.
We write{pn} ∼ {qn} if {pn}and {qn}are equivalent.
Exercise 4.6. Show that∼is an equivalence relation on the set of all Cauchy sequences inQ(cf. Definition 1.3).
We denote by [pn] the equivalence class of {pn} and define R to be the set of all equivalence classes of Cauchy sequences on Q, that is, with the notation of Definition 1.6, we have
R:={Cauchy sequences inQ}/∼.
We note that Q can be identified with a subset of R by identifying q ∈ Q with the equivalence class of the constant Cauchy sequence Cq := {q, q, q, q, . . .}. It is clear that Cp∼Cq if and only if p=q, so the functionQ→R mappingq to[Cq](the equivalence class ofCq) is injective. HenceQcan be identified with the image of this function, which is a subset of the newly defined setR.
We define two operations + and · on R that are compatible with the ones on the subset Q:
[pn] + [qn] := [pn+qn] and [pn]·[qn] := [pn·qn].
We have to prove that these operations are well-defined, that is, that they are indepen- dent of the choice of representative for the equivalence classes of Cauchy sequences. This is taken care of by the following:
Lemma 4.7. If {p0n} ∼ {pn} and {qn0} ∼ {qn}, then {p0n +qn0} ∼ {pn +qn} and {p0n·qn0} ∼ {pn·qn}.
Proof. By definition, {p0n} ∼ {pn} and {qn0} ∼ {qn} means that lim|p0n−pn|= 0 and lim|qn0 −qn|= 0. By the triangle inequality,
|(p0n+qn0)−(pn+qn)| ≤ |p0n−pn|+|q0n−qn|
whencelim|(p0n+qn0)−(pn+qn)|= 0. It follows that{p0n+q0n} ∼ {pn+qn}, as desired.
To prove the remaining part, we write
|p0nqn0 −pnqn|=|qn0(p0n−pn) +pn(qn0 −qn)| ≤ |qn0||p0n−pn|+|pn||qn0 −qn|, and us the fact that any Cauchy sequence is bounded (see Exercise 4.8 below) to conclude
again that lim|p0nq0n−pnqn|= 0.
Exercise 4.8. Prove that a Cauchy sequence (in an arbitrary metric space) is bounded.
One can verify that the set R with the operations + and · is a field, with neutral element for addition0Rbeing[C0], the equivalence class of the constant Cauchy sequence C0 ={0,0,0,0, . . .} and identity element (for multiplication) 1R being the equivalence class [C1]of the constant Cauchy sequence C1 ={1,1,1,1, . . .}. Similarly, the additive inverse element of [pn] is [−pn] and the multiplicative inverse of [pn] 6= [C0] is [1/pn].
(Here we need to be a bit careful due to the possibility thatpn= 0for some n. On the other hand, as[pn]6= [C0], we have that for any > 0, the inequality|pn| ≥ holds for large enoughn. Thus we may pick a representativeqn of the equivalence class such that qn6= 0 for alln.) The operations are compatible with the ones on Q, soRcontains Qas a subfield.
Exercise 4.9. Work out the details above to prove thatR with the operations+and· is indeed a field.
We now want to define an order on R compatible with the one on Q. We start by defining, for any two Cauchy sequences{pn} and {qn} inQ:
{pn}<{qn} ⇐⇒there is a integer N >0 and a rational numberr >0 such thatpn+r < qn for all n≥N . Lemma 4.10. Assume that {pn}<{qn}.
If {p0n} ∼ {pn} and{q0n} ∼ {qn}, then {p0n}<{qn0}.
Proof. Since we assume that{pn}<{qn}, then, by definition, there is an integerN1 >0 and a rational numberr >0such that
(6) qn> pn+r for all n≥N1.
Since{p0n} ∼ {pn} and{q0n} ∼ {qn} there are integers N2 and N3 such that
(7) −r
3 ≤pn−p0n≤ r
3 for all n≥N2
(8) −r
3 ≤qn−q0n≤ r
3 for all n≥N3.
Hence, for n≥N := max{N1, N2, N3}, we have by (6)-(8) that qn0 ≥qn−r
3 >(pn+r)−r
3 =pn+2r
3 ≥(p0n− r 3) + 2r
3 =p0n+r 3,
whence{qn0}>{p0n}by definition.
By the lemma, we may therefore define, for any[pn],[qn]∈R,
[pn]<[qn]⇐⇒ {pn}<{qn} for any representatives{pn} and{qn}.
Lemma 4.11. The relation <is an order relation [Ru, Def. 1.5] on Rsatisfying
• [pn]<[qn]⇒[pn] + [rn]<[qn] + [rn],
• [pn],[qn]>0R⇒[pn]·[qn]>0R, for all [pn],[qn],[rn]∈R.
Exercise 4.12. Prove Lemma 4.11.
Thus, we have madeRinto an ordered field [Ru, Def. 1.17] containingQas an ordered subfield.
We finally want to prove that R satisfies the least upper bound property [Ru, Def.
1.10]. This will conclude the proof of Theorem 4.1.
Assume therefore that E is a nonempty subset ofR that is bounded above.
Let [un]∈ R be an upper bound. Since {un}, being a Cauchy sequence, is bounded (cf. Exercise 4.8), there is a rational number U such that
un≤U−1 for all n.
Hence [CU] = [{U, U, U, U, . . .}]> [un] in R and [CU] is an upper bound of E as well.
(The point is to substitute the original upper bound with an upper bound represented by aconstantsequence.)
Since E is nonempty, there is at least one element [pn]∈E. Since {pn} is bounded, again as it is a Cauchy sequence, there is a rational numberL such that
pn≥L+ 1 for all n.
Hence [CL] = [{L, L, L, L, . . .}]<[pn] and [CL] is notan upper bound of E. Also note thatL < U by construction.
We now define two sequences{xn} and {yn} inQrecursively as follows.
We setx0 =U andy0 =L. Then, having definedx0, . . . , xnandy0, . . . yn, we proceed as follows: If
h Cxn+yn
2
i
is an upper bound of E, let xn+1 := xn+y2 n and yn+1 := yn. Otherwise, letyn+1:= xn+y2 n andxn+1 :=xn.
Hence, at each step, one ofxn andyn remains the same, whereas the other one takes the value of the intermediate point between the two. Therefore|xn+1−yn+1|= 12|xn−yn|, so that
(9) |xn−yn|= |U −L|
2n It is also easy to see that
(10) yn≤yn+k ≤xn+k≤xk for all n, k≥0.
Moreover, both{xn} and {yn} are Cauchy, since (using (10)) one has, form≥n:
|xm−xn| ≤ |xm−yn|+|yn−xn|= (xm−yn)+(xn−yn)≤(xn−yn)+(xn−yn) = 2(xn−yn), which tends to0asn→ ∞ by (9). Similarly for{yn}. By (9) again, and Definition 4.5, {xn} and{yn}are equivalent, thus define the same element [xn] = [yn]∈R.
We will prove that [xn]is the least upper bound of E.
To do so, we will need:
Lemma 4.13. For any k ≥ 0, one has that [Cxk] = [{xk, xk, xk, xk, . . .}] is an upper bound for E and[Cyk] = [{yk, yk, yk, yk, . . .}]is not.
Proof. We prove this by induction onk.
By construction,[Cx0] = [CU]is an upper bound ofE.
Assume now that [Cxk] is an upper bound of E, for k ≥ 0. If xk+1 = xk, then [Cxk+1] = [Cxk]is an upper bound of E. If xk+1 < xk, then, by the recursive definition of the sequence, we must have
xk+1 = xk+yk
2 and
h Cxk+yk
2
i
is an upper bound ofE, so that[Cxk+1] =
h Cxk+yk
2
i
is an upper bound of E.
The proof concerning[Cyk]is similar.
We will now prove that [xn]is an upper bound ofE.
If it were not, there would exist some [qn] ∈ E such that [qn] > [xn]. This means that there is an integer N1 and a rational number r > 0 such that qn > xn+r for all n ≥ N1. Since {xn} is Cauchy and decreasing by (10), there is an integer N2 so that 0≤xm−xn< 12rfor alln≥m≥N2. Hence, forn≥m≥N := max{N1, N2}, we have
qn> xn+r > xm+1 2r,
whence [qn] > [CxN] = [{xN, xN, xN, xN, . . .}], contradicting the fact that [CxN] is an upper bound ofE by Lemma 4.13.
Finally, we will prove that[xn]is aleast upper bound ofE.
If it were not, there would exist some upper bound [qn]of E such that [qn]<[xn] = [yn]. This means that there is an integerN1 and a rational numberr >0such thatyn>
qn+r for alln≥N1. Since{yn}is Cauchy and increasing by (10), there is an integerN2 so that0≤yn−ym < 12r for alln≥m≥N2. Hence, for n≥m≥N := max{N1, N2}, we have
qn< yn−r < ym−1 2r,
whence[qn]<[CyN] = [{yN, yN, yN, yN, . . .}], contradicting the fact that[CyN]is not an upper bound ofE by Lemma 4.13.
We have therefore concluded the proof that[xn]is a least upper bound of E, whence thatRhas the least-upper-bound property, thus finishing the proof of Theorem 4.1.
5. Completion of a metric space This section complements [Ru, 3.8-12].
The procedure of the preceding section of constructingRfromQby adding equivalence classes of Cauchy sequences can be generalized to arbitrary metric spaces. We will follow closely the treatment in [Sh] and [Li], adjusted to fit to the notation and contents of [Ru].
We start with the following definition, which is very important in its own right Definition 5.1. LetX and Y be metric spaces, with metrics dX anddY, respectively.
An isometryfromX to Y is a bijectionf :X→Y such that dY(f(x), f(y)) =dX(x, y).
(The inverse mappingf−1 :Y →X is automatically also an isometry.)
We say thatX andY areisometric if there exists an isometry between them.
The point is that an isometry, besides inducing a one-to-one-correspondence between X and Y as sets, also preserves all distances. Thus, we may regardX and Y to be “the same space” for all practical purposes.
Recall the definition of a complete metric space from [Ru, Def. 3.12] and of a dense subset from [Ru, Def. 2.18(j)].
Definition 5.2. IfX is a metric space, with metricd, acompletion ofX is a complete metric spaceX∗, with metric d∗, such that
(i) X is a subspace ofX∗; that is, X ⊂X∗ and the metric on X is induced by the one on X∗ (meaning that d(x, y) =d∗(x, y) for all x, y∈X);
(ii) X is dense inX∗.
We will prove that any metric space admits a completion and that all such completions are isometric:
Theorem 5.3. Every metric space admits a completion that is unique up to isometry (that is, any two completions of the same metric space are isometric).
By the uniqueness property we will talk aboutthe completion of a metric space.
By [Ru, Thm. 3.11(c)] we know that R (constructed by Dedekind cuts as in [Ru, App. to Chp. 1] or by Cauchy sequences as in the previous section) is complete, and by [Ru, Thm. 1.20(b)] it contains Qas a dense subset. Hence, by uniqueness, R is the completion of Q(in its “usual” metric).
The proof of Theorem 5.3 is interesting in itself because the construction behind is typical in mathematics.
We start by proving the uniqueness statement of Theorem 5.3.
Proposition 5.4. If Y andZ are completions of a metric space X, then Y and Z are isometric.
Proof. We have bothX ⊂Y and X⊂Z. Denote the metrics inX,Y andZ bydX,dY
and dZ, respectively.
SinceX is dense inY, any pointy∈Y is a limit point ofX or a point ofX. By [Ru, Thm. 3.2(d)], there is a sequence{xn}inX converging toy. This is a Cauchy sequence inX (by [Ru, Thm. 3.11]), whence also inZ ⊃X. AsZ is complete, the sequence{xn} converges to somez∈Z.
We define a mapping
ι:Y →Z by ι(y) =z.
This is well-defined: if we choose another sequence{x0n} inX converging to y, then dZ(xn, x0n) =dX(xn, x0n) =dY(xn, x0n)→0 asn→ ∞,
since xn → y and x0n → y in Y. Hence, by the triangle inequality and the fact that xn→z inZ, we have that
dZ(x0n, z)≤dZ(x0n, xn) +dZ(xn, z)→0,
so thatx0n→zinZ. Thus, our mapping does not depend on the choice of the sequence {xn} inX.
We next prove thatιpreserves distances. Lety, y0 ∈Y with{xn}and{x0n}sequences inXsuch thatxn→yandx0n→y0. By definition ofι, we havexn→ι(y)andx0n→ι(y0) inZ. Hence
dZ(ι(y), ι(y0)) = limdZ(xn, x0n) = limdX(xn, x0n) = limdY(xn, x0n) =dY(y, y0), so thatιpreserves distances.
Finally, we prove thatι is a bijection.
Since distance is preserved, it is immediate thatι is injective: indeed, ifι(y) =ι(y0), then0 =dZ(ι(y), ι(y0)) =dY(y, y0), whence y=y0 by the property of metrics.
To prove thatιis surjective, letz∈Z. AsXis dense in Z, there exists by [Ru, Thm.
3.2(d)] again a sequence {xn} inX converging to z. Since{xn}is a Cauchy sequence in X by [Ru, Thm. 3.11], it is a Cauchy sequence also in Y ⊃X. As Y is complete, the sequence{xn}converges to somey∈Y. By construction ofι, we have thatι(y) =z.
The following generalizes Definition 4.5
Definition 5.5. Two Cauchy sequences {pn}and {qn} in a metric spaceX are said to be equivalentif the sequence {d(pn, qn)} in R converges to 0. We write {pn} ∼ {qn} if {pn}and {qn} are equivalent.
Exercise 5.6. Show that∼is an equivalence relation on the set of all Cauchy sequences inX (cf. Definition 1.3).
As in the previous section, we denote the equivalence class of{pn}by[pn]. We define X∗:={Cauchy sequences in X}/∼, the set of all equivalence classes. This set will turn out to be the one fulfilling the conditions in Definition 5.2, once we define a metric on it. To do so, we need:
Lemma 5.7. If{pn}and{qn}are Cauchy sequences inX, then the sequence{d(pn, qn)}
converges inR.
Moreover, if {p0n} ∼ {pn} and{q0n} ∼ {qn}, then limd(pn, qn) = limd(p0n, qn0).
Proof. We have, by the triangle inequality, for any m, n:
d(pn, qn)≤d(pn, pm) +d(pm, qm) +d(qm, qn), whence
|d(pn, qn)−d(pm, qm)| ≤d(pn, pm) +d(qn, qm).
Since{pn} and{qn} are Cauchy, the sum on the right is arbitrarily small form, n large enough, whence also the sequence{d(pn, qn)}inRis Cauchy. As Ris complete by [Ru, Thm. 3.11(c)], it converges.
Assume now that{p0n} ∼ {pn} and{q0n} ∼ {qn}. Then
0≤d(pn, qn)≤d(pn, p0n) +d(p0n, qn0) +d(qn0, qn),
and the first and third term on the right tend to 0 by definition of equivalent Cauchy sequences. Hence limd(pn, qn)≤limd(p0n, q0n) and the reverse inequality is obtained by
symmetry.
By this lemma, the function
∆([pn],[qn]) = limd(pn, qn)
is well-defined on X∗×X∗ (that is, independent of the choice of representative in the equivalence class) and takes values inR.
Lemma 5.8. ∆is a metric on X∗. Exercise 5.9. Prove Lemma 5.8.
Next we want to prove thatX can be identified with a subspace of X∗. To do so, we mimic the way we identifiedQwith a subset ofRin the previous section, and define, for each point p ∈ X, the constant Cauchy sequence Cp := {p, p, p, p, . . . ,} and denote its equivalence class inX∗ by [Cp].
Lemma 5.10. The mappingp7→[Cp]fromX toX∗ is injective and distance-preserving.
Proof. The mapping is injective, since if [Cp] = [Cq], then by definition of equivalence of Cauchy sequences, limd(p, q) = 0, which happens only if p=q (as the sequences are constant).
The mapping preserves distances because
∆([Cp],[Cq]) = limd(p, q) =d(p, q),
again because the sequences are constant.
Hence we can identify X with a subspace of X∗ and treat X ⊂ X∗ as a subspace.
(Formally speaking, we should say thatX is isometric to a subspace ofX∗.) Lemma 5.11. X is dense in X∗.
Proof. We must show that any [pn]∈X∗ is a limit point ofX or a point of X, cf. [Ru, Def. 2.18(j)]. By definition of Cauchy sequence, for any >0, there is anN such that d(pm, pn)< if n, m≥N. Then,
∆([pn],[CpN]) = lim
n→∞d(pn, pN)≤.
This means that any -neighborhood of [pn]intersects X in at least one point, namely [CpN]. If[CpN] = [pn], then [pn]∈X and if[CpN]6= [pn], then [pn]is a limit point ofX,
by definition.
To finish the proof of Theorem 5.3 we only have left to prove thatX∗ is complete. To do so, we exploit the density ofX inX∗ and need the following lemma.
Lemma 5.12. Every Cauchy sequence in X⊂X∗ converges to an element in X∗. Proof. Let {ck} be a Cauchy sequence inX (k= 1,2,3, . . .), that is, recalling the iden- tification of X as a subspace of ofX∗ by Lemma 5.10, we have,ck = [Cpk], for a point pk ∈ X for any k, recalling that [Cpk] is the equivalence class of the constant Cauchy sequence with terms pk. Since, for anym, n we have by Lemma 5.10,
d(pm, pn) = ∆([Cpm],[Cpn]) = ∆(cm,cn),
it follows that {pn} is a Cauchy sequence inX, thus defining an element[pn]∈X∗. We claim that {ck}converges to [pn], as k→ ∞. Indeed,
∆(ck,[pn]) = ∆([Cpk],[pn]) = lim
n→∞d(pk, pn),
which tends to 0ask→ ∞, since {pk} is Cauchy.
Lemma 5.13. X∗ is complete.
Proof. Let {xn} be a Cauchy sequence in X∗. Since X is dense in X∗ (through the identificationp7→[Cp], the equivalence class of the constant Cauchy sequence atp), we have that any neighborhood of any xn ∈ X∗ contains a point of X. Hence, for any n, there is a point yn ∈ X such that ∆([Cyn], xn) < n1. The sequence {[Cyn]} inX∗ is a Cauchy sequence, since for any m, n:
∆([Cyn],[Cym]) ≤ ∆([Cyn], xn) + ∆(xn, xm) + ∆(xm,[Cym])
< 1
n+ ∆(xn, xm) + 1 m,
and we can make this sum arbitrarily small by choosingm and nlarge enough, as{xn} is Cauchy.
By Lemma 5.12 the sequence {[Cyn]} converges to an element in X∗, say P ∈ X∗. Then
∆(xn, P)≤∆(xn,[Cyn]) + ∆([Cyn], P)< 1
n+ ∆([Cyn], P),
and this can be made arbitrarily small for largen, since {[Cyn]}tends toP. Thus,{xn}
converges to P ∈X∗, as desired.
This concludes the proof of Theorem 5.3.
6. An alternative description of compact sets
This section complements [Ru, 2.31-2.41 and 3.5-3.6] (and also uses [Ru, 3.11]).
The aim of this section is to prove the following important result, which gives three equivalent notions of compactness. It will be used in the proof of the Arzelà-Ascoli theorem 7.7 below.
Theorem 6.1. Let X be a metric space. The following conditions are equivalent:
(i) X is compact;
(ii) every infinite subset of X has a limit point in X;
(iii) every sequence in X contains a convergent subsequence.
Before proving this result, we give a couple of results hidden in the proof of [Ru, Thm.
3.6(a)]. The first one clarifies the notion of a point in a metric space being asubsequential limitof a sequence, cf. [Ru, Def. 3.5]2. The second, which builds on the first, proves the implication "(ii)⇒(iii)” in Theorem 6.1.
Lemma 6.2. Let {pn} be a sequence andp a point in a metric space X.
Thenpis a subsequential limit of{pn}if and only if for every >0there are infinitely many indicesn such that d(pn, p)< .
Remark 6.3. Note the difference between this characterization ofpbeing asubsequential limitof a sequence and of the definition ofsequence converging to p (cf. [Ru, Def. 3.1]).
Proof of Lemma 6.2. We first prove the “only if” part.
Assume that p is a subsequential limit of {pn}, that is, there are integersn1 < n2 <
n3< . . . such that the subsequence{pnk}of {pn}converges top. Then, by definition of convergence, for every >0there is an integer N such thatd(pnk, p)< for all k≥N.
In particular, there are infinitely many indices nsuch thatd(pn, p)< . We then prove the “if” part.
Assume that for every >0 there are infinitely many indicesn such that d(pn, p)<
. Then, in particular, for any i ∈ Z+, there are infinitely many indices n such that d(pn, p) < 1i. Choose n1 such that d(pn1, p) < 1. Having chosen n1, n2, . . . , ni−1, we may choose ni such that ni > ni−1 and d(pni, p) < 1i. We thus obtain a sequence pn1, pn2, pn3, . . . such thatpnk →p, whence p is a subsequential limit of {pn}.
Lemma 6.4. Let X be a metric space enjoying the property that every infinite subset of X has a limit point inX.
Then every sequence in X contains a convergent subsequence.
2As an aside we also remark that Lemma 6.2 yields a simplified proof of [Ru, Thm. 3.7]:
Theorem. The subsequential limits of a sequence{pn}in a metric spaceX form a closed subset of X.
Proof. LetE∗be the set of all subsequential limits of {pn}and letq∈X be a limit point ofE∗. We want to prove thatq ∈ E∗. Fix an >0. By definition, N
2(q)∗∩E∗6=∅, whence contains a point p(different fromq, but we will not need this), which then satisfiesd(p, q)< 2. Sincep∈E∗, then by Lemma 6.2 there exists infinitely many indicesn such thatd(pn, p) < 2. Thus, there exists infinitely many indicesnsuch thatd(pn, q)≤d(pn, p) +d(p, q)< 2+2 =. It follows by Lemma 6.2 again that qis a subsequential limits of{pn}, whenceq∈E∗, as desired.
Proof. Let {pn} be a sequence in X and let E be the range of X, that is, the set of all pn inX (cf. [Ru, p. 48]).
If E is finite, then there exists a p ∈ E such that p = pn for infinitely many n. By Lemma 6.2, the sequence {pn} contains a subsequence converging to p. (In fact, this sequence is the constant sequence {p}.)
IfE is infinite, then by our assumptions E has a limit point p. By [Ru, Thm. 2.20], for every > 0, the set N(p)∩E contains infinitely many points, which means that d(p, pn) < for infinitely many indices n. By Lemma 6.2 again, the sequence {pn}
contains a subsequence converging top.
Proof of Theorem 6.1. The fact that (i) implies (ii) is [Ru, Thm. 2.37], and the fact that (ii) implies (iii) is Lemma 6.4.
We will now prove that (iii) implies (i).
Assume therefore thatX satisfies condition (iii) and that{Uα}is an open cover ofX.
We want to find a finite subcover. We will assume that X is nonempty, otherwise there is nothing to prove.
We give two different proofs of this.
PROOF ALTERNATIVE 1: This follows the lines in [Ru, Exc. 23, 24, 26, p. 45].
Step I: X has an at most countable dense (cf. [Ru, Def. 2.18(j)]) subset.
Indeed, fix δ > 0. Pick x1 ∈ X. If Nδ(x1) $ X, choose x2 ∈ X\Nδ(x1). Having chosenx1, . . . , xk−1 ∈X, pick
xk∈X\(Nδ(x1)∪ · · · ∪Nδ(xk−1)),
if possible, that is, if the set on the right is nonempty. If the process does not terminate, we would have a sequence{xk}inXsuch thatd(xi, xj)≥δfor alli6=j. No subsequence of this sequence can therefore be Cauchy, whence (by [Ru, Thm. 3.11]) no subsequence can converge, a contradiction to assumption (iii). Hence, the process must terminate, which means that there are finitely many pointsx1, . . . , xM ∈X so that
X =Nδ(x1)∪ · · · ∪Nδ(xM).
Both M and the points x1, . . . , xM depend on δ. Hence, using δ = 1/n, we obtain for anyn∈Z+ that there is an integerMn and finitely many pointsxn,1, . . . , xn,Mn ∈X so that
(11) X=N1/n(xn,1)∪ · · · ∪N1/n(xn,Mn).
The setE :={xn,i}n∈Z+,1≤i≤Mn is at most countable (by [Ru, Thm. 2.12]; our set may be finite due to repetitions among the xn,i) and we claim that it is dense: Indeed, if x∈X andδ >0are given, then pick anyn so that1/n < δ. By (11),x must lie in one of the neighborhoods N1/n(xn,i), whence d(x, xn,i) <1/n < δ for some i, which means that Nδ(x)∩E 6=∅. Hence, by definition, x is a limit point or a point of E, so thatE is dense in X by definition (see [Ru, Def. 2.18(j)]).
Step II: X can be covered by a countable subcollection of the Uαs.
To prove this, let x ∈ X. Then x ∈ Uαx for some index αx, and since Uαx is open, there is a positive rational numberrx such that
(12) N2rx(x)⊂Uαx.
LetE ⊂X be the at most countable dense subset of X that exists by Step I. Then, as E is dense, x is either a point ofE or a limit point of E (see [Ru, Def. 2.18(j)]. Hence, Nrx(x)∩E6=∅, which means that there is ap∈E such thatd(p, x)< rx. We have
(13) x∈Nrx(p)⊂Uαx,
since for any z ∈ Nrx(p), we have d(z, x) ≤ d(z, p) +d(p, x) < rx+rx = 2rx, so that z∈N2rx(x)⊂Uαx, using (12).
To summarize, we have for any x∈ X found a Uαx in{Uα}, a p ∈E and a rational numberrx such that (13) holds. Since
{Nr(p)}p∈E,r∈Q
is a countable collection of subsets of X by [Ru, Thm. 2.12], we see from (13) that a countable collection of the subsets in {Uα} are sufficient to coverX.
Step III: Conclusion of the proof.
By Step II, we can find a countable subcollection{U1, U2, U3, . . .} of sets in {Uα} so thatX =∪∞i=1Un. We want to prove thatX can be covered by finitely many of the open sets Un. If this were not the case, we would have that
Vn:=U1∪ · · · ∪Un$X for all n∈Z+, which means that the complements Fn:= (Vnc) =X\Vn satisfy (14) Fn+1⊂Fn and Fn6=∅ for all n.
Now we can construct a sequence {xn} in X by picking one xn ∈Fn for each n∈ Z+. By assumption (iii), this sequence contains a convergent subsequence {xnk}. Letp∈X be its limit. Then, as X is covered by the sets Un, there is an integer M such that p∈UM ⊂VM. AsVM is open, there is aδ such thatNδ(p)⊂VM. AsVM = (FM)c, we have that Nδ(p)∩FM =∅, whence by (14), also
Nδ(p)∩Fn=∅ for all n≥M,
which means thatNδ(p) can contain only finitely many xn, so that p cannot be a limit of any subsequence of {xn}(by Lemma 6.2), a contradiction.
PROOF ALTERNATIVE 2: This follows closely the approach in [Li].
We start by defining a function
f :X→[0,1]
by
f(x) = sup{r ∈R|0< r <1 and Nr(x)⊂Uα for some α}.
Step I: f is positive, that is, f(x)>0 for all x∈X.
Indeed, for allx∈X, there exists an indexα0such thatx∈Uα0, asX=∪αUα. Since Uα0 is open, there exists anr such that0< r <1andNr(x)⊂Uα0. Thus,f(x)≥r >0.
Step II: |f(x)−f(y)| ≤d(x, y) for all x, y∈X.
Indeed, as f(x) ≥ 0, the inequality is immediate if f(x), f(y) ≤ d(x, y). Hence, we may without loss of generality assume that f(x) ≥f(y) and f(x) > d(x, y). Then, by definition of f, there exists anr > d(x, y) and an indexα0 such that
(15) Nr(x)⊂Uα0.
For any z∈Nr−d(x,y)(y) we have
d(z, x)≤d(z, y) +d(y, x)<(r−d(x, y)) +d(x, y) =r;
it follows that
Nr−d(x,y)(y)⊂Nr(x)⊂Uα0, for any r satisfying (15), so that
f(y)≥r−d(x, y) for any r satisfying (15).
Hence, by definition off, one hasf(y)≥f(x)−d(x, y). As we assumed thatf(x)≥f(y), we obtain that
|f(x)−f(y)|=f(x)−f(y)≤d(x, y),
as desired.
Step III: f is continuous on X.
Indeed, f is even uniformly continuous (cf. [Ru, Def. 4.18]): given > 0, choose δ=; by Step II, if d(x, y)< δ, then|f(x)−f(y)| ≤d(x, y)< δ=.
Step IV: f attains a minimum r0 on X, that is, there is a p∈X such that 0< r0 =f(p)≤f(x) for all x∈X.
Let E := {f(x) |x ∈ X} ⊂ (0,1] be the range of E. Then by [Ru, Thm. 2.28 and 3.2(c)], there is a sequence{yn=f(xn)} inE converging to infE. By assumption (iii), there is a subsequence {xnk} of xn converging to some p ∈ X, and by the continuity of f (Step III), the sequence {f(xnk)} inE converges to f(p) by [Ru, Thm. 4.2]. But any subsequence of a convergent sequence has the same limit as the sequence, whence f(p) = limf(xn) = infE. Thus,f(x)≥f(p) for all x∈X, as desired.
Step V: There are finitely many pointsx1, . . . , xn∈X such thatX=Nr0(x1)∪
· · · ∪Nr0(xn), where r0 is as in Step IV.
Indeed, if this were not true, we could pick an x1 ∈ X, an x2 ∈ X\Nr0(x1), and continuing this way, having pickedx1, . . . , xn−1 we could pick
xn∈X\
Nr0(x1)∪ · · · ∪Nr0(xn−1) ,
and the process would never stop. Hence, for allm, n, we would have d(xm, xn)≥r0, so that no subsequence of{xn} is Cauchy, whence by [Ru, Thm. 3.11], no subsequence of {xn} converges, a contradiction to assumption (iii).
Step VI: Conclusion of the proof.
By definition off and ofr0as its minimal value, for each of the points x1, . . . , xn∈X of the previous step, there is an index αi such thatNr0(xi)⊂Uαi. Hence,
X=Nr0(x1)∪ · · · ∪Nr0(xn)⊂Uα1 ∪ · · · ∪Uαn,
and we have found a finite subcover, proving thatX is compact.
We note that we also proved the following, which is used in [Ru, pf. of Thm. 7.25]:
Proposition 6.5. Every compact metric space contains an at most countable dense subset.
Proof. In Step I of alternative 1 in the last proof, we proved that condition (iii) in Theorem 6.1 implies the existence of an at most countable dense subset.
We conclude this section with the following easy result, left out in [Ru], which will be used in the proof of Theorem 7.7 below.
Proposition 6.6. Every compact metric space is bounded.
Proof. Let X be a compact metric space and pick any x ∈ X (if X is empty, there is nothing to prove). Since any point inX lies within finite distance fromx, the collection of neighborhoods
{Nn(x)}n∈Z+
of X is an open cover of X. Since X is compact, there must exist a finite subcover, which means that there is anM ∈Z+ so that
X=∪Mi=1Ni(x) =NM(x).
This implies that d(x, y) < M for all y ∈ X, whence X is by definition bounded (see
[Ru, Def. 2.18(i)]).
7. Spaces of functions and the Arzelà-Ascoli theorem This section complements [Ru, 7.19-7.25].
There are several variants of the Arzelà-Ascoli theorem, introduced around the same time by the two Italian mathematicians Cesare Arzelà (1847-1912) and Giulio Ascoli (1843-1896) [Ar, As].
One weaker version of this theorem is [Ru, Thm. 7.25]. Another version is the one giving necessary and sufficient conditions for a subspace ofC(X) to be compact whenX is a compact metric space, given in [Ru, Exc. 19, p. 168]. The aim of this section is to give a proof of this result, in Theorem 7.7 below, as well as summarizing some important results from [Ru].
We recall that, for any metric space X, one defines the set
C(X) :={f :X →C|f is continuous and bounded on X},
cf. [Ru, Def. 7.14]. (The condition of being bounded is redundant when X is compact, by [Ru, Thm. 4.15].) One makesC(X)a metric space by the metric
d(f, g) :=||f −g||:= sup
x∈X
|f(x)−g(x)|,
calledthe supremum norm metric, cf. [Ru, p. 150–151]. Note that C(X) is complete by [Ru, Thm. 7.15].
Following [Ru], we want to describe:
• convergent sequences inC(X);
• closed subsets of C(X);
• bounded subsets of C(X);
• compact subsets ofC(X).
The first three are taken care of in [Ru], although not as explicitly as one might wish, so we will summarize the results here. The last item, concerning compact subsets, is taken care of by the Arzelà-Ascoli theorem 7.7 below (but only for compact X).
We recall the following important result, mentioned briefly on [Ru, p. 151]:
Proposition 7.1. A sequence in C(X) converges (with respect to the supremum norm metric) if and only if it converges uniformly as a sequence of functions on X.
Proof. By [Ru, Thm. 7.9], a sequence of functions{fn:X →C}converges uniformly to a functionf :X→Cif and only if
(16) sup
x∈X
|fn(x)−f(x)| −→0 as n→ ∞,
which by definition of the supremum norm metric is equivalent to saying that
||fn−f|| −→0 as n→ ∞.
Therefore, the “only if” part follows, while the “if” part follows once we prove that f ∈ C(X) if all fn∈ C(X), that is, if allfn are continuous and bounded, then so is the limit function f.
The fact thatf is continuous follows from [Ru, Thm. 7.12]. The fact that it is bounded follows as there by (16) is anN such that
sup
x∈X
|fN(x)−f(x)| ≤1 and, asfN is bounded onX, anM ∈Rso that
|fN(x)| ≤M for all x∈X.
Hence
|f(x)| ≤ |f(x)−fN(x)|+|fN(x)| ≤1 +M for all x∈X,
and f is bounded, as desired.
This result makes it possible to describe closed subsets ofC(X) explicitly. Indeed, we have the following general result valid on any metric spaceX:
Lemma 7.2. Let E be a subset of a metric space X. Then its closure E (recall [Ru, Def. 2.26]) can be described as
(17) E={p∈X |pis the limit of a sequence inE}.
In particular,E is closed if and only if it contains all limits of sequences in E.
Proof. We first prove the containment “⊂” in (17).
Ifp∈E, thenp∈E or pis a limit point ofE. Ifp∈E, it is the limit of the constant sequence {p, p, p, . . .}in E. Ifp is a limit point of E, then by [Ru, Thm. 3.2(d)], there exists a sequence in E converging to p.
We then prove the containment “⊃” in (17).
If p ∈X is the limit of a sequence {pn} inE, then by definition of convergence, for any >0, the neighborhoodN(p) contains all but finitely many of the terms of {pn}.
In particular,N(p)∩E 6=∅, so that eitherp∈E orpis a limit point ofE by definition.
Hencep∈E.
Since E = E if and only if E is closed, by [Ru, Thm. 2.27(b)], the last assertion
follows.
We recall the following definitions (cf. [Ru, Def. 7.28], where the definition is made only for algebras of functions):
Definition 7.3. A family of complex functionsF defined on a set Xisuniformly closed if it contains all complex functions f satisfying the property that there is a sequence {fn}of functions in F converging uniformly to f.
The uniform closure of F is the set of complex functions f satisfying the property that there is a sequence{fn}of functions in F converging uniformly to f.
Therefore, an immediate consequence of Proposition 7.1 and Lemma 7.2 is the follow- ing.
Proposition 7.4. A subset S ⊂ C(X) is closed (with respect to the supremum norm metric in C(X)) if and only if it is uniformly closed as a family of functions on X.
The closure of S equals its uniform closure as a family of functions.
To describe bounded subsets ofC(X), we recall the following definitions (cf. [Ru, Def.
7.19], where the definition is made only for countable sets, that is, for sequences):
Definition 7.5. A family of complex functionsFdefined on a setXispointwise bounded if the set {f(x)}f∈F is bounded inC for every x∈X, that is, for every x∈X, there is an M(x)∈R (depending onx) such that
|f(x)| ≤M(x) for allf ∈ F.
A family of complex functions F defined on a set X is uniformly bounded if the set {f(x)}x∈X,f∈F is bounded in C, that is, there is an M ∈Rsuch that
(18) |f(x)| ≤M for allx∈X and f ∈ F.
In particular, a uniformly bounded set of functions is automatically pointwise bounded.
We have:
Proposition 7.6. A subset S ⊂ C(X) is bounded (with respect to the supremum norm metric in C(X)) if and only if it is uniformly bounded as a family of functions on X.