ADDENDUM B: CONSTRUCTION OF R AND THE COMPLETION OF A METRIC SPACE

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1 ADDENDUM B: CONSTRUCTION OF R AND THE COMPLETION OF A METRIC SPACE ANDREAS LEOPOLD KNUTSEN Abstract. These notes are written as supplementary notes for the course MAT11- Real Analysis, taught at the University of Bergen, Fall 015. They are meant to supplement [Ru,.1-.3 and ]. 1. Relations and equivalence relations We start with a couple of denitions (almost) left out in [Ru]. We also recall the following terminology not used in [Ru]: We denote a function f from a set A to a set B by f : A B and say that f is injective or an injection if it is one-to-one; f is surjective or a surjection if it is onto; f is bijective or a bijection if it is both one-to-one and onto. We also recall that if f is bijective, there is a well-dened inverse mapping g : B A dened as g(y) = the unique x A such that f(x) = y. Applying f rst and then g simply brings x back to x. The inverse map g is often denoted by f 1, which should not be confused with the notation of inverse images as in [Ru, Def..]. Denition 1.1. Let X be a set. A relation R on X is 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 notation x y. Denition 1.. An equivalence relation is a relation satisfying the following properties: (i) Reexivity: x x for all x X. (ii) Symmetry: if x y, then y x, for all x, y X. (iii) Transitivity: if x y and y z, then x z, for all x, y, z X. Example 1.3. The order relation x y x < y on R dened 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 1.4. The relation on the set of all sets in [Ru, Def..3] dened by A B bijection f : A B (meaning in the language of [Ru, Def..3] that the sets A and B can be put in oneto-one correspondence) is an equivalence relation: Reexivity holds because the identity map id : A A mapping x to x is a bijection. Symmetry holds because a bijection f : A B has an inverse map 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 C dened by (g f)(x) = g(f(x)) 1

2 ANDREAS LEOPOLD KNUTSEN is a bijection. Example 1.5. Fix any m N. Let X = Z. The relation on Z dened by is an equivalence relation. Example 1.6. The relation on C given by is an equivalence relation. x y m divides x y z 1 z z 1 = z Exercise 1.7. Prove that the relations in the last two examples are indeed equivalence relations. Denition 1.8. Given an equivalence relation on a set X and x X, we dene the equivalence class [x] of x to be the set: [x] := {y X y x}. Any element in the equivalence class is called a representative of the equivalence class. We often denote the set of equivalence classes by X/. Using the properties of equivalence relations, it is not dicult to prove that any x X lies in precisely one equivalence class. Indeed, by reexivity, x x, so that x [x], whence x lies in at least one equivalence class. If x [y 1 ] and x [y ], then x y 1 and x y. By symmetry, we have y 1 x. But this, together with x y and transitivity, yields that y 1 y, that is, [y 1 ] = [y ]. Hence, X/ is a collection of disjoint subsets of X whose union is the whole of X. (Such a collection is called a partition.) Example 1.9. Consider Example 1.5 and let n Z. Then the equivalence class of n is [n] = {k Z m divides n k} = {n, n ± m, n ± m, n ± 3m,...}. Clearly [n] contains a unique number n 0 satisfying 0 n 0 m 1. Thus the set of equivalence classes Z/, which is often denoted by Z/mZ or Z m, is in one-to-one correspondence with the set {0, 1,,..., m 1}. Example Consider Example 1.6 and let z 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.. Cantor's construction of R from Q by means of Cauchy sequences In this section, we will go through Cantor's construction of the real numbers by using Cauchy sequences, and this idea will be generalized when we construct completions of arbitrary metric spaces in the next section. The aim is to give a dierent, independent proof of [Ru, Thm. 1.19]: Theorem.1. There exists an ordered eld R with the least-upper-bound property and containg Q as an ordered subeld.

3 CONSTRUCTION OF R AND THE COMPLETION OF A METRIC SPACE 3 Before starting the proof, we remark that we do not need to know beforehand what R is to talk about distances in Q and Cauchy sequences: Indeed, the number p q Q for any p, q Q and satises the denition of a metric. Moreover, we say that a sequence {p n } in Q is a Cauchy sequence if for any rational ɛ > 0, there is an integer N such that p m p n < ɛ whenever m, n N and that it converges to x Q if for any rational ɛ > 0, there is an integer N such that p n x < ɛ whenever n N. We recall the following: Proposition.. There are Cauchy sequences in Q that 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 satises the least-upper-bound nor the greatest-lower boundproperty. The association in [Ru, (3)] of a rational number q to any rational number p denes a sequence {p n } recursively by (1) p n+1 = p n p n p n + = p n + which is easily checked (by induction) to be increasing if p 1 < and decreasing if p 1 >. The sequence is a Cauchy sequence. If the limit exists, say L = lim p n, then taking limits in (1) shows that L =, whence the limit cannot exist in Q, by [Ru, Ex. 1.1]. Exercise.3. Prove that {p n } in the proof of Proposition. is a Cauchy sequence (without using the knowledge that it converges in R). Remark.4. There are other recursive relations yielding Cauchy sequences whose limits L, if they exist, must satisfy L =, for instance x n+1 = xn + 1 x n. The moral of Cantor's construction is the following: since Q lacks limits of Cauchy sequences in Q, 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 would be represented by the Cauchy sequence {p n } in the proof of Proposition.. However, we have to take into account that dierent Cauchy sequences may yield the same limits, for instance dierent starting points p 1 above yield dierent Cauchy sequences {p n } still representing, and the same applies for the sequences x n in Remark.4. It is to avoid this ambiguity that equivalences enter the picture: Denition.5. Two Cauchy sequences {p n } and {q n } in Q are said to be equivalent if the sequence { p n q n } in Q converges to 0. We write {p n } {q n } if {p n } and {q n } are equivalent. Exercise.6. Show that is an equivalence relation on the set of all Cauchy sequences in Q. We denote by [p n ] the equivalence class of {p n } and dene R to be the set of all equivalence classes of Cauchy sequences on Q, that is, with the notation of Denition 1.8, we have R := {Cauchy sequences in Q}/. We note that Q can be identied with a subset of R by identifying q Q with the constant Cauchy sequence C q := {q, q, q, q,...}. It is clear that C p C q if and only if p = q, so Q is in a natural way a subset of the newly dened set R. We dene two operations + and on R that are compatible with the ones on the subset Q: [p n ] + [q n ] := [p n + q n ] and [p n ] [q n ] := [p n q n ].

4 4 ANDREAS LEOPOLD KNUTSEN We have to prove that these operations are well-dened, that is, that they are independent of the choice of representative for the equivalence classes of Cauchy sequences. This is taken care of by the following: Lemma.7. If {p n} {p n } and {q n} {q n }, then {p n + q n} {p n + q n } and {p n q n} {p n q n }. Proof. By denition, {p n} {p n } and {q n} {q n } means that lim p n p n = 0 and lim q n q n = 0. By the triangle inequality, (p n + q n) (p n + q n ) p n p n + q n q n whence lim (p n + q n) (p n + q n ) = 0. It follows that {p n + q n} {p n + q n }, as desired. To prove the remaining part, we write p nq n p n q n = q n(p n p n ) + p n (q n q n ) q n p n p n + p n q n q n, and us the fact that any Cauchy sequence is bounded (see Exercise.8 below) to conclude again that lim p nq n p n q n = 0. Exercise.8. Prove that a Cauchy sequence (in any metric space) is bounded. One can verify that the set R with the operations + and is a eld, with neutral element for addition 0 R being the equivalence class of the constant Cauchy sequence C 0 and identity element (for multiplication) 1 R being the equivalence class of the constant Cauchy sequence C 1. Similarly, the additive inverse element of [p n ] is [ p n ] and the multiplicative inverse of [p n ] C 0 is [1/p n ]. (Here we need to be a bit careful due to the possibility that p n = 0 for some n. On the other hand, as [p n ] C 0, we have that for any ɛ > 0, the inequality p n ɛ holds for large enough n. Thus we may pick a representative q n of the equivalence class such that q n 0 for all n.) The operations are compatible with the ones on Q, so R contains Q as a subeld. Exercise.9. Work out the details above to prove that R with the operations + and is indeed a eld. We now want to dene an order on R compatible with the one on Q. We start by dening, for any two Cauchy sequences {p n } and {q n } in Q: {p n } < {q n } there is a integer N > 0 and a rational number r > 0 Lemma.10. Assume that {p n } < {q n }. If {p n} {p n } and {q n} {q n }, then {p n} < {q n}. such that p n + r < q n for all n N. Proof. Assume that {p n } < {q n }. Then by denition, there is an integer N 1 > 0 and a rational number r > 0 such that () q n > p n + r for all n N 1. Since {p n} {p n } and {q n} {q n } there are integers N and N 3 such that (3) r 3 p n p n r 3 for all n N (4) r 3 q n q n r 3 for all n N 3. Hence, for n N := max{n 1, N, N 3 }, we have by ()-(4) that q n q n r 3 > (p n + r) r 3 = p n + r 3 (p n r 3 ) + r 3 = p n + r 3, whence {q n} > {p n} by denition.

5 CONSTRUCTION OF R AND THE COMPLETION OF A METRIC SPACE 5 By the lemma, we may therefore dene, for any [p n ], [q n ] R, [p n ] < [q n ] {p n } < {q n } for any representatives {p n } and {q n }. Lemma.11. The relation < is an order relation [Ru, Def. 1.5] on R satisfying [p n ] < [q n ] [p n ] + [r n ] < [q n ] + [r n ], [p n ], [q n ] > 0 R [p n ] [q n ] > 0 R, for all [p n ], [q n ], [r n ] R. Exercise.1. Prove Lemma.11. Thus, we have made R into an ordered eld [Ru, Def. 1.17] containing Q as an ordered subeld. We nally would like to prove that R satises the least upper bound property [Ru, Def. 1.10]. This would conclude the proof of Theorem.1. Assume therefore that E is a nonempty subset of R that is bounded above. Let [u n ] R be an upper bound. Since the representative {u n } is a Cauchy sequence, there is an integer N such that, for all n N: u n u n u n u N + u N < 1 + u N Q. Hence, setting U := u n + Q, we have u n + 1 < U Q, so that C U [u n ] in R and C U is an upper bound of E as well. (The point is to substitute the original upper bound with an upper bound represented by a constant sequence.) Since E is nonempty, there is at least one element [p n ] E. As above, p n < 1 + p N for all n N for some integer N, whence p n 1 > p N =: L Q, so that C L < [p n ] and C L is not an upper bound of E. We now dene two sequences {x n } and {y n } in Q recursively as follows. We set x 0 = U and y 0 = L. Then, having dened x 0,..., x n and y 0,... y n, we proceed as follows: If C xn+yn Otherwise, let y n+1 := xn+yn is an upper bound of E, let x n+1 := xn+yn and y n+1 := y n. and x n+1 := x n. Hence, at each step, one of x n and y n remains the same, whereas the other one takes the value of the intermediate point between the two. Therefore x n+1 y n+1 = 1 x n y n, so that (5) x n y n = It is also easy to see that U L n (6) y n y n+k x n+k x k for all n, k 0. Moreover, both {x n } and {y n } are Cauchy, since (using (6)) one has, for m n: x m x n x m y n + y n x n = (x m y n )+(x n y n ) (x n y n )+(x n y n ) = (x n y n ), which tends to 0 as n by (5). Similarly for {y n }. By (5) again, and Denition.5, {x n } and {y n } are equivalent, thus dene the same element [x n ] = [y n ] R. We will prove that [x n ] is the least upper bound of E. To do so, we will need: Lemma.13. For any k 0, one has that C xk is an upper bound for E and C yk is not.

6 6 ANDREAS LEOPOLD KNUTSEN Proof. We prove this by induction on k. By construction, C x0 = C U is an upper bound of E. Assume now that C xk is an upper bound of E, for k 0. If x k+1 = x k, then C xk+1 = C xk is an upper bound of E. If x k+1 < x k, then, by the recursive denition of the sequence, we must have x k+1 = x k + y k and C x k +y k is an upper bound of E, so that C xk+1 = C x k +y k is an upper bound of E. The proof concerning C yk is similar. We will now prove that [x n ] is an upper bound of E. If it were not, there would exist some [q n ] E such that [q n ] > [x n ]. This means that there is an integer N 1 and a rational number r > 0 such that q n > x n + r for all n N 1. Since {x n } is Cauchy and decreasing by (6), there is an integer N so that 0 x m x n < 1 r for all n m N. Hence, for n m N := max{n 1, N }, we have q n > x n + r > x m + 1 r, whence [q n ] > C xn, contradicting the fact that C xn is an upper bound of E by Lemma.13. Finally, we will prove that [x n ] is a least upper bound of E. If it were not, there would exist some upper bound [q n ] of E such that [q n ] < [x n ] = [y n ]. This means that there is an integer N 1 and a rational number r > 0 such that y n > q n + r for all n N 1. Since {y n } is Cauchy and increasing by (6), there is an integer N so that 0 y n y m < 1 r for all n m N. Hence, for n m N := max{n 1, N }, we have q n < y n r < y m 1 r, whence [q n ] < C yn, contradicting the fact that C yn is not an upper bound of E by Lemma.13. We have therefore concluded the proof that [x n ] is a least upper bound of E, whence that R has the least-upper-bound property, thus nishing the proof of Theorem Completion of a metric space The procedure of the preceding section of constructing R from Q by 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 t to the notation and contents of [Ru]. We start with the following denition, which is very important in its own right Denition 3.1. Let X and Y be metric spaces, with metrics d X and d Y, respectively. An isometry from X to Y is a bijection f : X Y such that d Y (f(x), f(y)) = d X (x, y). (The inverse map f 1 : Y X is automatically also an isometry.) We say that X and Y are isometric 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 regard X and Y to be the same space for all practical purposes. Recall the denition of a complete metric space from [Ru, Def. 3.1] and of a dense subset from [Ru, Def..18(j)].

7 CONSTRUCTION OF R AND THE COMPLETION OF A METRIC SPACE 7 Denition 3.. If X is a metric space, with metric d, a completion of X is a complete metric space X, with metric d, such that (i) X is a subspace of X ; 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 in X. We will prove that any metric space admits a completion and that all such completions are isometric: Theorem 3.3. Any 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 about the 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. 1.0(b)] it contains Q as a dense subset. Hence, by uniqueness, R is the completion of Q (in its usual metric). The proof of Theorem 3.3 is interesting in itself because the construction behind is typical in mathematics. We start by proving the uniqueness statement of Theorem 3.3. Proposition 3.4. If Y and Z are completions of a metric space X, then Y and Z are isometric. Proof. We have both X Y and X Z. Denote the metrics in X, Y and Z by d X, d Y and d Z, respectively. Since X is dense in Y, any point y Y is a limit point of X or a point of X. By [Ru, Thm. 3.(d)], there is a sequence {x n } in X converging to y. This is a Cauchy sequence in X (by [Ru, Thm. 3.11]), whence also in Z X. As Z is complete, the sequence {x n } converges to some z Z. We dene a mapping ι : Y Z by ι(y) = z. This is well-dened: if we choose another sequence {x n} in X converging to y, then d Z (x n, x n) = d X (x n, x n) = d Y (x n, x n) 0 as n, since x n y and x n y in Y. Hence, by the triangle inequality and the fact that x n z in Z, we have, for any ɛ > 0, that d Z (x n, z) d Z (x n, x n ) + d Z (x n, z) 0, so that x n z in Z. Thus, our mapping does not depend on the choice of the sequence {x n } in X. We next prove that ι preserves distances. Let y, y Y with {x n } and {x n} sequences in X such that x n y and x n y. By denition of ι, we have x n ι(y) and x n ι(y ) in Z. Hence d Z (ι(y), ι(y ) = lim d Z (x n, x n) = lim d X (x n, x n) = lim d Y (x n, x n) = d Y (y, y ), so that ι preserves distances. Finally, we prove that ι is a bijection. Since distance is preserved, it is immediate that ι is injective: indeed, if ι(y) = ι(y ), then 0 = d Z (ι(y), ι(y )) = d Y (y, y ), whence y = y by the property of metrics. To prove that ι is surjective, let z Z. As X is dense in Z, there exists by [Ru, Thm. 3.(d)] again a sequence {x n } in X converging to z. Since {x n } 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 {x n } converges to some y Y. By construction of ι, we have that ι(y) = z.

8 8 ANDREAS LEOPOLD KNUTSEN The following generalizes Denition.5 Denition 3.5. Two Cauchy sequences {p n } and {q n } in a metric space X are said to be equivalent if the sequence {d(p n, q n )} in R converges to 0. We write {p n } {q n } if {p n } and {q n } are equivalent. Exercise 3.6. Show that is an equivalence relation on the set of all Cauchy sequences in X. As in the previous section, we denote the equivalence class of {p n } by [p n ]. We dene X := {Cauchy sequences in X}/, the set of all equivalence classes. This set will turn out to be the one fullling the conditions in Denition 3., once we dene a metric on it. To do so, we need: Lemma 3.7. If {p n } and {q n } are Cauchy sequences in X, then the sequence {d(p n, q n )} converges in R. Moreover, if {p n} {p n } and {q n} {q n }, then lim d(p n, q n ) = lim d(p n, q n). Proof. We have, by the triangle inequality, for any m, n: whence d(p n, q n ) d(p n, p m ) + d(p m, q m ) + d(q m, q n ), d(p n, q n ) d(p m, q m ) d(p n, p m ) + d(q n, q m ). Since {p n } and {q n } are Cauchy, the sum on the right is arbitrarily small for m, n large enough, whence also the sequence {d(p n, q n )} in R is Cauchy. As R is complete by [Ru, Thm. 3.11(c)], it converges. Assume now that {p n} {p n } and {q n} {q n }. Then 0 d(p n, q n ) d(p n, p n) + d(p n, q n) + d(q n, q n ), and the rst and third term on the right tend to 0 by denition of equivalent Cauchy sequences. Hence lim d(p n, q n ) lim d(p n, q n) and the reverse inequality is obtained by symmetry. By this lemma, the function ([p n ], [q n ]) = lim d(p n, q n ) is well-dened on X X (that is, independent of the choice of representative in the equivalence class) and takes values in R. Lemma 3.8. is a metric on X. Exercise 3.9. Prove Lemma 3.8. Next we want to prove that X can be identied with a subspace of X. To do so, we mimic the way we identied Q with a subset of R in the previous section, and dene, for each point p X, the constant Cauchy sequence {p, p, p, p,..., } and denote its equivalence class in X by C p. Lemma The map p C p from X to X is injective and distance-preserving. Proof. The map is injective, since if C p = C q, then by denition of equivalence of Cauchy sequences, lim(p q) = 0, which happens only if p = q (as the sequences are constant). The map preserves distances because again because the sequences are constant. (C p, C q ) = lim d(p, q) = d(p, q),

9 CONSTRUCTION OF R AND THE COMPLETION OF A METRIC SPACE 9 Hence we can identify X with a subspace of X and treat X X as a subspace. (Formally speaking, we should say that X is isometric to a subspace of X.) Lemma X is dense in X. Proof. We must show that any [p n ] X is a limit point of X or a point of X, cf. [Ru, Def..18(j)]. By denition of Cauchy sequence, for any ɛ > 0, there is an N such that d(p m, p n ) < ɛ if n, m N. Then, ([p n ], C pn ) = lim n d(p n, p N ) ɛ. This means that any ɛ-neighborhood of [p n ] intersects X in at least one point, namely C pn. If C pn = [p n ], then [p n ] X and if C pn [p n ], then [p n ] is a limit point of E, by denition. To nish the proof of Theorem 3.3 we only have left to prove that X is complete. To do so, we exploit the density of X in X and need the following lemma. Lemma 3.1. Any Cauchy sequence in X X converges to an element in X. Proof. Let {C k } be a Cauchy sequence in X (k = 1,, 3,...), that is, we have, C k = C pk, for a point p k X for any k, recalling that C pk is the equivalence class of the constant Cauchy sequence with terms p k. Since, for any m, n: d(p m, p n ) = (C pm, C pn ) = (C m, C n ), we have that {p n } is a Cauchy sequence in X, thus dening an element [p n ] X. We claim that {C k } converges to [p n ], as k. Indeed, (C k, [p n ]) = (C pk, [p n ]) = lim n d(p k, p n ), which tends to 0 as k, since {p k } is Cauchy. Lemma X is complete. Proof. Let {x n } be a Cauchy sequence in X. Since X is dense in X (through the identication p C p, the equivalence class of the constant Cauchy sequence at p), we have that any neighborhood of any x n X contains a point of X. Hence, for any n, there is a point y n X such that (C yn, x n ) < 1 n. The sequence {C y n } in X is a Cauchy sequence, since for any m, n: (C yn, C ym ) (C yn, x n ) + (x n, x m ) + (x m, C ym ) < 1 n + (x n, x m ) + 1 m, and we can make this sum arbitrarily small by choosing m and n large enough, as {x n } is Cauchy. By Lemma 3.13, the sequence {C yn } converges to an element in X, say P X. Then (x n, P ) (x n, C yn ) + (C yn, P ) < 1 n + (C y n, P ), and this can be made arbitrarily small for large n as {C yn } tends to P. Thus, {x n } converges to P X, as desired. This concludes the proof of Theorem 3.3.

10 10 ANDREAS LEOPOLD KNUTSEN References [Li] [Ru] [Sh] T. Lindstrøm, Mathematical Analysis, notes used in the course MAT400 at the University of Oslo. W. Rudin, Principles of mathematical analysis, International Series in Pure and Applied Mathematics, Third Edition, McGraw-Hill, G. E.Shilov, Elementary real and complex analysis. Revised English edition translated from the Russian and edited by Richard A. Silverman. Corrected reprint of the 1973 English edition. Dover Publications, Inc., Mineola, NY, Andreas Leopold Knutsen, Department of Mathematics, University of Bergen, Postboks 7800, 500 BERGEN, Norway address: