(C) The rationals and the reals as linearly ordered sets. Contents. 1 The characterizing results

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1 (C) The rationals and the reals as linearly ordered sets We know that both Q and R are something special. When we think about about either o these we usually view it as a ield, or at least some kind o algebraic structure. In this document we consider each merely as linearly ordered set. Even in this seemingly narrow context they are still something special. Contents 1 The characterizing results Basic properties Exercises: Basic properties Problems: Basic properties Embedding properties Exercises: Embedding properties Problems: Embedding properties The back-and-orth technique Exercises: The back-and-orth technique Problems: The back-and-orth technique The characteristic properties o R The characterizing results As indicated in the preamble, in these notes we consider the characterization o Q and R as linear sets. We prove look at two main results and these can be stated here. We will prove Theorem 1.1 but leave the proo o Theorem 2.2 or later. O course, in the statements o these results there are one or two technical words which have not yet been deined. I will indicate these, but leave the ormal deinitions until later. In the ollowing a line is a linear set with a certain extra property that both Q and R clearly have. 1.1 THEOREM. The rationals Q as a linear set has the ollowing properties. (i) Q is a countable line. (ii) Q is embedded in every line. (iii) Each countable linear set is embedded in Q. (iv) Each pair o countable lines are isomorphic. This is the characterization o Q. We also obtain a similar characterization o R. In the ollowing separable and complete are two technical properties that R has. 1.2 THEOREM. The reals R as a linear set has the ollowing properties. (i) R is a separable complete line. (ii) R is embedded in every complete line. 1

2 (iii) Each separable line is embedded in R. (iv) Each pair o separable complete lines are isomorphic. It is worth comparing these two characterizations. To do that let s state a cod result. In it sizable and string are used in a pseudo-technical sense. 1.3 THEOREM. The gadget G as a linear set has the ollowing properties. (i) G is a sizable string. (ii) G is embedded in every string. (iii) Each sizable linear set is embedded in G. (iv) Each pair o sizable strings are isomorphic. We can see that Theorems 1.1 and 1.2 have the orm o Theorem 1.3. We interpret the cod words as ollows. (1.1) countable line (1.2) separable complete line (1.3) sizable string (1.4) c.c.c. complete line In both cases the word string is a property that a linear ordered set may or may not have. However, there seem to be a disparity between the two meanings o sizable. For Q it is a cardinality condition, namely being countable. But or R it is a topological condition. This will be explained in more detailed later. But why do we need a topological condition? Can we replace being separable by some cardinality condition? There is an obvious condition to look at, that o being c.c.c (having the countable chain condition). This is weaker than being separable and making this change gives the ollowing. separable = c.c.c. 1.4 THEOREM. The reals R as a linear set has the ollowing properties. (i) R is a c.c.c. complete line. (ii) R is embedded in every complete line. (iii) Each c.c.c. line is embedded in R. (iv) Each pair o c.c.c. complete lines are isomorphic. Here part (i) is a consequence o the corresponding part o Theorem 1.2, and part (ii) is just the same. However, parts (iii, iv) are stronger than the corresponding parts o Theorem 1.2. Now the world becomes weired. It is know that parts (iii, iv) o Theorem 1.4 can never be proved or disproved without some extra hypothesis. The two statements are independent o the standard axioms o Axiomatic Set Theory. There are certain set theoretic universes in which the results are true, and other set theoretic universes in which the results are alse. We will not look at these independence results. The methods used are much to complicated to explain here. However, later we may look at the two conditions o being separable and having c.c.c., and observe that one implies the other almost trivially. 2

3 2 Basic properties These notes are concerned with linearly ordered sets in general, and the important role played by Q and R in this context. We need to set down the basic notions. 2.1 DEFINITION. A linear set (linearly ordered set) is a set A with a speciied linear comparison, that is a binary relation on A which is (Relexive) a a (Transitive) a b c = a c (Antisymmetric) a b & b a = a = b (Linear) a b or b a or all a, b, c A. We sometimes use the strict version < o, that is a < b a b and a b or a, b A. The term linear set is not standard, but it is useul since it is shorter than linearly ordered set. Let s look at some examples. Some o these are obvious and some not. The veriication o some aspects are let as exercises. 2.2 EXAMPLE. (a,b) Each o N, Z, Q, R is a linear set. There is a clear dierence between the pair N, Z and the pair Q, R. There is also a dierence between Q and R which is not so obvious. (c) Let A be the set o eventually zero members o [N 2]. Thus each member o A is a unction : N 2 with (m) = 0 or all suiciently large m. Consider the ollowing two relations on A. (last) (irst) g = g or ( m)[(m) < g(m) & ( n > m)[(n) = g(n)] g = g or ( m)[(m) < g(m) & ( n < m)[(n) = g(n)] These are the principles o last dierence and irst dierence, respectively. These two comparisons make A into a linear set in two dierent ways. In act, these are two well-known linear sets in disguise. (d) Let A be the set o all unctions [N 2] compared by irst dierence. This is a linear set and does have a more concrete description. Notice that it doesn t make sense to compare all such unction by last dierence. However, i we consider those unction which are eventually constant, then we can use the last dierence comparison. (e) Let A be any linear set. A lower section, initial section, is a subset X A such that y x X = y X holds. Let LA be the amily o all lower sections. This is a linear set. 3

4 As with any kind o structure, we need to decide how linear sets should be compared. One o the ollowing gives the correct notion. The other is also useul at times. 2.3 DEFINITION. Let A, B be a pair o linear sets. A unction : A B is monotone antitone i a 1 a 2 = (a 1 ) (a 2 ) a 1 a 2 = (a 2 ) (a 1 ) or all a 1, a 2 A. The standard comparison, a morphism, between linear sets is a monotone unction. But antitone unction are useul occasionally. Here we use a more restricted kind o comparison unction. 2.4 DEFINITION. (a) An embedding rom one linear set to another A B is a monotone unction that is also injective. In other words, it is a unction, as indicated, where a 1 a 2 (a 1 ) (a 2 ) or all a 1, a 2 A. (b) An isomorphism rom one linear set to another A B is a monotone unction which is also a bijection or which is inverse unction A g B is also monotone. In this document we mostly consider only embeddings between linear sets. However, monotone unctions do arise in other parts o the whole collection o documents. Also in this document we oten restrict out attention to a special kind o linear set. 2.5 DEFINITION. Let A be a linear set. This set is dense i or each comparable pair a < b o distinct elements there is a strictly intermediate elements, that is a < x < b or some element x. The set is without end points i or each element a we have y < a < z or some elements y, z. A line is a dense linear set without end points. 4

5 The use o line in this sense is not standard. In the literature you will oten see dense linearly ordered set without end points which is a bit long. The word line does seem to capture the intuitive idea a bit better. Let s have another look at the Examples EXAMPLES. (a) Neither N nor Z is a line, mainly because neither is dense. (b) Both Q and R are lines. (c) Let A be the set o eventually zero sequences. When compared by last dierence the linear set A is nowhere near a line. This is considered in one o the exercises. When compared by irst dierence this is not quite a line. It is dense and has no last point, but does have a irst point. Again this is considered in one o the exercises. (d) Let A be the set o all unctions : N 2, compared by irst dierence. This has a irst point, the constant 0 unction. It also has a last point, the constant 1 unction. Between these the linear set can look rather odd. (e) As a particular example consider the linear set LQ o initial sections o Q. This has end points, namely the empty set and the whole set. For each a Q the set is a member o LQ. The assignment a = {x Q x a} Q a LQ a is an embedding. Think about what else might occur in LQ. There is a copy o R and the rationals have a rather split personality. Linear sets are not quite as simple as we might think when we irst see them. Here we look at some o these aspects but concentrate on Q and R. 2.1 Exercises: Basic properties 2.1 (a) Consider three linear sets, and a composible pair o unctions A B g C between them. Suppose each o the unctions is either monotone or antitone (so there are our cases). What property does the composite have? (b) Show that or each antitone unction, i we take the opposite o the source or the target then we get a monotone unction. You will need to sort out what the opposite o a linear set is. 2.2 (a) Consider the two versions o embedding as given in Deinition 2.4. Show that they are equivalent. (b) Show that a monotone unction between to lines is an isomorphism precisely when it is a surjective embedding. 5

6 2.3 (a) Show that, as linear sets, each two open ended real intervals are isomorphic. And remember that ± can be end points. (b) Show there are ininitely many pairwise non-isomorphic lines o cardinality ℶ 1. (c) Recall that the dyadic rationals are those o the orm n/d where the denominator d is a power o 2. Show that the dyadic rationals orm a line. 2.4 Let A be the set o eventually zero members o [N 2], and consider the ollowing two relations on A. (last) (irst) g = g or ( m)[(m) < g(m) & ( n > m)[(n) = g(n)] g = g or ( m)[(m) < g(m) & ( n < m)[(n) = g(n)] These are the principles o last dierence and irst dierence, respectively. Show that each o these linearly orders A, and describe a more natural example o the corresponding order type. 2.2 Problems: Basic properties 2.5 Consider the set [N 2]. For a typical member x( ) o this set o unctions let x = x(r)2 r to obtain a real number x. (a) Show that 0 x 2. (b) Show that the assignment is surjective. (c) Show that or all x, y [0, 2]. (d) Show that r=0 [N 2] [0, 2] x( ) x x y ( r)[x(r) y(r)] φ [0, 2] [0, 3] x 2 x(r)3 r is a linear embedding, and describe its range. (e) As a linear set, where have you seen the range o φ beore. 2.6 Think about the unction Q a r=0 LQ a given by Exercise 2.6(e). (a) Show that this is an embedding. (b) Give a concrete description o the linear set LQ as a modiied version o R. 6

7 3 Embedding properties How might we prove Theorem 1.2? Part (i) is trivial (or those who can count). In this section we prove parts (ii) and (iii). To do that we set up an iteration procedure which produces an embedding bit by bit. To do that we need a notion o a partial isomorphism. 3.1 DEFINITION. Let A, B be a pair o linear sets. A partial isomorphism rom A to B is a pair o subsets U A V B together with a bijection : U V which is an isomorphism when U, V are viewed as linear sets, that is or a, x U. x a (x) (a) An ew examples never goes amiss. Also part o the ollowing will be useul later. 3.2 EXAMPLE. Notice that a ull isomorphism A B is also a partial isomorphism with U = A and V = B. A inite partial isomorphism rom A to B is a pair o lists a 1 < < a m b 1 < b m rom the two sets where these are the same length and the bijection a i b i or 1 i m is a partial isomorphism. We make use o inite partial isomorphism later. A partial isomorphism is an embedding as ar as it can be. In the notion above, the partial isomorphism embeds U into B. The trick is to ind a way o extending such a partial isomorphism a little bit urther, and then repeat the trick a lot. Here is the one step extension method. 3.3 LEMMA. Consider the situation on the let where A is a linear set, B is a line, and is a partial isomorphism between the inite sets U and V. Consider also any element a A. A B A B U V U {a} U g V {b} V Then there is an element b B together with a partial isomorphism g which extends and with g(a) = b. 7

8 Proo. I a U then we take g =. Thus we may assume a / U. This element a divides U into two parts X, Y given by x X x < a a < y y Y or x, y U. One o these parts may be empty, but it does no harm to assume that both are non-empty. Since U is inite we see that X has a largest member l, and Y has a smallest member s. Since is an isomorphism rom U to V we have (l) < (s) and these two elements split V into two parts. Since B is a line there is at least one element b with (l) < b < (s). With this we may set g(a) = b or the required extension. When one o X, Y is empty a minor variant o this construction works. With this we can obtain a result that covers both parts (ii) and (iii) o Theorem THEOREM. Let A be a countable linear set, and let B be a line. Then there is an embedding : A B. Proo. We irst produce an increasing chain i i < ω where U i i V i or each index i < ω, and where A = {U i i < ω} so that = { i i < ω} is the required embedding. In more detail we have U i U i+1 i = i+1 Ui or each index i. To produce this chain we irst select and a 0 A b 0 B and set U 0 = {a 0 } V 0 = {b 0 } with 0 (a 0 ) = b 0 and then we generate the rest o the chain by a certain recursion. 8

9 Since A is countable it has an enumeration (a i i < ω) where the order o this has nothing whatsoever to do with the given comparison in A. We use the irst element a 0 to set up 0, as above. Now suppose we have at stage i. We then set U i i V i where U i = {a 0,, a i } U i+1 = U i {a i+1 } and use Lemma 3.3 to obtain i+1. Since we use an enumeration o the whole o A we see that the resulting is an embedding o A into B. Two particular cases o this result gives us parts (ii, iii) o Theorem 1.2. You should work out how this is achieved. We haven t yet obtained part (iv) o Theorem 1.2. To obtain that we need a urther reinement o this embedding technique. That is the topic o the next section. 3.1 Exercises: Embedding properties 3.1 Using Theorem 3.4, how much o Theorem 1.1 can you prove. 3.2 A linear set A is homogeneous i or each inite linear set F and each pair o embeddings F g A rom F, there is an automorphism h o A with = h g. (a) Show that each non-trivial homegeneous linear set A is a line. (b) Show that both Q and R are homegeneous (c) Can you show that each line is homegeneous? In the ollowing two exercise we consider equ-structures (A, ) where is an equivalence relation on the set A. Recall that an equivalence relation is a binary relation that is relexive, transitive, and symmetric. For each a A we let [a] = {x A x a} be the block (equivalence class) in which a lives. Given two such equ-structures (A, ) (B, ) 9

10 a partial isomorphism rom A to B is a bijection : U V rom some U A to some V B such that a 1 a 2 (a 1 ) (a 2 ) or all a 1, a 2 A. I U = A then this is an embedding o A into B. I U = A and V = B then this is a (ull) isomorphism rom A to B. Finally, we say such a equ-structure (A, ) is chubby i it has ininitely many blocks, and or each a A the block [a] is ininite. 3.3 Consider a partial isomorphism rom a equ-structure A to a chubby equ-structure B, as on the let. And suppose U and V are inite. A B A B U {a} g V {b} U V U V Show that or each a A there is a partial isomorphism g rom A to B which extends and takes in a, as on the right. 3.4 By making use o Exercise 3.3, show that each countable equ-structure is embeddable in to each chubby equ-structure. 3.2 Problems: Embedding properties 3.5 Let A, B be a pair o linear sets and let A B g be a pair o embeddings where [A] is an initial section o B g[b] is a inal section o A respectively. By reining the proo o the CSB-result, show that A = B. 10

11 4 The back-and-orth technique In this section we extend the embedding methods o Section 3 to produce a method o showing that two linear sets are isomorphic. In act, this method can be used on many dierent algebraic systems, but here we concentrate on linear sets The general idea is to use a whole amily o partial isomorphisms which interact in a certain way. 4.1 DEFINITION. Let A, B be a pair o linear sets. (a) A back-and-orth system or the pair A, B is a non-empty set P o partial isomorphisms rom A to B with the ollowing two properties. (back) For each member o P : U V and each element y o B, there is an element x o A together with an extension + : U {x} V {y} o in P. (orth) For each member o P : U V and each element x o A, there is an element y o B together with an extension + : U {x} V {y} o in P. We oten abbreviate back-and-orth to b&. (b) We write A = p B and say A and B are potentially isomorphic i there is at least one back-and-orth system or the pair A, B. The name back-and-orth system is the common terminology. However, Robin Gandy used to say to-and-ro system, which perhaps is a better name. Robin Gandy was a student o Alan Turing, and had a big inluence on developing Mathematical Logic in Manchester. Notice that or each isomorphism : A B the singleton {} is a b& system. This ensures that A = B = A = p b holds. However, we are more interested in those systems P where each component is a small part o A, B. Mostly we use inite partial isomorphisms, A potential isomorphism does not require that the two linear sets have the same cardinality. Here is an important example o this. 11

12 4.2 THEOREM. Each two lines A, B are potentially isomorphic. Proo. Given the two lines A, B let P be the amily o all inite partial isomorphism rom A to B. We show that this P has the b& property. In act, by symmetry, it suices to show that P has the orth property. This is precisely what Lemma 3.3 does. As an example o this we have Q = p R and these two lines do not have the same cardinality. Trivially, each two isomorphic linear sets are potentially isomorphic but, as we have just seen, the converse is not true. However, when we restrict to the countable case the two notions are the same. 4.3 THEOREM. We have or all countable linear sets A, B. A = p B = A = B Proo. We are given a b& system P or the two countable linear sets. We use P to generate a ull isomorphisms rom A to B. Since both A and B are countable there are enumerations (x i i < ω) (y i i < ω) o these sets. These enumerations can be quite arbitrary with repetitions allowed. We produce an ascending chain 0 1 i i < ω o partial isomorphism in P and we arrange that or each i < ω. The union i : U i V i x i U i+1 y i V i+1 = { i i < ω} is then the required isomorphism. Let 0 be any member o P. Suppose we have produced i. By irst going orth (to deal with x i ) and then coming back (to deal with y i ) we obtain i+1. Theorems 4.2 and 4.3 gives us the remaining part o Theorem THEOREM. Each two countable lines are isomorphic. In particular, each countable line is isomorphic to Q. 12

13 Proo. Let A, B be two countable lines. By Theorem 4.2 we have A = p B, and then Theorem 4.3 gives A = B, as required. This back-and-orth technique is useul in several dierent areas. What I ind odd is that it seems to have been devised quite recently in the 1950s or 1960s (which is not very long ago in terms o the development o Mathematics). It would be nice to ind out when it irst appeared. (Hint: don t trust Wikipaedia. I m not even sure they can spell their own name.) In the Problems or this section there are two more examples o the use o this technique. 4.1 Exercises: The back-and-orth technique 4.1 This exercise extends Exercises 3.3 and 3.4. (a) Show that each pair o chubby equ-systems are potentially isomorphic (that is there are least one back-and-orth system between the pair). (b) Show that each two countable and chubby equ-systems are isomorphic. (c) Can you suggest a picture o the unique chubby (up to isomorphism) equ-system, say carried by N N. 4.2 Problems: The back-and-orth technique In the context o the next two problems a graph is a set A which carries a notion o being mates. We reer to the members o A as nodes (but some ancy-pants preer to call them vertices). For two nodes x, y we write x y x y to indicate that x, y are mates x, y are not mates respectively. We impose two conditions on being mates. The relation is symmetric, that is x y = y x or all x, y A. The relation is irrelexive, that is x x or all x A, so no node is a mate o itsel. Given two such graphs A, B a partial isomorphism rom A to B is a bijection A U B V 13

14 between subsets U A and V B such that x 1 x 2 (x 1 ) (x 2 ) x 1 x 2 (x 1 ) (x 2 ) or all x 1, x 2 U. O course, by taking contrapostives each o these equivalences implies the other. As usual, an embedding o A into B is a partial isomorphism where U = A. A ull isomorphism is a partial isomorphism with U = A and V = B. We are interested is a special kind o graph. Thus a graph A is random i it is ininite (or at least non-empty) and or each disjoint pair L, R A o inite sets o nodes we have ( x L)[x a] ( y R)[a y] or at least one node a / L R. It is sometimes convenient to write or this condition. L a R 4.2 Consider a inite partial isomorphism, as above, between two graphs A and B where B is random. Consider any a A. (a) Show there is some b B such that sending a b produces a partial isomorphism that extends. (b) Show that every countable graph is embedded in every random graph. (c) Show that each pair o countable random graphs are isomorphic. 4.3 This exercise produces a concrete example o a countable random graph. It makes use o the hereditary inite sets that occur in other places in these various documents. Let V (0) = V (r + 1) = PV (r) or each r < ω. This produces an ascending chain o inite sets. We let V (0) V (1) V (r) V = {V (r) r < ω} to produce a amily o sets. Each member x V is inite, each member o that y x V is inite, each member o that z Y X V is inite, and so on. For x, y V we let x y x y or y x x y x / y and y / x to produce a mate/non-mate relation on V. Let L, R be a disjoint pair o inite subsets o V. We show that L a R or ininitely many a V. To do that we consider a = L b or certain b V. 14

15 (a) Show that L a or each such b. (b) Show that i R b = then y R = y / a holds. (c) Show there are only initely many b such that a y R or at least one y V. (d) Show that a R or ininitely many b V. (e) Show that V is random. 5 The characteristic properties o R In this section we look at the characterization o R as given by Theorem 1.2. We can t quite prove that result. Not because it is too diicult, but because the proo is a bit long and we have other things to do. Theorem 1.2 contains two new notions that a linear ordered set may or may not have, that o being separable and that o being complete. The irst o these is easy to under stand. It is a reinement o being dense. To emphasize that we repeat the deinition o being dense here. 5.1 DEFINITION. Let A be a linear set (a linearly ordered set). The linear set A is dense i or each pair o elements l < r there is some element s A with l < s < r. The linear set A is separable i there is a countable subset S A such that or each pair o elements l < r there is some element s S with l < s < r. Trivially, each separable linear set is dense. Also a countable linear set is dense precisely when it is separable. However, in general being dense does not ensure being separable. The reals R are separable since each pair o real numbers is separated by a rational. Inormally, being complete means that each limiting process does produce a limit, in some sense or other. To make this precise there is a bit o a nuisance we have to sort out. Do we want our linear sets to have end points or not? Since here we are analysing Q and R it is best to insist that end points do not exists. To emphasize that we introduce a new word which is not standard terminology. 5.2 DEFINITION. A linswep is a linear set (linearly ordered set) without end points. Each linear set we deal with here will be a linswep. We come now to the crucial notion o being complete. 5.3 DEFINITION. Let A be a linswep. A cut o A is a pair L, U o non-empty subsets L, U o A such that L < U 15

16 that is x y or all x L and y U. The linswep A is complete (or D-complete, or Dedekind-complete) i or each cut L < U we have L a U that is L a U a or some a A. This gives us the two new properties o Theorem 1.2. They are designed to characterize R so we certainly need the ollowing. 5.4 THEOREM. As a linswep the reals R is complete. We won t prove this here but I will mention two possible proos. I you are Dedekind then this is the way R is deined. It is the unique completion o Q. I will explain what that means in more detail later, i we have time. I you are Cauchy then you prove it is complete. You show that each cauchy sequence has a limit, and you are done. This is not a conlict. We may deine R as the completion o Q, and then derive various general properties o R, as we might do here. However, or various particular calculations, in real analysis, use o dierential equations, and so on, this deinition is not very helpul. In such situations we work in terms o sequences and use cauchy completeness. We know that Q is inserted in R Q i R and we are aiming to determine the crucial properties o this unction. To do that we need the ollowing notions. 5.5 DEFINITION. Let A and B be a pair o linsweps. An embedding o A into B is a unction : A B such that a 1 a 1 (a 1 ) (a 1 ) or all a 1, a 2 A. An embedding, as above, is dense i or each b 1 < b 2 in B we have b 1 < (a) < b 2 or some a A. Observe that an embedding is an injection. I it also a surjection then it is an isomorphism. In act, an embedding is an isomorphism to its range. Observe also that i there is a dense embedding o A into B then B is dense (as a linear order). I A is countable the B is separable. O course, the standard example o such a dense embedding is the rationals Q in the reals R. As we will see, the ollowing generalizes the properties o this embedding. 16

17 A g B C where A, B, C are linsweps, g are embeddings Table 1: The wedge and the actorization problem The problem is to actorize g through, that is to produce an embedding B h C or which g = h. 5.6 DEFINITION. Let A be any linswep. A completion o A is a dense embedding A DA into a complete linswep. Observe that a completion o a linswep A is not just a special kind o linswep. It is such a linswep DA and and a dense embedding to connect the two. As a particular example the embedding Q i R is a completion o Q. We ought to sort out the existence and uniqueness o a completion. We can t go into the details, but we can state the appropriate results. The ollowing result deals partly with the required uniqueness and existence. 5.7 LEMMA. Consider the wedge o Table 1 (a) Suppose the embedding g is dense. Then there is at most one embedding h with g = h. (b) Suppose the embedding is dense, and the linswep C is complete. Then there is at least one embedding h with g = h. These help us to produce the ollowing important result. 5.8 THEOREM. Let A be a linswep (Existence) There is at least one dense embedding A DA into a complete linswep. That is A does have a completion. 17

18 (Uniqueness) Consider the situation o Table 1. Suppose that both, g are completions o A. Then there is a unique embedding B h C with g = h. Furthermore, this embedding is an isomorphism. Thus A has a unique completion up to isomorphism. With these general results we can veriy the our parts o Theorem 1.2. R is a separable complete line By Theorem 5.4 we observed that R is complete. The embedding o Q into R ensures that R is separable. R is embedded in every complete line Let C be any complete line. We irst set up a diagram as ollows. Q g At the top we use the canonical embedding o Q into R. We know this is a dense embedding, since Q is dense in R. Since C is a line, Theorem 1.1(ii) gives us an embedding g. With these Lemma 5.7(b) gives an embedding R C with g = h. This is what we want. R h C Each separable line is embedded in R Let B be any separable line. We irst produce a diagram Q g and then apply Theorem 5.8 to obtain an embedding o B into R. Here g is the canonical embedding o Q into R. The problem is to produce a dense embedding, and this is where the separability o B is used. By deinition, since B is separable, there is a countable subset X B which is dense in B, that is a < b = ( x X)[a < x < b] 18 B R

19 or all elements a, b B. The set X is certainly a linear set (using the restriction o the comparison on B). The separable property applied to elements o X shows that it is a line. But now X is a countable line, and hence isomorphic to Q by Theorem 1.1(iv). This gives the required dense embedding. Each pair o separable complete lines are isomorphic As we have just observed a line C being separable means there is an embedding Q C and so Theorem 5.8 ensures that any two complete and separable lines are isomorphic. At this point we will leave the analysis o Q and R. Later we may come back to ill in the missing proos to show how an arbitrary linswep can be completed. 19