3 Real Numbers 19It is obvious from the definition by set inclusion that at most one holds. Now suppose
that neitherA<BnorA=B. Then there existsa∈A\B. It follows from (i), applied
toB,thateveryb∈Bsatisfiesb<aand then from (i), applied toA,thatb∈A. Thus
B<A.
LetSbe any nonempty collection of cuts. A cutBis said to be anupper bound
forSifA≤Bfor everyA∈S,andalower boundforSifB≤Afor every
A∈S. An upper bound forSis said to be aleast upper boundorsupremumforS
if it is a lower bound for the collection of all upper bounds. Similarly, a lower bound
forSis said to be agreatest lower boundorinfimumforSif it is an upper bound for
the collection of all lower bounds. Clearly,Shas at most one supremum and at most
one infimum.
The setPhas the following basic property:
(P4)if a nonempty subsetShas an upper bound, then it has a least upper bound.
Proof LetCbe the union of all setsA∈S. By hypothesis there exists a cutBsuch
thatA⊆Bfor everyA∈S.SinceC⊆Bfor any suchB,andA⊆Cfor every
A∈S, we need only show thatCis a cut.
EvidentlyCis a nonempty proper subset ofP,sinceB=P. Supposec∈C.Then
c∈Afor someA∈S.Ifd∈Pandd<c,thend∈A,sinceAis a cut. Furthermore
c<a′for somea′∈A.SinceA⊆C, this proves thatCis a cut.
In the setPof positive rational numbers, the subsetTof allx ∈ Psuch that
x^2 <2 has an upper bound, but no least upper bound. Thus(P4)shows that there is a
difference between the total order onPand that onP.
We now define addition of cuts. For anyA,B∈P,letA+Bdenote the set of all
rational numbersa+b, witha∈Aandb∈B. We will show that alsoA+B∈P.
EvidentlyA+Bis a nonempty subset ofP. It is also a proper subset. For choose
c∈P\Aandd∈P\B. Then, by (i),a<cfor alla∈Aandb<dfor allb∈B.
Sincea+b<c+dfor alla∈Aandb∈B, it follows thatc+d∈/A+B.
Suppose now thata∈A,b∈Band thatc∈Psatisfiesc<a+b.Ifc>b,then
c=b+dfor somed∈P,andd<a. Hence, by (i),d∈Aandc=d+b∈A+B.
Similarly,c∈ A+Bifc>a. Finally, ifc≤aandc≤b, choosee∈Pso that
e<c.Thene∈Aandc=e+ffor somef ∈P.Thenf∈B,sincef<c,and
c=e+f∈A+B.
ThusA+Bhas the property (i). It is trivial thatA+Balso has the property (ii),
since ifa∈Aandb∈B, there existsa′∈Asuch thata<a′and thena+b<a′+b.
This completes the proof thatA+Bis a cut.
It follows at once from the corresponding properties of rational numbers that addi-
tion of cuts satisfies the commutative law(A2)and the associative law(A3).
We consider next the connection between addition and order.
Lemma 15For any cut A and any c∈P, there exists a∈A such that a+c∈/A.
Proof Ifc∈/A,thena+c∈/ Afor everya∈A,sincec<a+c. Thus we may
supposec∈A. Chooseb∈P\A. For some positive integernwe haveb<ncand
hencenc∈/ A.Ifnis the least positive integer such thatnc∈/A,thenn>1and
(n− 1 )c∈A. Consequently we can takea=(n− 1 )c.