6 I The Expanding Universe of Numbers
We now show that there exists such a mapφ.LetC be the collection of all
subsetsCofN×Asuch that( 1 ,a 1 )∈Candsuchthatif(n,a)∈C,thenalso
(S(n),T(a))∈C. The collectionCis not empty, since it containsN×A.Moreover,
since every set inCcontains( 1 ,a 1 ), the intersectionDof all setsC∈Cis not empty.
It is easily seen that actuallyD∈C. By its definition, however, no proper subset of
Dis inC.
LetMbe the set of alln∈Nsuch that(n,a)∈Dfor exactly onea∈Aand,
for anyn∈M,defineφ(n)to be the uniquea∈Asuch that(n,a)∈D.IfM=N,
thenφ( 1 )=a 1 andφ(S(n))=Tφ(n)for alln∈N. Thus we need only show that
M =N. As usual, we do this by showing that 1 ∈ M and thatn ∈ Mimplies
S(n)∈M.
We h av e ( 1 ,a 1 ) ∈ D. Assume( 1 ,a′) ∈ D for somea′ = a 1 .IfD′ =
D{( 1 ,a′)},then( 1 ,a 1 )∈ D′.Moreover,if(n,a)∈ D′then(S(n),T(a))∈ D′,
since (S(n),T(a))∈ Dand(S(n),T(a)) =( 1 ,a′). HenceD′∈C. But this is a
contradiction, sinceD′is a proper subset ofD. We conclude that 1∈M.
Suppose now thatn∈Mand letabe the unique element ofAsuch that(n,a)∈D.
Then(S(n),T(a)) ∈ D,sinceD ∈ C. Assume that(S(n),a′′)∈ Dfor some
a′′=T(a)and putD′′=D{(S(n),a′′)}.Then(S(n),T(a))∈D′′and( 1 ,a 1 )∈D′′.
For any(m,b)∈ D′′we have(S(m),T(b)) ∈ D.If(S(m),T(b)) =(S(n),a′′),
thenS(m)=S(n)andT(b)=a′′=T(a), which impliesm=nandb=a. Thus
Dcontains both(n,b)and(n,a), which contradictsn∈M. Hence(S(m),T(b))=
(S(n),a′′),andso(S(m),T(b))∈D′′.ButthenD′′∈C, which is also a contradic-
tion, sinceD′′is a proper subset ofD. We conclude thatS(n)∈M.
Corollary 2If the axioms(N1)–(N3)are also satisfied by a setN′wth element 1 ′and
map S′:N′→N′, then there exists a bijective mapφofNontoN′such thatφ( 1 )= 1 ′
and
φ(S(n))=S′φ(n) for every n∈N.Proof By takingA=N′,a 1 = 1 ′andT=S′in Proposition 1, we see that there
exists a unique mapφ:N→N′such thatφ( 1 )= 1 ′and
φ(S(n))=S′φ(n) for everyn∈N.By interchangingNandN′, we see also that there exists a unique mapψ:N′→N
such thatψ( 1 ′)=1and
ψ(S′(n′))=Sψ(n′) for everyn′∈N′.The composite mapχ=ψ◦φofNintoNhas the propertiesχ( 1 )=1andχ(S(n))=
Sχ(n)for everyn∈N. But, by Proposition 1 again,χis uniquely determined by these
properties. Henceψ◦φis the identity map onN, and similarlyφ◦ψis the identity
map onN′. Consequentlyφis a bijection.
We can also use Proposition 1 to define addition and multiplication of natural num-
bers. By Proposition 1, for eachm∈Nthere exists a unique mapsm:N→Nsuch
that
sm( 1 )=S(m), sm(S(n))=Ssm(n) for everyn∈N.