5 Recurrence 487Suppose we have defined pointsx 1 ,...,xk, positive integersn 1 ,...,nk,and
ε 1 ,...,εk∈( 0 ,ε 0 )such that, fori= 1 ,...,k,
d(Tnixi,xi− 1 )<εi− 1 ,...,d(Tpnixi,xi− 1 )<εi− 1 ,and d(x,xi)<εiimplies
d(Tnix,xi− 1 )<εi− 1 ,...,d(Tpnix,xi− 1 )<εi− 1.By (iv) there existxk+ 1 ∈Xandnk+ 1 ≥1 such that
d(Tnk+^1 xk+ 1 ,xk)<εk,...,d(Tpnk+^1 xk+ 1 ,xk)<εk,and we can then chooseεk+ 1 ∈( 0 ,ε 0 )so that d(x,xk+ 1 )<εk+ 1 implies
d(Tnk+^1 x,xk)<εk,...,d(Tpnk+^1 x,xk)<εk.Thus the process can be continued indefinitely.
By taking successivelyi=j−1,j− 2 ,...we see that, ifi<j,then
d(Tni+^1 +···+nj−^1 +njxj,xi)<εi,...,d(Tp(ni+^1 +···+nj−^1 +nj)xj,xi)<εi.SinceXis compact, it is covered by a finite numberrof open balls with radiusε 0 /2.
Hence there existi,jwith 0≤i < j ≤rsuch that d(xi,xj)<ε 0. If we put
n=ni+ 1 +···+nj− 1 +njthen, sinceεi<ε 0 , we obtain from the triangle inequality
d(Tnxj,xj)< 2 ε 0 ,...,d(Tpnxj,xj)< 2 ε 0.Butε 0 >0 was arbitrary.
It may be deduced from Proposition 26, by means ofBaire’s category theorem,
that under the same hypotheses there exists a pointz∈Xand an increasing sequence
(nk) of positive integers such thatTinkz→zask→∞(i= 1 ,...,p).However,as
we now show, Proposition 26 already suffices to proves van der Waerden’s theorem.
The setX∗of all infinite sequencesx=(x 1 ,x 2 ,...),wherexi ∈{ 1 , 2 ,...,r}
for everyi ≥1, can be given the structure of a compact metric space by defining
d(x,x)=0andd(x,y)= 2 −kifx=yandkis the least positive integer such that
xk=yk.Theshiftmapτ:X∗→X∗,definedbyτ((x 1 ,x 2 ,...))=(x 2 ,x 3 ,...),is
continuous, since
d(τ(x),τ(y))≤2d(x,y).With the partitionN=S 1 ∪···∪Srin the statement of van der Waerden’s theorem
we associate the infinite sequencex∈X∗defined byxi=jifi∈Sj.
LetXdenote the closure of the set(τnx)n≥ 1 .ThenXis a closed subset ofX∗
which is invariant underτ. By Proposition 26, there exists a pointz∈Xand a positive
integernsuch that
d(τnz,z)< 1 / 2 , d(τ^2 nz,z)< 1 / 2 ,..., d(τpnz,z)< 1 / 2 ;