Number Theory: An Introduction to Mathematics

6 Mahler’s Compactness Theorem 355

‖ykj−xkj‖<δ (j= 1 ,...,m).

Sinceykj−xkj∈Λk, this implies that, for all largek,ykj=xkj(j= 1 ,...,m). Thus
every facet vector ofΛkis an integral linear combination ofbk 1 ,...,bknand hence,
by Proposition 16, every vector ofΛkis an integral linear combination ofbk 1 ,...,bkn.
Sincebk 1 ,...,bknare linearly independent, this shows that they are a basis ofΛk.
LetWbe the Voronoi cell ofΛ. We wish to show thatV =W.Ifv∈V,then
there existvk∈Vksuch thatvk→v. Assumev/∈W.Then‖v‖>‖z−v‖for some
z∈Λ,andso

‖v‖=‖z−v‖+ρ,

whereρ>0. There existzk∈Λksuch thatzk→z. Then, for all largek,

‖v‖>‖zk−v‖+ρ/ 2

and hence, for all largek,

‖vk‖>‖zk−vk‖.

But this contradictsvk∈Vk.
This proves thatV⊆W. On the other hand,Vhas volume

λ(V)= lim

k→∞

λ(Vk)= lim

k→∞

d(Λk)

= lim

k→∞

|det(bk 1 ,...,bkn)|

=|det(b 1 ,...,bn)|=d(Λ)=λ(W).

It follows that every interior point ofWis inV, and henceW=V. Corollary 17 now
shows that the same latticeΛwould have been obtained if we had restricted attention
to some other subsequence of{Λk}.
Leta 1 ,...,anbe any basis ofΛ. We are going to show that, for the sequence{Λk}
originally given, there existaki∈Λksuch that

aki→aiask→∞ (i= 1 ,...,n).

If this is not the case then, for somei ∈{ 1 ,...,n}and someε>0, there exist
infinitely manyksuch that

‖x−ai‖>ε for allx∈Λk.

From this subsequence we could as before pick a further subsequenceΛkv →Λ.
Then everyy∈Λis the limit of a sequenceyv∈Λkv.Takingy=ai, we obtain a
contradiction.
It only remains to show thatak 1 ,...,aknis a basis ofΛkfor all largek.Since

lim

k→∞

|det(ak 1 ,...,akn)|=|det(a 1 ,...,an)|

=d(Λ)=λ(V)=lim

k→∞

λ(Vk),