354 VIII The Geometry of Numbers
The latticeΛkhas at most 2( 2 n− 1 )facet vectors, by Proposition 15. Hence, by
restriction to a subsequence, we may assume that allΛkhave the same numbermof
facet vectors. Letxk 1 ,...,xkmbe the facet vectors ofΛkand choose the notation so
thatxk 1 ,...,xknare linearly independent. Since they all lie in the ballB 2 R, by restric-
tion to a further subsequence we may assume that
xkj→xj ask→∞ (j= 1 ,...,m).Evidently‖xj‖≥δ(j = 1 ,...,m)since, fork ≥k 0 , all nonzerox ∈Λkhave
‖x‖≥δ.
The setΛof all integral linear combinations ofx 1 ,...,xmis certainly an additive
subgroup ofRn. MoreoverΛis discrete. For supposey∈Λand‖y‖<δ.Wehave
y=α 1 x 1 +···+αmxm,whereαj∈Z(j= 1 ,...,m).If
yk=α 1 xk 1 +···+αmxkm,thenyk→yask→∞and hence‖yk‖<δfor all largek.Sinceyk∈Λk, it follows
thatyk=Ofor all largekand hencey=O.
Since the latticeΛ′kwith basisxk 1 ,...,xknis a sublattice ofΛk,wehave
d(Λ′k)≥d(Λk)=λ(Vk)≥λ(Bδ/ 2 ).Since d(Λ′k)=|det(xk 1 ,...,xkn)|, it follows that also
|det(x 1 ,...,xn)|≥λ(Bδ/ 2 )> 0.Thus the vectorsx 1 ,...,xnare linearly independent. HenceΛis a lattice.
Letb 1 ,...,bnbe a basis ofΛ. Then, by the definition ofΛ,
bi=αi 1 x 1 +···+αimxm,whereαij∈Z( 1 ≤i≤n, 1 ≤j≤m).Put
bki=αi 1 xk 1 +···+αimxkm.Thenbki∈Λkandbki→biask→∞(i= 1 ,...,n). Hence, for all largek,the
vectorsbk 1 ,...,bknare linearly independent. We are going to show thatbk 1 ,...,bkn
is a basis ofΛkfor all largek.
Sinceb 1 ,...,bnis a basis ofΛ,wehave
xj=γj 1 b 1 +···+γjnbn,whereγji∈Z( 1 ≤i≤n, 1 ≤j≤m). Hence, if
ykj=γj 1 bk 1 +···+γjnbkn,thenykj∈Λkandykj→xjask→∞(j= 1 ,...,m). Thus, for all largek,