356 VIII The Geometry of Numbers
for all largekwe must have
0 <|det(ak 1 ,...,akn)|< 2 λ(Vk).But ifak 1 ,...,aknwere not a basis ofΛkfor all largek, then for infinitely manykwe
would have
|det(ak 1 ,...,akn)|≥2d(Λk)= 2 λ(Vk). Proposition 20 has the following counterpart:Proposition 21Let{Λk}be a sequence of lattices inRnand let Vkbe the Voronoi cell
ofΛk. If there exists a latticeΛsuch thatΛk→Λas k→∞, and if V is the Voronoi
cell ofΛ,thenVk→V in the Hausdorff metric as k→∞.
Proof By hypothesis, there exists a basisb 1 ,...,bnofΛand a basisbk 1 ,...,bknof
eachΛksuch thatbkj→bjask→∞(j= 1 ,...,n). ChooseR>0 so that the
fundamental parallelotope ofΛis contained in the ballBR={x∈Rn:‖x‖≤R}.
Then, for allk≥k 0 , the fundamental parallelotope ofΛkis contained in the ballB 2 R.
It follows that, for allk≥k 0 , every point ofRnis distant at most 2Rfrom some point
ofΛkand henceVk⊆B 2 R.
Consequently, by Blaschke’s selection principle, the sequence{Vk}has a subse-
quence{Vkv}which converges in the Hausdorff metric to a compact convex setW.
Moreover,
λ(W)= lim
v→∞
λ(Vkv)=lim
v→∞
d(Λkv)=d(Λ) > 0.Consequently, sinceWis convex, it has nonempty interior. It now follows from Propo-
sition 20 thatW=V.
Thus any convergent subsequence of{Vk}has the same limitV. If the whole
sequence{Vk}did not converge toV, there would existρ>0 and a subsequence
{Vkv}such that
h(Vkv,V)≥ρ for allv.By the Blaschke selection principle again, this subsequence would itself have a con-
vergent subsequence. Since its limit must beV, this yields a contradiction.
SupposeΛk ∈ LnandΛk → Λask →∞. We will show that not only
d(Λk)→d(Λ),butalsom(Λk)→m(Λ)ask→∞. Since everyx∈Λis the limit
of a sequencexk∈Λk,wemusthavelimk→∞m(Λk)≤m(Λ). On the other hand, by
Proposition 19, ifxk∈Λkandxk→x,thenx∈Λ. Hence limk→∞m(Λ)≥m(Λ),
sincex=0ifxk=0forlargek.
Suppose now that a subsetFofLnhas the property that any infinite sequenceΛk
of lattices inFhas a convergent subsequence. Then there exist positive constantsρ,
σsuch that
m(Λ)≥ρ^2 , d(Λ)≤σ for allΛ∈F.