TOPOLOGIES ON GROUPS DETERMINED BY SEQUENCES: ANSWERS TO SEVERAL QUESTIONS OF I.PROTASOV AND E.ZELENYUK

TOPOLOGIES ON GROUPS DETERMINED BY SEQUENCES:
ANSWERS TO SEVERAL QUESTIONS
arXiv:1011.4554v1 [math.GN] 20 Nov 2010




OF I.PROTASOV AND E.ZELENYUK




Taras Banakh


Abstract. Answering questions of Protasov and Zelenyuk we prove the following
results:
1. For every increasing function f : N → N with limn→∞ f (n + 1) − f (n) = ∞ and
every metrizable totally bounded group topology τ on Z there exists a convergent
to zero sequence (an )n∈ω in (Z, τ ) such that limn→∞ fa(n)
n
= 1.
a
2. For every real r > 1 there exists a sequence (an )n∈ω ⊂ Z such that limn→∞ n+1
an
=
r but there is no ring topology τ on Z such that (an )n∈ω converges to zero in
(Z, τ ).
3. There exists a countable topological Abelian group G determined by a T -sequence
and containing a closed subgroup H which is not determined by a T -sequence but
is homeomorphic to G.
4. There exist two group topologies τ1 , τ2 determined by T -sequences on Z such that
the topology τ1 ∨ τ2 is not complete and thus is not determined by a T -sequence.
5. There exists a countable Abelian group admitting a group topology τ determined
by a T -sequence and a metrizable group topology τ ′ such that the topology τ ∨ τ ′
is not discrete but contains no non-trivial convergent sequence.




In this note we give answers to several problems posed by I.Protasov and E.Zelenyuk
in [PZ1 ] and [PZ2 ]. Following [PZ2 ] we define a sequence (an )n∈ω of elements of
a group G to be a T -sequence if (an )n∈ω converges to zero in some non-discrete
Hausdorff group topology on G. Given a T -sequence (an )n∈ω in G we denote by
(G|(an )) the group G endowed with the strongest topology in which the sequence
(an ) converges to zero. We say that a topological group G is determined by a
T -sequence if G = (G|(an )) for some T -sequence (an )n∈ω in G.

1991 Mathematics Subject Classification. 22A05, 26A12, 54H11.
Research supported in part by grant INTAS-96-0753.

Typeset by AMS-TEX
1

, 2 TARAS BANAKH

1. There is no restriction on the growth of T -sequences in Z
All T -sequences of integers constructed in [PZ1 ] have exponential growth. This
led I.Protasov and E.Zelenyuk to the following question (see [PZ2 ] and [PZ1 , Ques-
tion 2.2.3]): is there a monotone T -sequence of integers having polynomial growth?
First our result answers this question affirmatively. We recall that a group topology
τ on a group G is called totally bounded if for every neighborhood U ∈ τ of zero in
G there exists a finite subset F ⊂ G with G = F · U .
Theorem 1.
(1) If (an )∞
n=1 ⊂ Z is an increasing T -sequence, then limn→∞ (an+1 − an ) = ∞.
(2) Suppose f : N → N and ε : N → [0, ∞) are functions such that limn→∞ ε(n) =
∞ and limn→∞ f (n + 1) − f (n) = ∞. For every metrizable totally bounded
group topology τ on Z there exists a converging to zero sequence (an )n∈ω ⊂
(Z, τ ) such that limn→∞ fa(n)
n
= 1 and |an − f (n)| ≤ ε(n) for every n ∈ ω.

Proof. 1. Suppose (an )n∈ω ⊂ Z is an increasing T -sequence with limn→∞ an+1 −
an 6= ∞. This means that for some C ∈ N and every n ∈ N we can find m ≥ n with
am+1 − am ≤ C. Let τ be a non-discrete Hausdorff group topology on Z such that
(an )∞
n=1 converges to zero in τ . Pick a τ -open neighborhood U ⊂ Z of zero such
that U ∩ (i + U ) = ∅ for every 1 ≤ i ≤ C and find n0 ∈ N such that an ∈ U for every
n ≥ n0 . By the choice of the constant C, there exists m ≥ n0 with am+1 − am ≤ C.
Then letting i = am+1 −am , we get am+1 = am +i ∈ (i+U )∩U = ∅, a contradiction.
2. Suppose functions f and ε satisfy thephypotheses of the theorem. Without loss
of generality, ε(1) = 0 and ε(n) ≤ 12 min{ f (n), f (n + 1) − f (n), f (n) − f (n − 1)}
for n > 1.
Let τ be any metrizable totally bounded group topology on Z and Z = U0 ⊃
U1 ⊃ U2 ⊃ . . . be a countable base of neighborhoods of zero in (Z, τ ). For every
n ∈ ω let k(n) = max{i ∈ ω : Ui ∩ [f (n) − ε(n), f (n) + ε(n)] 6= ∅} and an be
any point in Uk(n) ∩ [f (n) − ε(n), f (n) + ε(n)] (the number k(n) is finite since
the topology τ is Hausdorff). Evidently, |f (n) − an | ≤ ε(n) for every n ∈ ω and
0 ≤ limn→∞ fa(n) n
− 1 ≤ limn→∞ fε(n)(n) ≤ limn→∞
√ 1 = 0.
f (n)
It remains to verify the convergence of the constructed sequence (an )n∈ω to zero
in the topology τ . This will follow as soon as we prove that limn→∞ k(n) = ∞.
Fix any number m ∈ N. We have to find n0 ∈ N such that k(n) ≥ m for every
n ≥ n0 . Using the total boundedness of the topology τ , find l ∈ N such that
S
|i|<l (i + Um ) = Z. Since limn→∞ ε(n) = ∞, there exists n0 ∈ N such that
ε(n) > l for all n ≥ n0 . It follows that for every n ≥ n0 there exists i ∈ Z such that
|i| < l < ε(n) and i + Um ∋ f (n). Consequently, Um ∩ [f (n) − ε(n), f (n) + ε(n)] 6= ∅
and hence k(n) ≥ m. 
Remark 1. The requirement of the metrizability of the topology τ in Theorem 1 is
essential: according to [PZ1 , §5.1], there exists a totally bounded group topology τ
on Z such that the space (Z, τ ) contains no nontrivial convergent sequence.