Вiсник Харкiвського нацiонального унiверситету iменi В.Н. Каразiна
Серiя "Математика, прикладна математика i механiка"
УДК 517.521
№ 826, 2008, с.134–140
On the coincidence of the Limit Point Range and the Sum
Range along a filter of filter convergent series
Alexander Leonov
Kharkov National University, Ukraine
We study filters F for which the equality LP R = SRF holds. We
characterize filters F for which every F-convergent series has a null
subsequence. This property with the one which we call the unbounded gap
property of F imply that LP R = SRF . The unbounded gap property of F
(existence of A ∈ F, A = (an ) such that the sequence of gn = an+1 − an
is unbounded) is a necessary condition for this equality. The equality does
not hold for any ultrafilter.
2000 Mathematics Subject Classification 40A05, 54A20
1. Introduction
The sum range of a series on real line (the set of sums of all its convergent
rearrangements) is described by the famous Riemann’s theorem:
Riemann’s theorem Let ∞
k=1 xk be a conditionally convergent series of real
numbers.
Then
for
every
s
∈
R
∪ {±∞} there exists a permutation π such that
∞
k=1 xπ(k) = s.
There is a number of generalizations of the Riemann’s theorem to more general
series (vectors, functions, etc. [4]), or more general types of convergence [1], [2],
[5]. The authors of [2] considered two generalizations of the usual convergence: the
statistical convergence and the convergence along even numbers. They obtained
the full description of the sum range along both the statistical convergence
filter and the 2n-filter. Surprisingly, the simplicity of a filter does not mean the
simplicity of the problem. And the description of the sum range along 2n-filter
appeared to be much more complicated than along the statistical convergence
filter. Besides the two usual cases of one point or the whole real line, the sum range
can be an arithmetic progression a + λZ, a, λ ∈ R (for the statistical convergence
filter)
land shifted additive subgroup of the form a + {c1 z1 + · · · + cl zl | zk ∈ E, ci ∈
Z,
k=1 ck is even}, where E is an ε-separated set (for the 2n-filter).
134
On the equality LP R = SRF for F convergent series
135
In this paper we continue the study originated in [2], but instead of considering
concrete filters we study classes of filters for which the description of the sum range
can be given explicitly. We study a class of filters F with the following property:
for every F-convergent series the sum range along F of it coincides with its limit
point range.
Recall that a filter F on N is a non-empty collection
of subsets of N satisfying
the following axioms: ∅ ∈
/ F; if A, B ∈ F then A B ∈ F; and for every A ∈ F
if B ⊃ A then B ∈ F.
A sequence (xn ), n ∈ N in a topological space X is said to be F-convergent
to x if for every neighborhood U of x the set {n ∈ N : xn ∈ U } belongs to F.
The filter convergence of series is naturally
defined as filter convergence of the
s if the sequence
sequence
of its partial sums, i.e. a series
xk is F-convergent to = ( nk=1 xk ) is F-convergent to s (and we write s = F- k xk or simply
(sn ) s = F xk , when there is only one possible summing index).
In particular if one takes as F the filter of sets with finite complements (the
Fréchet filter), then F-convergence coincides with the ordinary one.
The natural ordering on the set of filters on N is defined as follows: F1 F2
if F1 ⊃ F2 . A maximal in the natural ordering filter is called an ultrafilter. The
Zorn lemma implies that every filter is dominated by an ultrafilter. A filter F on
N is an ultrafilter if and only if for every A ⊂ N either A or N \ A belongs to F.
A filter F on N is said to be free if it dominates the Fréchet filter. Below
when we say “filter” we mean a free filter on N. In particular every ordinary
convergent sequences and series will be automatically F-convergent. More about
filters, ultrafilters and their applications one can find in every modern General
Topology textbook, for example in [6].
Definition 1 A point s belongs to thesum range along F of the series
xk if
xπ(k) = s. The set of all such points
is
there exists a permutation π such that F called the sum range along F of the series
xk and is denoted by SRF ( xk ).
We need also the following definition from [4].
Definition 2 A point s belongs to the limit point range of the series
xk if
there exists apermutation π and an increasing sequence of naturals (mk ) such
k
xπ(j) = s. The set of all such
points is called the limit-point
that limk→∞ m
j=1 range of the series
xk and is denoted by LP R( xk ).
It is easy to see that LP R( xk ) is aclosed set and SRF ( xk ) ⊂
LP R( xk ). In [3] it was proved that LP R( xk ) is a shifted closed additive
subgroup
of the space in which the series lives. In particular, on the real line
LP R( xk ) can be either R or ∅ or an arithmetic progression of the form a + λZ.
More about series rearrangements whose terms are elements of a Banach space or
of other topological linear spaces one can find in [4].
In this paper we consider series whose terms lie in the field of reals. When we
write a subset of naturals A ⊂ N in the form of sequence (an ), we mean that it is
increasing.
136
Alexander Leonov
2. Sufficient and necessary conditions for the equality LP R = SRF
It is the specificity of the usual convergence that every conditionally convergent
series after rearrangement converges to every given in advance real number (the
sum range is R).
Proposition 1 Let F be a filter on N. The following conditions are equivalent
1. For every sequence (xk ) such that series
xk F-converges SRF ( xk ) is
either R or one point x ∈ R.
xk F-converges it follows that
2. For every sequence (xk ) such that series
xk →k→∞ 0.
3. F is the Fréchet filter.
Proof. Implication (3)⇒(1) is the usual Riemann rearrangement theorem; (3)⇒(2)
is the necessary condition for the usual convergence of series. To show the inverse
implications (1)⇒(3) and (2)⇒(3) let us suppose the contrary. F is not the Fréchet
filter means that there is some infinite C ⊂ N such that A = N \ C ∈ F. Denote
xk is 1 for k ∈ E,
by E the set A ∩ (C + 1) and let (xk ) be the following sequence:
−1 for k ∈ (E −1)
and 0 for the rest k ∈ N\(E ∪(E −1)). Then xk F-converges
to 0 but SRF ( xk ) ⊂ Z and xk →k→∞ 0. Thus, the sequence convergence to 0 is the characteristic property of the usual
convergence of series. Let us describe a class of filters with a weaker property.
Definition 3 A filter F is said to have 1-shift property if for every A ∈ F there
is s ∈ A such that s + 1 ∈ A.
Proposition 2 Let F be a filter on N. The following conditions are equivalent
1. For every sequence (xk ) such that
xk F-converges there is a null
subsequence (xkn ).
2. F has 1-shift property.
Proof. (2)⇒(1). Let F xk = x and suppose that (xk ) has no null subsequence.
Then there is δ > 0 and N ∈ N such that |xk | > δfor k > N . For ε = δ/4 let us
k
xn | < ε for every k ∈ N.
find A = (ak ) ∈ F, with a1 > N , such that |x − an=1
Take s ∈ A such that s + 1 ∈ A then we come to a contradiction:
ε > |x −
s+1
n=1
xn | ≥ |xs+1 | − |x −
s
xn | > δ − ε > 2ε.
n=1
(1)⇒(2). Suppose that there is A = (ak ) ∈ F such that s ∈ A implies that
s + 1 ∈ A, that is ak + 1 < ak+1 for all k ∈ N. Denote dk = ak − ak−1 , where
On the equality LP R = SRF for F convergent series
137
a0 = 0, and let (xk ) be such sequence: xn is 1 for n ∈ A, in the case of odd dk set
xak−1 +1 = −1, in the case of even dk set xak−1 +1 = −2, and set xn = (−1)n−ak−1
for ak−1 + 1 < n < ak . Then
xk F-converges but (xk ) has no null subsequence.
Definition 4 A filter F is said to have the unbounded gap property if there is
A ∈ F, A = (an ) such that the sequence of gn = an+1 − an is unbounded.
Theorem 1 Let F be a filter which has the 1-shift
property and the unbounded
gap property.
Then for every F-convergent series
xk the equality SRF ( xk ) =
LP R( xk ) holds.
Proof. We proceed the same way as it was made in [2, Theorem 2.2.1.] for statistical
convergence.
that xkn → 0 it follows
From the existence of a subsequence (xkn ) such
∞
that we can select a subsequence (xkni ) such that i=1 xkni < ∞. Note, that
absolutely convergent series does not change its sum (and hence F-sum) under
any permutation σ, so we can write
xσ(k) =
F
vσ(k) +
F
∞
zk ,
k=1
where (vk ) is the sequence (xk ) with 0 substituted for all xkni and zk = xk − vk .
So without loss of generality we may assume that there are infinitely many zeros
among the original series terms xk .
Let y be an arbitrary element of LP R( xk ) with π and (mk ) being the
permutation and the sequence corresponding
to y from the definition of LP R. To
obtain the permutation τ for which y = F xτ (k) let us take A ∈ F, A = (an )
such that (gn ) = (an+1 −an ) is unbounded and find a subsequence (gnk ) such that
gnk ≥ mk − mk−1 for all k ∈ N with m0 assigned to be 0. We arrange elements of
our series in the following way:
an2 −an1 −m1
an1
0 + . . . + 0 +xπ(1) + . . . + xπ(m1 ) + 0 + . . . + 0 +
xπ(m1 +1) + . . . + xπ(m2 ) + 0 + .
. . + 0 +xπ(m2 +1) + . . . + xπ(m3 ) + . . . .
an3 −an2 −m2
For this permutation τ the series
xτ (k) evidently F-converge to y.
Theorem 2 Let F be a filter which is not Fréchet. The unbounded
gap property
(
x
)
=
LP R( xk )
is a necessary condition in order
to
fulfill
the
equality
SR
F
k
for every F-convergent series
xk .
138
Alexander Leonov
Proof. Assuming that the unbounded gap property
does not hold
we construct
F-summable to 0 sequence (xk ) such that SRF ( xk ) = LP R( xk ).
The idea of the construction is the following. Let 1 be the desired
limiting
point which is not the F-limit. For the series terms, to make 1 ∈ LP R( xk ), we
take some integers yk from which we can make presentations of 1 as a sum of a
fixed number of terms. The monotone growth of the number of
these terms and
the uniqueness of each presentation will result in that 1 ∈ SRF ( xk ). Adjoining
(−yk ) to (yk ) and infinitely many zeros, if needed (to provide F-convergence to
0), we will get the sequence (xk ) we need.
First, before we chose yk , let us ascertain that F xk = 0. Since the filter
we consider is not the Fréchet filter, there is A ∈ F with infinite compliment
C. This allows us to rearrange our series grouping yk and −yk in pairs. Denote
E = A ∩ (C + 1), E = (e1 , e2 , . . .) and let xek = yk , xek −1 = −yk and the rest set
zero: xk = 0 for k ∈ N \ (E ∪ (E − 1)). We obtain the F-convergent to 0 series
xk .
We take yk ∈ Z \ {0} such that (|yk |) is a rapidly increasing sequence. The
elements yk form a tree. Its structure is as follows:
(a) for every s ∈ N there is d(s) such that ys =
d(s)+s
k=d(s) yk ;
(b) d(1) = 2, d(s + 1) = d(s) + s + 1.
(c) y1 = 1;
(d) yd(s)+i = 2
d(s)+i−1
k=1
(e) yd(s)+s = −
|yk | + 1, 0 ≤ i < s;
d(s)+s−1
k=d(s)
and from (a)
yk + ys .
We use the following terminology: {yk }∞
k=1 is the decomposition tree of 1, the
d(s)+s
set {yk }k=d(s) is the decomposition of ys , first level of the tree is y1 , n-th level of
the tree is the union of decompositions of all the elements of (n − 1)-th level. The
sum of all elements of every level equals 1 for the decomposition tree of 1 and
∞
equals −1 for the decomposition
tree of −1 (the set {−yk }k=1 ).
Observe that 1 ∈ LP R( xk ). One can reorder and sum (xk ) for example in
the following way: first take 1, then add 0, then add −1 with the decomposition
of 1, then add 0, then add decomposition of −1 with the second level of the
decomposition tree of 1, etc (if we do not have infinitely many zeros then we just
do not add them).
Let us show that any permutation π can not make F xπ(k) = 1. Supposing
1
bi+1
xπ(k) = 1 and k=b
x
=
the contrary we find B = (bi ) ∈ F such that bk=1
i +1 π(k)
0 for every i ∈ N. Consider two possible cases.
bi+1
The first case: there is N ∈ N such that with every element x ∈ {xπ(k) }k=b
i +1
b
bN
i+1
the −x ∈ {xπ(k) }k=b
for all i ≥ N . In this case
i +1
the set {xπ(k) }k=1 has to contain
with every y the −y in its turn. And we get F xπ(k) = 0.
On the equality LP R = SRF for F convergent series
139
The second case: there are infinitely many i ∈ N for which we can
bi+1
bi+1
find x ∈ {xπ(k) }k=b
that −x ∈ {xπ(k) }k=b
. Take the modulus largest
i +1
i +1
b
b
i+1
i+1
x ∈ {xπ(k) }k=b
that −x ∈ {xπ(k) }k=b
. It is either an element of the
i +1
i +1
decomposition tree of 1 or of −1. Without loss of generality let us reckon that
x = yp , i.e. belongs to the decomposition tree of 1. Consider the decomposition
d(s)+s
{yk }k=d(s) to which yp belongs. If not all s + 1 elements of this decomposition is
bi+1
bi+1
contained in {xπ(k) }k=b
then
+1
k=bi +1 xπ(k) = 0, because yk are chosen in such
i d(s)+s
a way that any sum k=d(s) θk yk , θk ∈ {0, 1, −1} is ether ±ys or is modulusd(s)−1
greater then k=1 |yk |. It follows that there are infinitely many i ∈ N for which
bi+1
exists a decomposition that is contained in {xπ(k) }k=b
. This contradicts our
i +1
assumption that F does not have the unbounded gap property. Though we do not know whether the 1-shift property is a necessary condition,
it is essential. To see this let us consider an example.
Proposition 3 If F contains an arithmetic progression, i.e. an element A of the
form qN
is a sequence (xk ) such that
+ n, where q > 1 and 0 ≤ n< q, then there
series
xk F-converges and SRF ( xk ) = LP R( xk ).
Proof. Let us take F-convergent to 0 series of the following sequence
q
q
n
(xk ) = 0, 0, . . . , 0, 1, 1, . . . , 1, −(q − 1), 1, 1, . . . , 1, −(q − 1), . . . .
For every permutation σ such that
xσ(j) F-converges we can find B ∈ F,
B = (bi ) ⊂ A such that max1≤j≤n σ(j) < b1 , i.e. all 0-s are already summed.
i
Then every sum bj=1
xσ(j) can be represented as a sum of summands with q nonzero terms in each. Every summand is the sum ofm “−(q − 1)”-s and q − m
“1”-s,
(
x
)
=
qZ
and
1
∈
SR
(
xk )
0 ≤ m ≤ q which
is
(1−m)q.
Thus
we
have
SR
F
F
k
but 1 ∈ LP R( xk ). In particular we have that neither the set of all odd numbers
nor the setof all
even numbers can belong to a filter with the property SRF ( xk ) = LP R( xk ).
Therefore every ultrafilter does not have the studied property.
Acknowledgment. The author would like to express his gratitude to his
scientific advisor, V. Kadets, who has posed this problem to him, for fruitful
communications, and for his help in preparing this text.
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140
Alexander Leonov
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