#### Abstract

In this work we introduce new spaces of double sequences defined by a double sequence of modulus functions, and we study some properties of this space.

#### 1. Introduction

In this work, by and , we denote the spaces of single complex sequences and double complex sequences, respectively. and denote the set of positive integers and complex numbers, respectively. If, for all , there is such that where and , then a double sequence is said to be converge (in terms of Pringsheim) to . A real double sequence is nondecreasing, if for . A double series is infinity sum and its convergence implies the convergence by of partial sums sequence , where (see [13]).

For , denote the space of sequences such that (see [4]).

A double sequence is said to be bounded if and only if . The space of all bounded double sequences is denoted by . It is known that is Banach space (see [5, 6]).

A double sequence space is said to be normal if whenever for all and .

The double sequence spaces in the various forms were introduced and studies by Khan and Tabassum in [714], by Khan in [15], and by Khan et al. in [16, 17].

Now let be a family of subsets having most elements in . Also denote the class of subsets in such that the elements of and are most and , respectively. Besides is taken as a nondecreasing double sequence of the positive real numbers such that (see [18]).

Let be a double sequence. A set is defined by A double sequence space is said to be symmetric if and whenever and .

A BK-space is a Banach sequence space in which the coordinate maps are continuous.

A function is said to be a modulus function if it satisfies the following:(1) if and only if ;(2) for all ;(3) is increasing;(4) is continuous from right at .

It follows that is continuous on . The modulus function may be bounded or unbounded. For example, if we take , then is bounded. But, for , is not bounded.

The BK-spaces , introduced by Sargent in [19], is in the form

Sargent studied some properties of this space and examined relationship between this space and -space.

The space was extended to by Tripathy and Sen [20] as follows: Recently, Raj et al. [21] introduced and studied the following sequence space .

Let be a sequence of modulus functions. Then the space is defined by

In this work we introduce sequence spaces defined by where is a double sequence of modulus functions.

#### 2. Main Results

The result stated in the first theorem is not hard. So, we give it without proof.

Theorem 1. The sequence space is a linear space.

Theorem 2. The sequence spaces are complete.

Proof. Let be a double Cauchy sequence in such that for all . Then for some and for all . For each , there exists a positive integer such that for all . Hence for some and for all . This implies that for all and for each fixed . Hence is a Cauchy sequence in .
Then, there exists such that as and let us define . From (10), for each fixed , for some , for all and .
Letting , we get for some , for all , and . Thus we obtain for some and for all . This implies that for all .
Hence . By (14), for all . This means that as . Hence is a Banach space.

Theorem 3. The space is a BK-space.

Proof. Suppose that with as . For each there exists such that for all . Thus for some and for all . Hence we obtain for all and for all . This implies as . This completes the proof.

Corollary 4. The space is a symmetric space.

Proof. Let and let . Then we can write . Thus we obtain

Corollary 5. The space is a normal space.

Proof. It is obvious.

Theorem 6. Consider

Proof. Suppose that . Then we have Thus for each fixed and for , for some . Hence for some . This implies that . Hence .

Theorem 7. if and only if .

Proof. Let . Then for all . If , then for some . Thus for some . Hence . This shows that . Conversely, let . We define . Let . Then there exists a subsequence of such that as . Then for we have for some . This is a contradiction as and this completes the proof.

Proposition 8. Consider

Proof. Clearly, , where for when and by nondecreasing . Then, by Theorem 7, first inclusion is obtained. Suppose . Then for some . Hence we obtain for all . Thus and proof is completed.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.