# Sequence spaces – Serlo

The sequence space is a vector space consisting of infinitely long tuples ${\displaystyle (x_{1},x_{2},x_{3},\ldots )}$. The operations on the sequence space are component-wise addition and scalar multiplication.

## Motivation

We have already learned about the coordinate spaces ${\displaystyle K^{n}}$ for a field ${\displaystyle K}$ as an example for vector spaces. Here, each element consists of ${\displaystyle n}$ distinct, that is, finitely many, entries from ${\displaystyle K}$. For example, ${\displaystyle (1,4,2,-1)}$ is an element of ${\displaystyle \mathbb {R} ^{4}}$. We can also consider infinite tuples. For example, ${\displaystyle (1,4,9,16,\dots )=(n^{2})_{n\in \mathbb {N} }}$ is such an "infinite tuple". A better name for infinite tuples is "sequence". If ${\displaystyle K}$ is the field of real or complex numbers, these are exactly the already known sequences from calculus.

How do we define the vector space operations on the sequences? On ${\displaystyle K^{n}}$ we have defined the operations component-wise. We already know that we can also add and scale sequences component-wise. Therefore, we can also define addition and scalar multiplication on infinite sequences over arbitrary fields. This leads us to the conjecture that the set of all sequences with entries in ${\displaystyle K}$ should form a vector space. We call it the sequence space over ${\displaystyle K}$.

We will first define the sequence space precisely and then prove that it is indeed a vector space. Then, in the section Subspaces of the sequence space, we will consider examples of subspaces of the sequence spaces over the real and complex numbers, which are important for advanced calculus.

## Notation

Let ${\displaystyle K}$ be a field.

We always write ${\displaystyle (x_{i})_{i}}$ instead of ${\displaystyle (x_{i})_{i\in \mathbb {N} }}$ in this article for sequences with elements from ${\displaystyle K}$.

## Definition of a sequence space

Definition (The sequence space as a set)

We define the set

${\displaystyle \omega :=\left\{\,(x_{i})_{i}\,{\bigg \vert }\,x_{i}\in K{\text{ for all }}i\in \mathbb {N} \,\right\}}$

We call it the set of all sequences over ${\displaystyle K}$, or the sequence space over ${\displaystyle K}$.

In analogy to the coordinate space, we can also define an addition and a scalar multiplication on ${\displaystyle \omega }$:

Definition (Vector space operations on ${\displaystyle \omega }$)

The addition ${\displaystyle \boxplus \colon \omega \times \omega \to \omega }$ is defined by

{\displaystyle {\begin{aligned}(x_{i})_{i}\boxplus (y_{i})_{i}:=(x_{i}+y_{i})_{i}.\end{aligned}}}

Similarly we define scalar multiplication ${\displaystyle \boxdot \colon K\times \omega \to \omega }$ by

${\displaystyle \lambda \boxdot (x_{i})_{i}=(\lambda \cdot x_{i})_{i}.}$

## The sequence space is a vector space

Theorem (${\displaystyle \omega }$ is a vector space)

${\displaystyle (\omega ,\boxplus ,\boxdot )}$ is a ${\displaystyle K}$-vector space.

How to get to the proof? (${\displaystyle \omega }$ is a vector space)

We proceed as in the article Proofs for vector spaces. Because the sequence space is defined similarly to the coordinate space, we use the same strategy as for the proof that the coordinate Space is a vector space.

Proof (${\displaystyle \omega }$ is a vector space)

We have to check the eight vector space axioms.

Let ${\displaystyle x=(x_{i})_{i},y=(y_{i})_{i},z=(z_{i})_{i}\in \omega }$. Then:

{\displaystyle {\begin{aligned}(x\boxplus y)\boxplus z&=((x_{i})_{i}\boxplus (y_{i})_{i})\boxplus (z_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(x_{i}+y_{i})_{i}\boxplus (z_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=((x_{i}+y_{i})+z_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associativity of the addition in }}K\right.}\\[0.3em]&=(x_{i}+(y_{i}+z_{i}))_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(x_{i})_{i}\boxplus (y_{i}+z_{i})_{i}=x\boxplus (y\boxplus z).\end{aligned}}}

This shows the associativity of the addition.

Let ${\displaystyle x=(x_{i})_{i},y=(y_{i})_{i}\in \omega }$. Then:

{\displaystyle {\begin{aligned}x\boxplus y&=(x_{i})_{i}\boxplus (y_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(x_{i}+y_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{commutativity of the addition }}K\right.}\\[0.3em]&=(y_{i}+x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(y_{i})_{i}\boxplus (x_{i})_{i}=y\boxplus x.\end{aligned}}}

This establishes the commutativity of the addition.

Proof step: Neutral element of addition

We now have to show that there is a neutral element ${\displaystyle 0_{\omega }\in \omega }$, that is, ${\displaystyle x\boxplus 0_{\omega }=x}$ for all ${\displaystyle x\in \omega }$. Since we trace all properties back to the properties in ${\displaystyle K}$, we choose

${\displaystyle 0_{\omega }=(0)_{i}=(0_{K},0_{K},\ldots )}$

as an approach for the neutral element.

Let ${\displaystyle x=(x_{i})_{i}\in \omega }$. Then:

${\displaystyle x\boxplus 0_{\omega }=(x_{i}+0)_{i}=(x_{i})_{i}=x.}$

Thus we have shown that ${\displaystyle 0_{\omega }\in \omega }$ is the neutral element of addition.

Proof step: Inverse with respect to addition

Let ${\displaystyle x=(x_{i})_{i}\in \omega }$. We need to show that there is a ${\displaystyle y\in \omega }$ such that ${\displaystyle x\boxplus y=0_{\omega }}$ . As with the neutral element of addition, we use the corresponding counterpart from ${\displaystyle K}$ as a starting point. That is, we choose for ${\displaystyle y}$ the sequence ${\displaystyle (-x_{i})_{i}}$. Then

${\displaystyle x\boxplus y=(x_{i})_{i}\boxplus (-x_{i})_{i}=(x_{i}+(-x_{i}))_{i}=(0)_{i}=0_{\omega }.}$

Thus we have shown that for any ${\displaystyle x\in \omega }$ there is a ${\displaystyle y\in \omega }$ with ${\displaystyle x\boxplus y=0_{\omega }}$.

Proof step: Scalar distributive law

Let ${\displaystyle \lambda ,\mu \in K}$ and ${\displaystyle x=(x_{i})_{i}\in \omega }$. Then:

{\displaystyle {\begin{aligned}(\lambda +\mu )\boxdot x&=(\lambda +\mu )\boxdot (x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=((\lambda +\mu )\cdot x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law in }}K\right.}\\[0.3em]&=(\lambda \cdot x_{i}+\mu \cdot x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(\lambda \cdot x_{i})_{i}\boxplus (\mu \cdot x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \boxdot (x_{i})_{i})\boxplus (\mu \boxdot (x_{i})_{i})\\[0.3em]&=(\lambda \boxdot x)\boxplus (\mu \boxdot x).\end{aligned}}}

Thus the scalar distributive law is shown.

Proof step: Vector Distributive Law

Let ${\displaystyle \lambda \in K}$ and ${\displaystyle x=(x_{i})_{i},y=(y_{i})_{i}\in \omega }$. Then:

{\displaystyle {\begin{aligned}\lambda \boxdot (x\boxplus y)&=\lambda \boxdot ((x_{i})_{i}\boxplus (y_{i})_{i})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=\lambda \boxdot (x_{i}+y_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \cdot (x_{i}+y_{i}))_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law in }}K\right.}\\[0.3em]&=(\lambda \cdot x_{i}+\lambda \cdot y_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(\lambda \cdot x_{i})_{i}\boxplus (\lambda \cdot y_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \boxdot (x_{i})_{i})\boxplus (\lambda \boxdot (y_{i})_{i})\\[0.3em]&=(\lambda \boxdot x)\boxplus (\lambda \boxdot y).\end{aligned}}}

Thus the vectorial distributive law is shown.

Proof step: Associativity with respect to multiplication

Let ${\displaystyle \lambda ,\mu \in K}$ and ${\displaystyle x=(x_{i})_{i}\in \omega }$. Then:

{\displaystyle {\begin{aligned}(\lambda \cdot \mu )\boxdot x&=(\lambda \cdot \mu )\boxdot (x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=((\lambda \cdot \mu )\cdot x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{Aassociative law for multiplication in }}K\right.}\\[0.3em]&=(\lambda \cdot (\mu \cdot x_{i}))_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \boxdot (\mu \cdot x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \boxdot (\mu \boxdot (x_{i})_{i})\\[0.3em]&=\lambda \boxdot (\mu \boxdot x).\end{aligned}}}

This establishes the associative law for multiplication.

Proof step: Existence of a unit

Let ${\displaystyle x=(x_{i})_{i}\in \omega }$. Then:

${\displaystyle 1_{K}\boxdot x=1_{K}\boxdot (x_{i})_{i}=(1_{K}\cdot x_{i})_{i}=(x_{i})_{i}=x.}$

Thus we have shown the unitary law.

Thus we have established all eight vector space axioms and ${\displaystyle (\omega ,\boxplus ,\boxdot )}$ is a ${\displaystyle K}$ vector space.

## Subspaces of the sequence space

The sequence space has some frequently used subspaces. Most of these subspaces can be defined only over the fields ${\displaystyle \mathbb {R} }$ and ${\displaystyle \mathbb {C} }$. They have many applications in functional analysis, where they are part of an important class of examples. In the field of linear algebra over arbitrary fields, the space of sequences with finite support serves as an example in many places. It is the simplest example of a infinite-dimensional vector space and thus can be used as a good example where statements cease to hold, as the vector space if "too large".

### The subspace of sequences with finite support

Definition (Set of sequences with finite support)

We define

${\displaystyle c_{00}:=\left\{\,(x_{i})_{i}\in \omega \,{\bigg \vert }\,{\text{there is an }}i_{0}\in \mathbb {N} ,{\text{ such that }}x_{i}=0{\text{ for all }}i\geq i_{0}\,\right\}}$.

Theorem (The sequences with finite support form a subspace)

${\displaystyle c_{00}\subseteq \omega }$ is a subspace.

Proof (The sequences with finite support form a subspace)

We check the three subspace criteria.

Proof step: ${\displaystyle 0_{\omega }\in c_{00}}$

The zero element of ${\displaystyle \omega }$ is the sequence which is constant ${\displaystyle 0}$. Thus, for ${\displaystyle i_{0}=1}$, for all ${\displaystyle i\geq i_{0}}$, it holds that ${\displaystyle x_{i}=0}$, viz. ${\displaystyle 0_{\omega }\in c_{00}}$.

Proof step: ${\displaystyle c_{00}}$ is closed under addition.

Let ${\displaystyle x=(x_{i})_{i},y=(y_{i})_{i}\in c_{00}}$. By assumption, there exist ${\displaystyle i_{1},i_{2}in\mathbb {N} }$ with the property that ${\displaystyle x_{i}=0}$ for all ${\displaystyle i\geq i_{1}}$ and ${\displaystyle y_{i}=0}$ for all ${\displaystyle i\geq i_{2}}$. We set ${\displaystyle i_{0}=\operatorname {max} (i_{1},i_{2})}$. Then, for all ${\displaystyle i\geq i_{0}}$ we have that ${\displaystyle (x\boxplus y)_{i}=x_{i}+y_{i}=0+0=0}$. This shows ${\displaystyle x\boxplus y\in c_{00}}$.

Proof step: ${\displaystyle c_{00}}$ is closed under scalar multiplication.

Let ${\displaystyle x=(x_{i})_{i}\in c_{00}}$ and ${\displaystyle \lambda \in K}$. By assumption, there exists ${\displaystyle i_{0}\in \mathbb {N} }$ with the property that ${\displaystyle x_{i}=0}$ for all ${\displaystyle i\geq i_{0}}$. Then for all ${\displaystyle i\geq i_{0}}$, it holds that ${\displaystyle (\lambda \boxdot x)_{i}=\lambda \cdot x_{i}=\lambda \cdot 0=0}$. Hence ${\displaystyle \lambda \boxdot x\in c_{00}}$.

Thus we have shown that ${\displaystyle c_{00}}$ is a subspace of ${\displaystyle \omega }$.

For example, the notation ${\displaystyle c_{00}}$ for the space of sequences with finite support can be derived like this: This vector space is a subspace of the space of zero sequences over the fields ${\displaystyle \mathbb {R} }$ or ${\displaystyle \mathbb {C} }$. The latter subspace is usually denoted by ${\displaystyle c_{0}}$. The ${\displaystyle c}$ stands for convergence and the ${\displaystyle 0}$ for the fact that we put only zero sequences of the convergent sequences into the vector space. When talking about convergence, the condition that the sequence eventually becomes ${\displaystyle 0}$ is of course significantly stronger than the condition to converge against ${\displaystyle 0}$. Therefore the space of sequences with finite support ${\displaystyle c_{00}}$ gets an additional zero into the index.

### Subspaces from calculus

In the following, we assume that ${\displaystyle K\in \{\,\mathbb {R} ,\mathbb {C} \,\}}$.

Exercise (Vector space of bounded sequences)

The subset

${\displaystyle \ell ^{\infty }:=\left\{\,(x_{i})_{i}\in \omega \,{\bigg \vert }\,\sup _{i\in \mathbb {N} }|x_{i}|<\infty \,\right\}\subseteq \omega }$

is a subspace of the sequence space. The space ${\displaystyle \ell ^{\infty }}$ is called vector space of bounded sequences.

Solution (Vector space of bounded sequences)

We check the three subspace criteria.

Proof step: ${\displaystyle 0_{\omega }\in \ell ^{\infty }}$

The zero element of ${\displaystyle \omega }$ is the sequence which is constant ${\displaystyle 0}$. This is of course bounded (e.g. by ${\displaystyle 1}$) and thus lies in ${\displaystyle \ell ^{\infty }}$.

Proof step: ${\displaystyle \ell ^{\infty }}$ is closed under addition

Let ${\displaystyle x=(x_{i})_{i},y=(y_{i})_{i}\in \ell ^{\infty }}$. Then ${\displaystyle x\boxplus y=(x_{i}+y_{i})_{i}}$. From the triangle inequality ${\displaystyle |x_{i}+y_{i}|\leq |x_{i}|+|y_{i}|}$ for all ${\displaystyle i\in \mathbb {N} }$ we obtain

{\displaystyle {\begin{aligned}\sup _{i\in \mathbb {N} }|x_{i}+y_{i}|\leq \sup _{i\in \mathbb {N} }\,(|x_{i}|+|y_{i}|)\leq \underbrace {\sup _{i\in \mathbb {N} }|x_{i}|} _{<\infty }+\underbrace {\sup _{i\in \mathbb {N} }|y_{i}|} _{<\infty }<\infty .\end{aligned}}}

Thus ${\displaystyle x\boxplus y\in \ell ^{\infty }}$.

Proof step: ${\displaystyle \ell ^{\infty }}$ is closed under scalar multiplication.

Let ${\displaystyle x=(x_{i})_{i}\in \ell ^{\infty }}$ and ${\displaystyle \lambda \in K}$. Then ${\displaystyle \lambda \boxdot x=(\lambda \cdot x_{i})_{i}}$. Since ${\displaystyle |\lambda \cdot x_{i}|=|\lambda |\cdot |x_{i}|}$ for all ${\displaystyle i\in \mathbb {N} }$ we obtain

{\displaystyle {\begin{aligned}\sup _{i\in \mathbb {N} }|\lambda \cdot x_{i}|=\sup _{i\in \mathbb {N} }\,(|\lambda |\cdot |x_{i}|)=|\lambda |\cdot \underbrace {\sup _{i\in \mathbb {N} }|x_{i}|} _{<\infty }<\infty .\end{aligned}}}

Hence ${\displaystyle \lambda \boxdot x\in \ell ^{\infty }}$.

Thus we have shown that ${\displaystyle \ell ^{\infty }}$ is a subspace of ${\displaystyle \omega }$.

Exercise (Vector space of convergent sequences)

The subset

${\displaystyle c:=\left\{\,(x_{i})_{i}\in \omega \,{\bigg \vert }\,\lim _{i\to \infty }x_{i}{\text{ exists}}\,\right\}\subseteq \omega }$

is a subspace of ${\displaystyle \omega }$. It is called vector space of convergent sequences.

Solution (Vector space of convergent sequences)

We check the three subspace criteria.

Proof step: ${\displaystyle 0_{\omega }\in c}$

The zero element of ${\displaystyle \omega }$ is the sequence which is constant ${\displaystyle 0}$. This obviously converges to ${\displaystyle 0}$ and thus lies in ${\displaystyle c}$.

Proof step: ${\displaystyle c}$ is closed under addition

Let ${\displaystyle x=(x_{i})_{i},y=(y_{i})_{i}\in c}$. Then, ${\displaystyle x\boxplus y=(x_{i}+y_{i})_{i}}$. We have ${\displaystyle \lim _{i\to \infty }(x\boxplus y)_{i}=\lim _{i\to \infty }(x_{i}+y_{i})=\lim _{i\to \infty }x_{i}+\lim _{i\to \infty }y_{i}}$. The two limits on the right hand side exist by assumption, so the limit on the left hand side exists. In particular, ${\displaystyle x\boxplus y}$ converges, so it is true that ${\displaystyle x\boxplus y\in c}$.

Proof step: ${\displaystyle c}$ is closed under scalar multiplication.

Let ${\displaystyle x=(x_{i})_{i}\in c}$ and ${\displaystyle \lambda \in K}$. Then, ${\displaystyle \lambda \boxdot x=(\lambda \cdot x_{i})_{i}}$. We have ${\displaystyle \lim _{i\to \infty }(\lambda \boxdot x)_{i}=\lim _{i\to \infty }(\lambda \cdot x_{i})=\lambda \lim _{i\to \infty }x_{i}}$. The two limits on the right hand side exist by assumption, so the limit on the left hand side also exists. Hence ${\displaystyle \lambda \boxdot x\in c}$.

Thus we have shown that ${\displaystyle c}$ is a subspace of ${\displaystyle \omega }$.

Exercise (Vector space of zero sequences)

The subset

${\displaystyle c_{0}:=\left\{\,(x_{i})_{i}\in \omega \,{\bigg \vert }\,\lim _{i\to \infty }x_{i}=0\,\right\}\subseteq \omega }$

is a subspace of ${\displaystyle \omega }$. This subspace is called vector space of zero sequences.

Solution (Vector space of zero sequences)

We check the three subspace criteria.

Proof step: ${\displaystyle 0_{\omega }\in c_{0}}$

The zero element of ${\displaystyle \omega }$ is the sequence which is constant ${\displaystyle 0}$. This naturally converges to ${\displaystyle 0}$ and thus lies in ${\displaystyle c_{0}}$.

Proof step: ${\displaystyle c_{0}}$ is closed under addition

Let ${\displaystyle x=(x_{i})_{i},y=(y_{i})_{i}\in c_{0}}$. Then ${\displaystyle x\boxplus y=(x_{i}+y_{i})_{i}}$. We have ${\displaystyle \lim _{i\to \infty }(x\boxplus y)_{i}=\lim _{i\to \infty }(x_{i}+y_{i})=\lim _{i\to \infty }x_{i}+\lim _{i\to \infty }y_{i}=0+0=0}$. Consequently, ${\displaystyle x\boxplus y}$ converges to ${\displaystyle 0}$. Thus ${\displaystyle x\boxplus y\in c_{0}}$.

Proof step: ${\displaystyle c_{0}}$ is closed under scalar multiplication

Let ${\displaystyle x=(x_{i})_{i}\in c_{0}}$ and ${\displaystyle \lambda \in K}$. Then ${\displaystyle \lambda \boxdot x=(\lambda \cdot x_{i})_{i}}$. We have ${\displaystyle \lim _{i\to \infty }(\lambda \boxdot x)_{i}=\lim _{i\to \infty }(\lambda \cdot x_{i})=\lambda \cdot \lim _{i\to \infty }x_{i}=\lambda \cdot 0=0}$. Hence ${\displaystyle \lambda \boxdot x\in c_{0}}$.

Thus we have shown that the space of zero sequences ${\displaystyle c_{0}}$ is a subspace of ${\displaystyle \omega }$.

Exercise (Vector space of absolutely summable sequences)

The subset

${\displaystyle \ell ^{1}:=\left\{\,(x_{i})_{i}\in \omega \,{\bigg \vert }\,\sum _{i=1}^{\infty }|x_{i}|<\infty \,\right\}}$

is a subspace of ${\displaystyle \omega }$. It is called the vector space of absolutely summable sequences.

Solution (Vector space of absolutely summable sequences)

We check the three subspace criteria.

Proof step: ${\displaystyle 0_{\omega }\in \ell ^{1}}$

The zero element of ${\displaystyle \omega }$ is the sequence which is constant ${\displaystyle 0}$. For this sequence ${\displaystyle \sum _{i=1}^{\infty }|0|=0<\infty }$ holds. So ${\displaystyle 0_{\omega }\in \ell ^{1}}$.

Proof step: ${\displaystyle \ell ^{1}}$ is closed under addition

Let ${\displaystyle x=(x_{i})_{i},y=(y_{i})_{i}\in \ell ^{1}}$. Then ${\displaystyle x\boxplus y=(x_{i}+y_{i})_{i}}$. Because of the triangle inequality ${\displaystyle |x_{i}+y_{i}|\leq |x_{i}|+|y_{i}|}$ for all ${\displaystyle i\in \mathbb {N} }$ we obtain ${\displaystyle \sum _{i=1}^{\infty }|x_{i}+y_{i}|\leq \sum _{i=1}^{\infty }(|x_{i}|+|y_{i}|)}$. With the sum rule for series we get ${\displaystyle \sum _{i=1}^{\infty }(|x_{i}|+|y_{i}|)=\sum _{i=1}^{\infty }|x_{i}|+\sum _{i=1}^{\infty }|y_{i}|<\infty }$. Hence ${\displaystyle x\boxplus y\in \ell ^{1}}$.

Proof step: ${\displaystyle \ell ^{1}}$ is closed under scalar multiplication

Let ${\displaystyle x=(x_{i})_{i}\in \ell ^{1}}$ and ${\displaystyle \lambda \in K}$. Then ${\displaystyle \lambda \boxdot x=(\lambda \cdot x_{i})_{i}}$. Since ${\displaystyle |\lambda \cdot x_{i}|=|\lambda |\cdot |x_{i}|}$ for all ${\displaystyle i\in \mathbb {N} }$ it follows that ${\displaystyle \sum _{i=1}^{\infty }\lambda \cdot x_{i}|=\sum _{i=1}^{\infty }(|\lambda |\cdot |x_{i}|)=|\lambda |\cdot \sum _{i=1}^{\infty }|x_{i}|<\infty }$. Hence ${\displaystyle \lambda \boxdot x\in \ell ^{1}}$.

Thus we have shown that ${\displaystyle \ell ^{1}}$ is a subspace of ${\displaystyle \omega }$.

### Relationships between the subspaces

We have now learned about some subspaces of the sequence space for ${\displaystyle K\in \{\mathbb {R} ,\mathbb {C} \}}$. This raises the question of what relations exist between them. Most of the conditions we used to construct the subspaces are conditions from calculus. Fortunately, there are already results in calculus that describe implications between the individual conditions. If we translate these implications into the world of sets and vector spaces, we get the following result:

Exercise (Inclusions between the subspaces)

It is true that ${\displaystyle c_{00}\subsetneq \ell ^{1}\subsetneq c_{0}\subsetneq c\subsetneq \ell ^{\infty }\subsetneq \omega }$.

Solution (Inclusions between the subspaces)

We will now show the four inclusions one after the other. Thereby we also prove that in each case no equality is valid.

Proof step: ${\displaystyle c_{00}\subsetneq \ell ^{1}}$

Proof step: ${\displaystyle c_{00}\subseteq \ell ^{1}}$

Let ${\displaystyle x=(x_{i})_{i}\in c_{00}}$. By definition there exists some ${\displaystyle i_{0}\in \mathbb {N} }$, such that ${\displaystyle x_{i}=0}$ for all ${\displaystyle i\geq i_{0}}$. Thus ${\displaystyle \sum _{i=1}^{\infty }|x_{i}|=\sum _{i=1}^{i_{0}}|x_{i}|<\infty }$ and we obtain ${\displaystyle x\in \ell ^{1}}$. This shows ${\displaystyle c_{00}\subseteq \ell ^{1}}$.

Proof step: ${\displaystyle c_{00}\neq \ell ^{1}}$

Consider the sequence ${\displaystyle x=\left({\frac {1}{2^{i}}}\right)_{i\in \mathbb {N} }\in \omega }$. We have ${\displaystyle x\in \ell ^{1}}$, since ${\displaystyle \sum _{i=1}^{\infty }\left|{\frac {1}{2^{i}}}\right|=1<\infty }$. But ${\displaystyle x\notin c_{00}}$, since ${\displaystyle {\frac {1}{i}}\neq 0}$ for all ${\displaystyle i\in \mathbb {N} }$. This shows ${\displaystyle c_{00}\neq \ell ^{1}}$.

Proof step: ${\displaystyle \ell ^{1}\subsetneq c_{0}}$

Proof step: ${\displaystyle \ell ^{1}\subseteq c_{0}}$

Let ${\displaystyle x=(x_{i})_{i}\in \ell ^{1}}$. This means, ${\displaystyle \sum _{i=1}^{\infty }|x_{i}|<\infty }$. Then, by means of the term test, we have that ${\displaystyle \lim _{i\to \infty }|x_{i}|=0}$. That is, ${\displaystyle \lim _{i\to \infty }x_{i}=0}$ and thus ${\displaystyle x\in c_{0}}$. This shows ${\displaystyle \ell ^{1}\subseteq c_{0}}$.

Proof step: ${\displaystyle \ell ^{1}\neq c_{0}}$

Consider the sequence ${\displaystyle x=\left({\frac {1}{i}}\right)_{i\in \mathbb {N} }\in \omega }$. We have ${\displaystyle x\in c_{0}}$, since ${\displaystyle \lim _{i\to \infty }{\frac {1}{i}}=0<\infty }$. But ${\displaystyle x\notin \ell ^{1}}$, since ${\displaystyle \sum _{i=1}^{\infty }\left|{\frac {1}{i}}\right|}$ diverges by means of the theorem on the divergence of the harmonic series. This shows ${\displaystyle \ell _{1}\neq c_{0}}$.

Proof step: ${\displaystyle c_{0}\subsetneq c}$

Proof step: ${\displaystyle c_{0}\subseteq c}$

This is clear since every sequence converging to zero converges.

Proof step: ${\displaystyle c_{0}\neq c}$

Consider the sequence ${\displaystyle x=(1)_{i}=(1,1,\dots )\in \omega }$. We have ${\displaystyle x\in c}$, since ${\displaystyle \lim _{i\to \infty }1=1}$. But ${\displaystyle x\notin c_{0}}$, since ${\displaystyle \lim _{i\to \infty }1=1\neq 0}$. This shows ${\displaystyle c_{0}\neq c}$.

Proof step: ${\displaystyle c\subsetneq \ell ^{\infty }}$

Proof step: ${\displaystyle c\subseteq \ell ^{\infty }}$

From calculus we know, that every convergent sequence is bounded. So the inclusion holds true.

Proof step: ${\displaystyle c\neq \ell ^{\infty }}$

Consider the sequence ${\displaystyle x=\left((-1)^{i}\right)_{i\in \mathbb {N} }\in \omega }$. We have ${\displaystyle x\in \ell ^{\infty }}$, since ${\displaystyle \left|(-1)^{i}\right|=1}$ for all ${\displaystyle i\in \mathbb {N} }$. But ${\displaystyle x\notin c}$, since ${\displaystyle \lim _{i\to \infty }(-1)^{i}}$ does not exist. This shows ${\displaystyle c\neq \ell ^{\infty }}$.

Proof step: ${\displaystyle \ell ^{\infty }\subsetneq \omega }$

Proof step: ${\displaystyle \ell ^{\infty }\subseteq \omega }$

We already know this because ${\displaystyle \ell ^{\infty }}$ is a subspace of ${\displaystyle \omega }$.

Proof step: ${\displaystyle \ell ^{\infty }\neq \omega }$

Consider the sequence ${\displaystyle x=(n)_{n\in \mathbb {N} }\in \omega }$. We have ${\displaystyle x\notin \ell ^{\infty }}$, since ${\displaystyle \lim _{n\to \infty }n=\infty }$. This shows ${\displaystyle \ell ^{\infty }\neq \omega }$.