Showing vector space

**surjective** · April 5th 2010, 09:02 AM

Hey,

I have given the space:

$l^{\infty}(\mathbb{N})=\Big\{\{x_{k}\}_{k=1}^{\inf ty} \vert x_{k}\in \mathbb{C}, \forall k\in \mathbb{N}, \text{and only finitely many} x_{k} \text{are nonzero} \Big\}$

How do I show that this space is a vector space. Should I run through various rules which must be fulfilled in order for a set to be a vector-space or is there another way of going about it?

Appreciate it.

**Enrique2** · April 5th 2010, 09:18 AM

Originally Posted by surjective

Hey,

I have given the space:

$l^{\infty}(\mathbb{N})=\Big\{\{x_{k}\}_{k=1}^{\inf ty} \vert x_{k}\in \mathbb{C}, \forall k\in \mathbb{N}, \text{and only finitely many} x_{k} \text{are nonzero} \Big\}$

How do I show that this space is a vector space. Should I run through various rules which must be fulfilled in order for a set to be a vector-space or is there another way of going about it?

Appreciate it.

It is confusing this notation. $l_\infty(\mathbb{N})$ is denoted usually by $l_\infty$ and it is the space of all bounded sequences. The space of sequences with only finite coordinates non-zero is the countable direct sum of copies of $\mathbb{C}$ and it is usually denoted by $\varphi$

To show that a set is a vector space simply show that the sum of two vectors defined in a natural way is in the set and also the multiplication by a scalar, it is simply algebra, and both, $l_\infty$ and $\varphi$ are vector spaces, the sum of two sequences which are bounded or eventually null is bounded or eventually null and the same with the multiplication by scalars.

**FancyMouse** · April 5th 2010, 03:57 PM

Clearly the direct sum is a subset of direct product. So you'll only need to check that it's actually a subspace.

**surjective** · April 5th 2010, 04:08 PM

Hey,

Thanks for the reply. Yes I know that a vector space is a nonempty set V that has been equipped with two operations, namely addition and multiplication. The thing is that in order for a set to qualify as a vector space, the mentioned operations must adhere to certain axioms. They are quite many. Therefore I would, as far as possible, like to avoid showing all of them.

But as far as I have understood from your post it would be sufficient to choose two arbitrary bounded sequences in $l^{\infty}(\mathbb{N})$ and show that their sum is bounded as well as multiplying a scalar with an arbitrary bounded sequence in $l^{\infty}(\mathbb{N})$ and show that it is bounded. Would that really be sufficient to show that the space mentioned is a vector-space.

Appreciate the help.

**Drexel28** · April 5th 2010, 04:23 PM

Originally Posted by surjective

Hey,

Thanks for the reply. Yes I know that a vector space is a nonempty set V that has been equipped with two operations, namely addition and multiplication. The thing is that in order for a set to qualify as a vector space, the mentioned operations must adhere to certain axioms. They are quite many. Therefore I would, as far as possible, like to avoid showing all of them.

But as far as I have understood from your post it would be sufficient to choose two arbitrary bounded sequences in $l^{\infty}(\mathbb{N})$ and show that their sum is bounded as well as multiplying a scalar with an arbitrary bounded sequence in $l^{\infty}(\mathbb{N})$ and show that it is bounded. Would that really be sufficient to show that the space mentioned is a vector-space.

Appreciate the help.

Where's the confusion coming from? If I am understanding your question correctly the above space is a subset of a bigger vector space (which?) and to show that it's a vector space we must thus show it's closed under scalar mult. and addition.

But, if $x_n,y_n$ are eventually zero of order $k,\ell$ respectively then $x_n+y_n$ is eventually zero of order $\max\{k,\ell\}$ . etc.

**surjective** · April 6th 2010, 10:11 AM

Hey,

Thanks for your patience.

Let me try again. Looking at the previous posts I saw that I had written the space incorrectly. I have the space:

$l^{\infty}(\mathbb{N})=\Big\{\{x_{k}\}_{k=1}^{\inf ty} \vert x_{k}\in \mathbb{C}, \sup_{k \in \mathbb{N}}|x_{k}| < \infty \Big\}$

It does not state that $l^{\infty}(\mathbb{N})$ ´is a subset of a larger space!!!?

My confusion has to do with the textbook I'm using (and have been using since I started the course on Real Analysis). It is terrible and the formulation is not clear at all.

I know that a subspace of a vector-space $V$ is a subset $S\subseteq V$ such that when the addition and scalar multiplication of $V$ are used to add and scalar-multiply the elements of $S$ , then $S$ is a vector-space. Hence by showing that $l^{\infty}(\mathbb{N})$ is closed under addition and multiplication then that would show the intended (as mentioned several times now).

Appreciate the help.

**Drexel28** · April 6th 2010, 11:47 AM

Originally Posted by surjective

Hey,

Thanks for your patience.

Let me try again. Looking at the previous posts I saw that I had written the space incorrectly. I have the space:

$l^{\infty}(\mathbb{N})=\Big\{\{x_{k}\}_{k=1}^{\inf ty} \vert x_{k}\in \mathbb{C}, \sup_{k \in \mathbb{N}}|x_{k}| < \infty \Big\}$

It does not state that $l^{\infty}(\mathbb{N})$ ´is a subset of a larger space!!!?

My confusion has to do with the textbook I'm using (and have been using since I started the course on Real Analysis). It is terrible and the formulation is not clear at all.

I know that a subspace of a vector-space $V$ is a subset $S\subseteq V$ such that when the addition and scalar multiplication of $V$ are used to add and scalar-multiply the elements of $S$ , then $S$ is a vector-space. Hence by showing that $l^{\infty}(\mathbb{N})$ is closed under addition and multiplication then that would show the intended (as mentioned several times now).

Appreciate the help.

Where's the difficulty with this too? If $x_n\in\ell_\infty(\mathbb{N})$ then $\sup_{n\in\mathbb{N}}|\alpha x_n|=\alpha\sup_{n\in\mathbb{N}}|x_n|<\infty\impli es \{\alpha x_n\}\in\ell_{\infty}(\mathbb{N})$ . Similarly, if $\{x_m\},\{y_n\}\in\ell_\infty(\mathbb{N})$ then $\sup_{n\in\mathbb{N}}|x_n+y_n|\leqslant\sup_{n\in\ mathbb{N}}|x_n|+\sup_{n\in\mathbb{N}}|y_n|<\infty\ implies \{x_n+y_n\}\in\ell_\infty(\mathbb{N})$

**surjective** · April 6th 2010, 12:06 PM

Yes that is also what I was able to extract from the previous posts. Some general questions:

1) Is $l^{\infty}(\mathbb{N})$ a subset of a larger vector-space?

2) When you have a set equipped with the operations of addition and multiplication, is it then a given that these operations satisfy the wellknown axioms of a vector-space?

3) Does every supspace fulfill the properties of its larger vector-space?

$\text{thanks}^{\infty}$

**Drexel28** · April 6th 2010, 12:25 PM

Originally Posted by surjective

Yes that is also what I was able to extract from the previous posts. Some general questions:

1) Is $l^{\infty}(\mathbb{N})$ a subset of a larger vector-space?

Probably, you can probably embed it in a larger vector space. Is there a canonical choice? Probably not. Does it matter though? For if you are trying to embed it so you only have to prove closure of scalar mult. and addition (since it's a subspace) you'd still have to prove that the space you embedded it in is a vector space

2) When you have a set equipped with the operations of addition and multiplication, is it then a given that these operations satisfy the wellknown axioms of a vector-space?

What do you mean? Not every mult. and addition satisfies the vector space axioms.

3) Does every supspace fulfill the properties of its larger vector-space?

It inherits associativity and distributivity of the mult. and addition, etc. But it need not be closed under mult. and addition.

**surjective** · April 6th 2010, 12:40 PM

I meant the following:

This is what my textbook says:

A vectorspace is a non-empty set V that has been equipped with two operations, called addition and scalar-multiplication, statisfying certain rules.

Notice the underlined. What you have helped me to show so far is that the space $l^{\infty}(\mathbb{N})$ is closed under addition and multiplication. But what about the "certain rules" mentioned? Is is not a requirement to show them also?

**Drexel28** · April 6th 2010, 12:47 PM

Originally Posted by surjective

I meant the following:

This is what my textbook says:

A vectorspace is a non-empty set V that has been equipped with two operations, called addition and scalar-multiplication, statisfying certain rules.

Notice the underlined. What you have helped me to show so far is that the space $l^{\infty}(\mathbb{N})$ is closed under addition and multiplication. But what about the "certain rules" mentioned? Is is not a requirement to show them also?

Yeah, but they are trivial. They are that...well I'm not going to write them all. Here they are. Personally, I would just say "it's apparent that they are satisfied" but I wouldn't do that if I were you.

**surjective** · April 6th 2010, 01:02 PM

I'm suspecting more problems will show up

but thank a lot for now.

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