Proving Subspace: Basic Proof for $V(A)$

[nacho-man]

Hi everyone, would really appreciate if someone could help me with the
attached question (its the one in the red box).

View attachment 3826

My start:

Assume $A$ is a subspace. We need to show that
$V(A):= \{ \{x_n\}_n \in V_0 : \lim_{{n}\to{\infty}}x_n \in A \} $

By definition, a subspace is closed under scalar multiplication and addition, and also contains the $0 \in A$ in our case.

So my assumption is I would start off by proving each of these three properties takes place in $V(A)$.

For closure under addition:
Take $x_1, x_2 \in V_0$ then $x_1 + x_2 \in A \in V(A)$

For closure under scalar multiplication:
For any $x_n \in V_0$ and $\lambda \in \mathbb{R}$ it holds $x \lambda \in A \in V(A)$

Finally $\because A$ is a subspace, by definition $0 \in A \in V(A)$.
$\therefore V(A)$ is a subspace.

Would this be considered, by and large, a sufficient proof?

 

[Evgeny.Makarov]

By definition, a subspace is closed under scalar multiplication and addition, and also contains the $0 \in A$ in our case.

The fact that every subspace of a linear vector space contains 0 follows from the closure under scalar multiplication (multiply any vector by 0).

For closure under addition:
Take $x_1, x_2 \in V_0$ then $x_1 + x_2 \in A \in V(A)$

This does not make type sense. If $x_1,x_2\in V_0$, then $x_1,x_2\in(\Bbb R^n)^{\Bbb N}$ (the set of functions from $\Bbb N$ to $\Bbb R^n$), but $A\subseteq\Bbb R^n$. So, $x_i$ have a different type than elements of $A$. Also, $A$ cannot be an element of $V(A)$ since the latter set contains functions (from $\Bbb N$ to $\Bbb R^n$), not sets like $A$.

For closure under scalar multiplication:
For any $x_n \in V_0$ and $\lambda \in \mathbb{R}$ it holds $x \lambda \in A \in V(A)$

A similar remark applies.

 

[nacho-man]

thanks for your reply, I see exactly what you mean about where my proof didn't make sense.

So how exactly would I show that $V(A)$ is a subspace?

I don't see how working with the assumption that $A$ is a subspace is of any benefit to me?

If $A$ is a subspace, its closed under addition and scalar multiplication.

I have the definitions of $V$, $V_0$ and $V(A)$ at my disposal.

Logically, where would one go from here?

 

[Evgeny.Makarov]

You assume that $\{x_n\}_n\in V(A)$ and $\{y_n\}_n\in V(A)$. You need to show that $\{x_n\}_n+\{y_n\}_n\in V(A)$. By definition, this means that there exists an $a\in A$ such that $\lim_{n\to\infty}(x_n+y_n)=a$. Can you use the definition of $V(A)$ and the fact that $\{x_n\}_n\in V(A)$ and $\{y_n\}_n\in V(A)$ to find such $a$?

 

[nacho-man]

So assume $A$ is a subspace.
we take $\{x_n\}_n + \{y_n\}_n \in V(A)$
$$\because \{x_n\}_n + \{y_n\}_n = x + y = a \in A$$
$$\therefore \{x_n\}_n + \{y_n\}_n \in V(A)$$ (not sure if I have sufficiently justified this)
Hence the set is closed under addition.

Then,
$$\forall \lambda \in \mathbb{R}^n $$
$\lambda V(A) \in \mathbb{R}^n$, $\therefore$ closed under scalar multiplication.

Hence the properties we need are satisfied, and $V(A)$ is a subspace if $A$ is a subspace.

Have I sufficiently proved this? How could I improve my reasoning and logic?

 

[Evgeny.Makarov]

So assume $A$ is a subspace.
we take $\{x_n\}_n + \{y_n\}_n \in V(A)$

Not sure what you mean by "take", but you can't assume that $\{x_n\}_n + \{y_n\}_n \in V(A)$. This is what you need to prove.

$$\because \{x_n\}_n + \{y_n\}_n = x + y = a \in A$$

You have not properly introduced $x$ and $y$, so it's nor clear what can be said about them.

I personally think that using "because" and "therefore" in a row is potentially confusing and should be avoided when learning to write proofs. A proof should be a sequence of statements each of which follows from previous ones, together with an explanation why it follows.

$$\forall \lambda \in \mathbb{R}^n $$

I believe the vector space in this problem is over the field $\Bbb R$, not $\Bbb R^n$, so $\lambda\in\Bbb R$.

$\lambda V(A) \in \mathbb{R}^n$, $\therefore$ closed under scalar multiplication.

What do you mean by multiplying a real number $\lambda$ by a vector space $V(A)$?

You may need to review the definition and some examples of linear vector spaces.

 

[nacho-man]

Not sure what you mean by "take", but you can't assume that $\{x_n\}_n + \{y_n\}_n \in V(A)$. This is what you need to prove.

You have not properly introduced $x$ and $y$, so it's nor clear what can be said about them.

I personally think that using "because" and "therefore" in a row is potentially confusing and should be avoided when learning to write proofs. A proof should be a sequence of statements each of which follows from previous ones, together with an explanation why it follows.

I believe the vector space in this problem is over the field $\Bbb R$, not $\Bbb R^n$, so $\lambda\in\Bbb R$.

What do you mean by multiplying a real number $\lambda$ by a vector space $V(A)$?

You may need to review the definition and some examples of linear vector spaces.

Sorry I could have worded my previous post a lot better, let me re-iterate what I was trying to say

When I say "take" I meant just select some $\{x_n\}_n$ and $\{y_n\}_n$ in $V(A)$ and add them together.

In regards to not properly introducing the $x$ and $y$ terms, I was just going off the definition given in the question. Is that considered sloppy when writing proofs?

Also I don't know why I wrote $\lambda * V(A)$ it should had been $\lambda * $ some $\{x_n\}_n \in V(A)$

I will re-read the definitions and post another attempt again. Thanks.

edit: My third attempt:

Our goal is to show that for $V(A)$ to be a subspace we require that:
$ \forall \{x_n\}_n, \{y_n\}_n \in V(A)$
$$ \{x_n\}_n + \{y_n\}_n \in V(A)$$
and also that it's closed under scalar multiplication

To do this, let's first assume that $A$ is a subspace, ie
$$ \forall u,v \in A, u+v \in A$$ and
$ \lambda \in \mathbb{R}$ and $u \in A$
$$\lambda \cdot u \in A$$

let $\\lim_{{n}\to{\infty}} x_n = x $ and $\\lim_{{n}\to{\infty}} y_n = y$
and let $\{x_n\}_n \{y_n\}_n$ be convergent sequences.

$V(A) := \{\{x_n\}_n \in V_0 : x \in A\}$

$\{x_n\}_n + \{y_n\}_n = \{x_n+y_n\}_n \in V(A)$ (I feel very unsure here, and in general how to convey/justify that two elements in $V(A)$ will remain in the space when summed.

And for the scalar multiplication, I would make a similar assertion as before.

 

[Evgeny.Makarov]

In regards to not properly introducing the $x$ and $y$ terms, I was just going off the definition given in the question.

If by $x$ you mean the function from $\Bbb N$ to $\Bbb R^n$, i.e., $\{x_n\}_n$ itself, then that's OK. Sorry for not catching this the first time. But then you can't say that $x+y\in A$ because $x+y\in V(A)$. If you mean something else by $x$ and $y$, then I don't know what it is. Also, you did not introduce $a$. In my suggestion $a$ was an element of $A$ and the limit of $x_n+y_n$. The question was why such limit exists.