Question about my approach to linear algebra proof, Axler Ch.2A #14 I need some help verifying my approach because the answers that I found online use a different approach.
Problem statement:
Prove that a vector space V is infinite-dimensional if and only if there is a sequence of vectors $v_1, v_2, ...$ in $V$ such that $v_1,..,v_m$ is linearly independent for every positive integer $m$.
Forward direction.
Premise: Suppose $V$ is infinite-dimensional.
We want to show that a linearly independent list of vectors $v_1,..,v_m$ exists for every positive integer $m$.
Proof by contradiction.
Suppose no linearly independent list exists for some list of $m$ vectors. This implies that the spanning list is less than $m$. To see this, consider that a list of vectors will always be dependent if and only if the size of the list is greater than a linearly independent list that also spans the vector space.
So, there is some list size $<m$ that spans V. This contradicts the original premise that V is infinite dimensional. Therefore, a linearly independent list must exist for every positive integer $m$.
Reverse direction.
Premise: Suppose a linearly independent list exists for some $v_1,..,v_m$ for every positive integer $m$. We want to show that $V$ is infinite-dimensional.
Proof by contradiction.
Suppose $V$ is finite dimensional. Then, suppose some arbitrary list of vectors of size $m-1$ spans this finite-dimensional $V$. But, $v_1,..,v_m$ must be a linearly independent list also exists in $V$ according to our premise. A spanning list cannot be less than the length of a linearly independent list. So, $V$ must be infinite dimensional.
Therefore, a vector space $V$ is infinite dimensional if and only if there is a sequence of vectors $v_1, v_2, ...$ in $V$ such that $v_1,..,v_m$ is linearly independent for every positive integer $m$.
Thanks.
 A: Here are some running comments in your first proof: 

Premise: Suppose  is infinite-dimensional. 
We want to show that a linearly independent 
list of vectors _1,..,v_ exists for every 
positive integer .

Good and clear premise. 

Proof by contradiction.
Suppose no linearly independent list exists for some list of  vectors. 

No, the contradiction should not start by assuming you have a list of $m$ vectors. It should start by assuming you have a positive integer $m$ for which no linearly independent list of $m$ vectors exists. 

 This implies that the spanning list is less than 

What do you mean by "the spanning list"? What spanning list?  Spanning list of what? 

To see this, consider that a list of 
vectors will always be dependent if and 
only if the size of the list is greater 
than a linearly independent list that 
also spans the vector space.

The "if and only if" is incorrect.  You are trying to state the general fact that if (but not only if) a collection of vectors is larger than a collection of vectors that span the space, then the first collection cannot be linearly independent.  However, while that is a general fact, you have not set up your proof in a way that clearly makes use of this fact.  What list/lists are you talking about? You have not fixed any list, you only make vague claims about nonexistence of a list. You need to instantiate the objects you use with phrases such as "let xyz be a pdq with property abc" or "There exists an object abc with property xyz." 

So, there is some list size m
that spans V. 

How do you conclude this? 

This contradicts 
the original premise that V is infinite 
dimensional. Therefore, a linearly 
independent list must exist for every 
positive integer .
To see this, consider that a list of 
vectors will always be dependent if 
and only if the size of the list is 
greater than a linearly independent 
list that also spans the vector space.

