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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.

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    $\begingroup$ In your proof of the reverse direction, you might want to assume the finite dimension of $V$ is $n.$ No list of more than $n$ vectors can be linearly independent in $V$, which contradicts your premise. $\endgroup$ – Chris Leary Jul 24 '19 at 16:37
  • $\begingroup$ The above comment by Chris (+1) follows my own general critique (see answer below) that you need to instantiate the objects you use: Suppose $V$ is finite dimensional, then there is a positive integer $n$ and a list of vectors $\{v_1, ..., v_n\}$ such that $Span\{v_1, ..., v_n\}=V$. You see, now we have instantiated an explicit integer $n$ and explicit vectors $v_1,...,v_n$ that we can talk about. $\endgroup$ – Michael Jul 24 '19 at 16:42
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Here are some running comments in your first proof:

Premise: Suppose 𝑉 is infinite-dimensional. 
We want to show that a linearly independent 
list of vectors 𝑣_1,..,v_π‘š exists for every 
positive integer π‘š.

Good and clear premise.

Proof by contradiction.
Suppose no linearly independent list exists for some list of π‘š 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 π‘š

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 π‘š.
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.
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  • $\begingroup$ While the premise is clear, a higher-level critique wonders if the premise is sufficient for the forward direction: It seems the question asks for a single infinite sequence $\{v_n\}_{n=1}^{\infty}$ rather than a finite set of vectors that exists for each $m$ (which might use completely different vectors for each $m$, i.e., $m=2$ list is $\{v_1,v_2\}$; $m=3$ list is $\{w_1,w_2,w_3\}$, ...). $\endgroup$ – Michael Jul 24 '19 at 16:46
  • $\begingroup$ It may be better to change your premise/approach: Try recursively building your list $\{v_n\}_{n=1}^{\infty}$: Let $v_1$ be any nonzero vector in $V$. Now form $v_2$. In general you start with $\{v_1, ..., v_k\}$ and then find another $v_{k+1}$. $\endgroup$ – Michael Jul 24 '19 at 16:53
  • $\begingroup$ For the forward direction, I want to state the fact that an arbitrary collection of vectors will always be dependent if and only if that arbitrary collection of vectors has a size greater than another collection of vectors that is both linearly independent and spans the vector space $V$. Since that linearly independent set spans $V$, it contradicts the premise that $V$ is infinite-dimensional. Point taken about instantiation! $\endgroup$ – Richard K Yu Jul 24 '19 at 16:58
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    $\begingroup$ The "if and only if" regarding the size of your list is incorrect: The vectors $\{(0,0,1), (0,0,2)\}$ are linearly dependent but $\{(0,0,1), (0,1,0), (1,0,0)\}$ are linearly independent. You want to use if rather than if and only if. $\endgroup$ – Michael Jul 24 '19 at 17:00
  • $\begingroup$ So I mean to say that any list of vectors with length greater than 3 in $R^3$ will necessarily be linearly dependent (Any list in the set of all lists of vectors greater than 3 in $R^3$ is linearly dependent). In your example, while a list of vectors length 2 can be linearly dependent, it is not always linearly dependent (i.e. there is a linearly independent list of length 2 as well). Is the if and only if true in that sense? $\endgroup$ – Richard K Yu Jul 24 '19 at 17:47

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