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I am currently self-studying Axler's Linear Algebra Done Right. I came up with a proof for the question listed in the title, but I want to check if my approach is correct. I looked at Question about my approach to linear algebra proof, Axler Ch.2A #14 which offers helpful suggestions, but takes a slightly different direction than my proof. Here is my proof:

$\triangle$ $Proof:$

Assume V is an infinite-dimensional vector space. To prove that there is a sequence of vectors such that $v_1, v_2, ..., v_m$ is linearly independent for every $m \in \mathbb{N}$, I will utilize induction. For m=1, just choose any non-zero vector, and you have a linearly independent list (I proved this in an earlier problem, so I will omit the proof here). Now, before going to the inductive step, consider the following lemma

$\textit{Lemma: Given a list of vectors $v_1,v_2,...,v_n$ in an infinite-dimensional vector space,}$ $\textit{V, there exists a vector w $\in V$ such that w $\notin span(v_1,v_2,...,v_n)$}$.

To prove this, assume for contradiction that there is a list of vectors $v_1,v_2,...,v_n$ in V such that for every arbitrary vector $w\in V$, $w\in span(v_1,v_2,...,v_n)$. Since vector spaces are closed under addition and scalar multiplication, $span(v_1,v_2,...,v_n) \subseteq V$. However, since we assumed $w\in V$ and $w\in span(v_1,v_2,...,v_n)$, we also have that $V \subseteq span(v_1,v_2,...,v_n)$. Thus, $V=span(v_1,v_2,...,v_n)$. This is a contradiciton since we assumed $V$ was infinite dimensional, and therefore it cannot be spanned by a list of vectors. Thus, there must exist some $w\in V$ such that $w \notin span(v_1,v_2,...,v_n)$.

Now, apply the inductive step. Assume that we have a list of $m-1$ linearly independent vectors $v_1,v_2,..,v_{m-1}$ where $m \in \mathbb{N}$. We can apply our lemma which tells us that there must exist $v_m \in V$ such that $v_m \notin span(v_1,v_2,..,v_{m-1})$. This tells us that the list $v_1,v_2,...,v_{m-1},v_m$ is linearly independent (Note that this should be proven, but I did so in an earlier question, so I take the result as given). Thus, we have proved the first direction of our statement.

Now, for the other direction. Assume that there is a sequence of vectors $v_1,v_2,...,v_m$ in V that is linearly independent for every $m \in \mathbb{N}$. For contradiction, assume that V is finite dimensional. This implies that there is a list of vectors of length $x\in \mathbb{N}$ that spans V. However, by our initial assumption, we have that there exists a sequence of vectors in V of length $x+1$ where each vector is linearly independent. This is a contradiction since a linearly independent list of vectors cannot have greater length than a spanning list of vectors. Thus, V must be infinite dimensional.

Thanks for taking the time to read. Any suggestions or corrections are much appreciated!

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  • $\begingroup$ Your argument works but in my opinion makes unnecessary use of proofs by contradiction. Using the other facts you already know, it's easier to argue that if $V$ is finite-dimensional (i.e. spanned by a finite list), then there can't be arbitrarily long linearly independent lists; and conversely if there aren't arbitrarily long linearly independent lists, then a maximally long one must span, so $V$ is finite-dimensional. $\endgroup$
    – blargoner
    Oct 23, 2023 at 0:25

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