I am reading a theorem in Linear Algebra Done Right that proves
Length of a Linearly Independent List $\le$ Length of a Space-Spanning List
Here is the theorem
$\quad\,$
I think that the reasoning in the red box is not true because it assumes that $m \le n$ since it says after step $m$! The only way that we can reach step $m$ is that we don't run out of $w_i,\,\,i=1,2,...,n$ which already means $m \le n$.
I think that we should assume that the negation is true, i.e. $m>n$, and achieve a contradiction. But I cannot find the paradox! :)
Before this theorem, things introduced in the book are the notion of fields $\Bbb{F}$, lists, $\Bbb{F}^n$, vector spaces, subspace, sums and direct sums of subspaces, linear combinations, span, linear In/dependence. So please write an answer according to theses concepts only not including basis and dimension.
I agree that the statement "After step $m$" is begging the question, but you can just remove that sentence from the proof and it works.
But here's a little more exposition:
Let's assume for the sake of contradiction that $m> n$ and then consider the $n$th step. After this step there are no more $w$'s to remove. But we know that after every step we have a list that spans $V$. Thus if we were to add one more $u$, that $u$ must necessarily be a linear combination of $u_1, \dots, u_n$ (by the definition of spanning). But that contradicts the assumption in the statement of the theorem that $u_1, \dots, u_m$ are linearly independent. Thus $m\not>n$, so $m\ge n \implies m=n$.