# We can find at most $x$ linearly independent vectors that span a subspace $V$

Consider vectors $\vec{v_1},\ldots,\vec{v_p}$ and $\vec{w_1},\ldots,\vec{w_q}$ in a subspace $V$ of $\mathbb{R}^n$. If the vectors $\vec{v_1},\ldots,\vec{v_p}$ are linearly independent, and the vectors $\vec{w_1},\ldots,\vec{w_q}$ span $V$, then $q \ge p$.

I can see how this would be true, but I'm not sure how it can be proven. What type of rigor would be expected for a proof like this? I was thinking that, conceptually, we know that $\vec{w_1},\ldots,\vec{w_q}$ span $V$, so they must contain a subset of linearly independent vectors that also span $V$, because we can eliminate all of the redundant vectors in $\vec{w_1},\ldots,\vec{w_q}$ and still obtain the same span.

$v_1,\ldots,v_p$ linearly independent in $V$ then $p\leq \dim V$.
$w_1,\ldots,w_q$ span $V$ then $\dim V\le q$.
so $p\le \dim V\le q$.