Proof that spanning set is greater than linearly independent subset Let G be a set that generates V containing n vectors and H be a linearly independent subset of V containing m vectors. Prove that $m \le n$.
My attempt:
We will use a proof by induction with base case $m=0$.
$m=0$ implies that $H =∅$ and since m cannot be less than $0$, $m \le n$.

Our induction hypothesis is that the theorem holds for some $n&gt;0$. We need to show that it holds for $m+1$, ie. $m+1 \le n$.

Suppose the opposite, that $m+1 \ge n$ $\implies m&gt;n$ which cannot be true but I am not sure how to prove it.
Any help is appreciated!
If $G = \{ v_1, \cdots, v_n\}$ generates $V$, then $n \geq\dim V $ by the definitions of basis and dimension. If $H = \{w_1, \cdots, w_m \}$ has $m$ linearly independent vectors, again by the definitions of basis and dimension we have $m \leq \dim V$. So $m \leq \dim V \leq n$, as we wanted to prove.
If $G = \{ v_1, \cdots, v_n\}$ generates $V$, then $n \geq\dim V $ by the definitions of basis and dimension. If $H = \{w_1, \cdots, w_m \}$ has $m$ linearly independent vectors, again by the definitions of basis and dimension we have $m \leq \dim V$. So $m \leq \dim V \leq n$, as we wanted to prove.