Size of Linearly Independent Subset is at Most Size of Finite Generator/Proof 1

Theorem
Let $G$ be a $K$-vector space which is generated by a finite set of $p$ vectors.

Then every linearly independent subset of $G$ is finite and contains at most $p$ vectors.

Hence if $T$ is a linearly independent subset of $G$, then $\left|{T}\right| \le \dim_K \left({G}\right)$.

Proof
Let $H$ be a finite linearly independent subset of $G$ containing $m$ vectors.

First we prove that every finite generator $F$ for $G$ contains at least $m$ vectors.

Let $S$ be the set of $n \in \N^*$ such that:
 * For every finite set $F$ of generators for $G$, if the relative complement $H \setminus F$ has $n$ vectors then $F$ contains at least $m$ vectors.

If $H \setminus F = \varnothing$, then $H \subseteq F$, so $F$ contains at least $m$ vectors.

Thus $0 \in S$.

Now assume that $n \in S$.

Let $F$ be a finite set of generators for $G$ such that $H \setminus F$ has $n + 1$ vectors.

Let $a$ be a vector in $H \setminus F$.

Then $\left({H \cap F}\right) \cup \left\{{a}\right\}$ is a subset of $H$, and so by Subset of Linearly Independent Set is linearly independent.

So by Linearly Independent Subset of Basis of Vector Space, there is a basis $B$ of $G$ such that $\left({H \cap F}\right) \cup \left\{{a}\right\} \subseteq B \subseteq H \cup \left\{{a}\right\}$.

Then $H \setminus B = \left({H \setminus F}\right) \setminus \left\{{a}\right\}$, so $H \setminus B$ has $n$ vectors.

Consequently, as $n \in S$, $B$ has at least $m$ vectors.

Since $a$ is a linear combination of $F$ but $a \notin F$, $F \cup \left\{{a}\right\}$ is linearly dependent.

Therefore $B$ is a proper subset of $F \cup \left\{{a}\right\}$.

Thus if $b$ is the number of vectors in $B$ and if $h$ is the number of vectors in $F$, we have $m \le b < h + 1$.

So $m \le h$.

Hence $n + 1 \in S$ and so by induction $S = \N^*$.

Thus, if $H$ is a finite linearly independent subset of $G$, then $H$ contains no more than $p$ vectors.

If there existed an infinite linearly independent subset $H$ of $G$, then $H$ would contain a subset having $p + 1$ vectors, which would again be linearly independent, which is a contradiction.

Thus every linearly independent subset of $G$ is finite and contains at most $p$ vectors.

By the definition of dimension, it follows that $\dim_K \left({G}\right) \ge m$.