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.

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 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 and 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.