Theorem of Existence:
Let $E \neq\{0\}$ be a finite dimensional vector space and $G$ a spanning set. Let us consider $L\subseteq G$ a linearly independent subset of $E$.
Then there exists a basis $B$ of E such that $L\subseteq B \subseteq G$.
Proof
Let $G=${$v_1$, $v_2$,…, $v_p$} be a spanning set and $L_1$ be a linearly independent subset contained in G.
Let's suppose $L_1=\{v_1, v_2,\dotsc,v_r\}$. If $L_1$ is a spanning set of $E$ the the Theorem has been proven.
Let us now suppose that $L_1$ is not a span.
a) Let's show that there exists $w\in \{v_{r+1}, v_{r+2},\dotsc, v_p\}$ such that $L_2=\{v_1, v_2,\dotsc, v_r, w\}$ is linearly independant.
If there is not such a vector $w$, each vector of $\{v_{r+1}, v_{r+2},\dotsc, v_p\}$ would be a linear combination of $L_1$'s vectors.
And thats where I have a problem. Why would each vector of $\{v_{r+1},v_{r+2},\dotsc, v_p\}$ be a linear combination of $L_1$'s vectors if $L_2$ wasn't linearly independant.
Best Answer
Because you know that $L_1$ is linearly independent and you add a new vector $v_{i_1}$ in $L_1$ to get $L_2$ and you picked $v_{i_1}$ from some set, so if $L_2$ is not linearly independent than that means that no matter what vector $v_{i_1}$ you chose, you can represent it as a linear combination of vectors from $L_1$.