$V$ is finite dimensional vector space. $G$ is a spanning set of $V$, $I \subseteq G$ is linearly independent

Question

I'm totally new to the subjects like spanning set, bases and dimension. Here is a theorem that I could not understand.

Let $V$ be a finite dimensional vector space. If $G$ is a finite spanning set of $V$ and if $I \subseteq G$ is linearly independent, then there is a basis $B$ of $V$ such that $I \subseteq B \subseteq G$.

If $\text{Span}(I)= V$ then $B = I$ and we have nothing left to prove.

I know that if $\text{Span}(I)= V$ and since $\text{Span}(B)= V$ follows $\text{Span}(I)= \text{Span}(B)$ but how do we know $I=B$ and why we have nothing left to prove?

Suppose $\text{Span}(I) \ne V$ then $I \ne G$ and $G \setminus \text{Span}(I) \ne \emptyset$ (if $G \subseteq \text{Span}(I)$ we would have $V=\text{Span}(G) \subseteq \text{Span}(I)$).Thus $\exists g_1 \in G$ such that $g_1 \notin \text{Span}(I)$. Then $I'= I \cup \left\{g_1 \right\} \subseteq G$ is linearly independent.

Why $\text{Span}(I) \ne V$ does imply that $I \ne G$ and also $I'= I \subseteq V$?

If $\text{Span}(I')=V$ we are done. If not,$\exists g_2 \in G$ such that $g_2 \notin \text{Span}(I')$ and we get a linearly independent set $I''= I' \cup \left\{g_1 ,g_2\right\} $. If $\text{Span}(I'')=V$ we are done. If not, we continue. Since $G$ is a finite set, this process will come to an end.Finally we will get a linearly independent set $I'= I \cup \left\{g_1 ,g_2,\ldots,g_r\right\} \subseteq G$ which spans $V$ . This is the basis $B$ and obviously $I \subseteq B \subseteq G$.

I know that $G$ is a spanning set of $V$ but why a subset of $G$ like $I'= I \cup \left\{g_1 ,g_2,\ldots,g_r\right\} $ is also a spanning subset of $G$?

Answer 1

1. We don't know that $B=I$; we define $B$ as being equal to $I$.
2. Since $\operatorname{Span}(G)=V$, if $\operatorname{Span}(I)\ne V$, then $I\ne G$. So, $I\varsubsetneq G$. And $I'\subseteq G$ because $I'=I\cup\{g_1\}$, where $I$ is a subset of $G$ and $g_1\in G$.
3. Because otherwise $I'\varsubsetneq G$ and then we could take an element $g$ of $G$ and add it to $I'$ so that $I'\cup\{g\}$ would still be linearly independent. But we cannot keep on doing that forever, since $G$ is finite. So, sooner or later, that process must come to an end.