Exercise 1.7.7

Let $F$ be the family of all linearly independent subset of $\beta$ such that union of it and $S$ is independent. Then for each chain $C$ of $F$ we may choose $U$ as the union of all the members of $C$. We should check $U$ is a member of $F$. So we should check wether $S\cup U$ is independent. But this is easy since if ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{a}_i{v}_i+{\sum }_{j=1}^m{b}_j{u}_j=0$ with $v_i\in S$ and $u_j\in U$, say $u_j\in U_j$ where $\{U_j\}$ are a members of $C$, then we can pick the maximal element, say $U_1$, of $\{U_j\}$. Thus we have $u_j\in U_1$ for all $j$. So $S\cup U$ is independent and hence $a_i=b_j=0$ for all $i$ and $j$.

Next, by Maximal principle we can find a maximal element $S_1$ of $C$. So $S\cup S_1$ is independent. Furthermore by the maximality of $S_1$ we know that $S\cup S_1$ can generate $\beta$ and hence can generate $V$. This means $S\cup S_1$ is a basis for $V$.