= Solution
{c}
For a finite set $S=\{s_1,\ldots,s_k\}$ of <vectors> in a real <vector space>, its <span> is the set of all <linear combinations>:
$$
\boxed{\operatorname{Sp}(S)=\left\{\sum_{i=1}^ka_is_i:a_i\in\mathbb R\right\}.}
$$
The <span> of the empty set is $\{0\}$. The set $S$ is <linearly independent> if $\sum_i a_is_i=0$ implies $a_i=0$ for every $i$. It is a <spanning set> for $V$ if $\operatorname{Sp}(S)=V$. A <basis> is both a <linearly independent> set and a <spanning set>.
For <basis extension>, start with the given <linearly independent> set. If it does not span $V$, choose $v\notin\operatorname{Sp}(S)$ and adjoin it. The enlarged set remains <linearly independent>: a relation $av+\sum_i a_is_i=0$ with $a\ne0$ would express $v$ as a <linear combination> of $S$, a contradiction; if $a=0$, independence of $S$ forces all other coefficients to be zero.
Repeat whenever the current <span> is proper. Each step increases the set's size by one, while the allowed bound says its size cannot exceed $\dim V$. Thus the process terminates with a <linearly independent> set spanning $V$, hence a <basis> containing the original set. This also covers an initially empty set and the zero-dimensional <vector space>.
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