Continuously extending a set of independent vectors to a basis.

Solution 1:

Let's call the continuous functions $v_1, \dotsc, v_k$ linearly independent (resp. a basis) if the vectors $v_1(t),\dotsc,v_k(t)$ are linearly independent (resp. a basis) in $\Bbb{R}^n$ for every $t \in I$.

We shall proceed by induction. If $n = 1$ then $k = 0$, so you can choose $v_1$ to be the constant function with value $e \in \Bbb{R}^*$. Now that the base case is cleared we assume that $v_1,\dotsc,v_{n-1}$ are linearly independent functions and we are going to complete them to a basis.

For each $t \in I$ consider the space $V_t \subset \Bbb{R}^n$ spanned by $v_1(t),\dotsc,v_{n-1}(t)$. Since the $v_j$ are continuous in $t$ we can also see $V_t$ as the section of a surface $V \subset X := I \times \Bbb{R}^n$ by the hyperplane $\{x_0 = t\}$. It isn't hard to see that $V$ is homeomorphic to $I \times \Bbb{R}^{n-1}$: for example consider the map $$ (t, a_1v_1(t) + \dotsb + a_{n-1}v_{n-1}(t)) \mapsto (t,a_1e_1(t) + \dotsb + a_{n-1}e_{n-1}(t)) $$ where $e_1,\dotsc,e_{n-1}$ are the constant functions with values in the standard basis of $\Bbb{R}^{n-1}$. In particular, this means that $V$ divides $X$ in two halves. You can choose as $v_n$ any continuous curve which takes exactly one value for every $t \in I$ and which is entirely contained within one of those halves (without touching $V$).