Continuously extending a set of independent vectors to a basis. 
Question: Let $I=(a,b)$ be an interval and let
  $$v_i:I\to\mathbb{R}^n,\quad i=1,\ldots,k$$
  be continuous curves such that $v_1(t),\ldots,v_k(t)$ are linearly independent in $\mathbb{R}^n$ for every $t\in I$. Can we always find continuous curves
  $$v_{k+1},\ldots,v_n:I\to\mathbb{R}^n$$
  such that $v_1(t),\ldots,v_n(t)$ forms a basis for $\mathbb{R}^n$ for each $t\in I$?

I know that if $v_1(t),\ldots,v_k(t)$ is a set of linearly independent vectors, then we can extend it to a basis $v_1(t),\ldots,v_n(t)$ for $\mathbb{R}^n$. But how do we ensure that $v_{k+1},\ldots,v_n$ will be continuous?
It is intuitively very clear that there must exist such continuous curves, because there are a lot of possible choices for extending the set to a basis. Surely, there must be no problem of doing it continuously, but I cannot find a way to prove it rigorously.
It is easy to do this in $\mathbb{R}^3$ when we have two curves $v_1,v_2$. In that case, we can just take the cross product
$$v_3(t)=v_1(t)\times v_2(t)\in\mathbb{R}^3.$$
But in general, I don't know any explicit expression for producing linearly independent vectors $v_{k+1},\ldots,v_n$ in terms of the previous ones. Is there any?
 A: 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 $V$).
