The definition of continuous mapping on a vector space In the book of Linear Algebra by Werner Greub, at page 136, it is given that 

Let $a_v$ and $b_v (v= 1, ..., n)$ be two bases of $E$. Then the basis
  $a_v$ is called deformable into the basis $b_v$ if there exist n
  continuous mappings $$x_v:t \to x_v(t) \quad t_o \leq t \leq t_1$$
  satisfying the conditions
1-) $x_v(t_o)=a_v \quad and \quad x_v(t_1)=b_v$
2-)The vectors $x_v(t)(v= 1,..., n)$ are linearly independent for
  every fixed $t$.

Note that $E$ is a real vector space, and the map $x_v$ is defined on $E$, so how can we talk about the continuity of $x_v$. I mean what is meant by "continuity" of $x_v$ since $E$ is an arbitrary real vector space ?
 A: Each finite-dimensional real vector space $E$ has a canonical topology defined in either of the following ways:
Define an arbitrary norm on $E$, and talk about continuity using this specific norm. All norms are equivalent to each other on a finite-dimensional vector space (see for Kreyszig's Introductory Functional Analysis with Applications, page 75). By norm equivalence of two norms $\lVert\cdot\rVert_1,\lVert\cdot\rVert_2$, we mean that there exists two positive constants $c,C\gt0$ such that for all vectors $x\in E$, $c\lVert x\rVert_1\le\lVert x\rVert_2\le C\lVert x\rVert_1$. Because of this theorem, the notion of a continuous function from $[t_0,t_1]$ to $E$ is independent of the choice of norm.
The following way could be unfamiliar to you right now, but you should be able to understand once you learn about definition of topology and notion of continuity in terms of topology:
Pick any vector space isomorphism $L:\Bbb R^n\to E$ from $\Bbb R^n$ to $E$. Define a topology $\tau_L$ on $E$ by $\tau_L=\{L(U)\subseteq E:U\text{ is open in }\Bbb R^n\}$, where openness of a subset of $\Bbb R^n$ is to be understood with the Euclidean metric. This $\tau_L$ is a topology on $E$, and all such $\tau_L$ are equal no matter which isomorphism $L$ we pick (i.e. $\tau_L=\tau_{L'}$ for any two isomorphisms $L,L':\Bbb R^n\to E$). So we can talk about continuous functions from $[t_0,t_1]$ to $E$ using this topology $\tau_L$.
Note also that the above two definitions of continuous functions using norms and using topology are equivalent.
A: You have learnt in advanced calculus that a vector valued function $${\bf f}:\quad{\mathbb R}\to{\mathbb R}^n,\qquad t\mapsto\bigl(f_1(t),f_2(t),\ldots,f_n(t)\bigr)$$
is continuous iff the $n$ coordinate functions $t\mapsto f_i(t)$ are continuous.
In the case at hand we have $n$ such vector valued functions $t\mapsto{\bf x}_\nu(t)$ $\>(1\leq\nu\leq n)$. In order to apply the above principle we choose a basis of $E$ once and for all, say the "given starting basis" $({\bf a}_1,{\bf a}_2,\ldots,{\bf a}_n)$, and describe the complete dynamical process envisaged by Greub in this basis. Each function $t\mapsto{\bf x}_\nu(t)$ then appears as
$$t\mapsto {\bf x}_\nu(t)=\bigl(x_{\nu,1}(t),x_{\nu,2}(t),\ldots,x_{\nu,n}(t)\bigr)\qquad(t_0\leq t\leq t_1)\ .$$
The vector valued functions $t\mapsto{\bf x}_\nu(t)$ are continuous iff all $n^2$ coordinate functions $$t\mapsto x_{\nu,i}(t)\qquad(1\leq\nu\leq n, \ 1\leq i\leq n)$$ are continuous.
This notion of continuity does not depend on the chosen basis, since changing the basis amounts to replacing the $x_{\nu,i}(\cdot)$ by  constant linear combinations of the $x_{\nu,i}(\cdot)$.
