Proof related to $C([a,b],\mathbb{R}^n)=\{ f:[a,b]\to \mathbb R^n:f=(f_1\ldots f_n)/f \text{ is continuous on }[a,b]\}$ where $[a,b]\subset \mathbb R$ Question
Let $[a,b] \subset \mathbb{R}$,
We define the set $C([a,b], \mathbb{R}^n) = \{ f : [a,b] \rightarrow \mathbb{R}^n:f=(f_1\ldots f_n)/f \text{ is continuous on } [a,b]\}$
Prove that $C([a,b], \mathbb{R}^n)$ forms a linear space in relation to the estimation operations of two functions in $C([a,b], \mathbb{R}^n)$, respectively of multiplication with real scalars with functions from $C([a,b], \mathbb{R}^n)$.
My Attempt
I know that we have to prove the $10$ axioms of vector spaces, I just don't know how to take the functions, I mean in what form to take them, so that I can prove those axioms in $\mathbb{R}^n.$ Please, I just need a hint on how to take those two functions $f$ and $g$ in what form to prove those $10$ axioms.
 A: Some of the axioms are readily verified and don't require much attention, seeing as they are more or less pretty obvious. Perhaps the more sensible ones would be showing that the addition of two continuous functions is again continuous, as well as that any scalar multiple of a continuous function is indeed continuous, so I will prove these for you here.
Let $f, g \in C([a, b], \mathbb{R}^n)$ and $x \in [a, b]$ fixed. Let $\varepsilon > 0$ and $\lambda \in \mathbb{R} \setminus \{0\}$ be fixed (that the function which is identically zero is continuous is immediate). Since $f$ and $g$ are continuous at $x,$ there exist $\delta_\varepsilon^f, \delta_\varepsilon^g > 0$ such that $\Vert f(x) - f(y) \Vert < \min \{\frac{\varepsilon}{2}, \frac{\varepsilon}{\vert \lambda \vert} \} > \Vert g(x) - g(y) \Vert$ whenever $\vert x - y \vert < \delta_\varepsilon^f$ and $\vert x - y \vert < \delta_\varepsilon^g,$ respectively, where I think it's pretty clear how they correspond to each other ($\Vert \cdot \Vert$ denotes the standard euclidean norm on $\mathbb{R}^n$). It follows that for all $y \in [a, b]$ with $\vert x - y \vert < \min \{\delta_\varepsilon^f, \delta_\varepsilon^g\},$ we have that $$\Vert f(x) + g(x) - f(y) - g(y) \Vert \leq \Vert f(x) - f(y) \Vert + \Vert g(x) - g(y) \Vert \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$ and $$\Vert \lambda f(x) - \lambda f(y) \Vert = \vert \lambda \vert \Vert f(x) - f(y) \Vert \leq \vert \lambda \vert \frac{\varepsilon}{\vert \lambda \vert} = \varepsilon,$$ which proves continuity of $f + g$ and $\lambda f.$
