"considering functions on $\mathbb{R}^n$ produces no loss of generality"

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This is in the context of defining the derivative. Here is the full quote from Calculus on Manifolds, pg 17:

The definition of $Df(a)$ could be made if $f$ were defined only in some open set containing $a$. Considering only functions defined on $\mathbb{R}^n$ streamlines the statement of theorems and produces no real loss of generality.

Why is it no loss of generality? What is the formal statement that is being made?

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The corresponding notion on $\Bbb R$ is something like this:

"We could define differentiability for a function on an open interval, but we'll be defining it for functions on $\Bbb R$ without loss of generality. The definition involves limits, for example, and so rather than writing something like

$$ \lim_{x \in (a, b), x \to c} \ldots $$ we can simply write $$ \lim_{x \to c} \ldots . $$

Why? Because for every function $f$ on the interval $(a, b)$, we can define a function $\hat{f}$ on $\Bbb R$ via $$ \hat{f}(x) = \begin{cases} f(x) & a \le x \le b \\ 0 & \text{otherwise} \end{cases}. $$

And for any $c \in (a, b)$, the limits of $f$ and $\hat{f}$, as we approach $c$ will be identical, since limits involve only local properties."