Why $Dom(f)$ needs to be open for $Df(a)$ to work?

Question

M. Spivak in Calculus on Manifolds defined differentiability as:

$f:\mathbb R^n\to\mathbb R^m$ is differentiable at $a\in\mathbb R^n$ if there exists a linear transformation $\lambda:\mathbb R^n\to\mathbb R^m$ such that $$\lim_{h\to0}\dfrac{|f(a+h)-f(a)-\lambda(h)|}{|h|}=0$$

$\lambda$ is denoted by $Df(a)$ which is called the derivative of $f$ at $a.$

He made the following remark regarding the above definition:The definition of $Df(a)$ could be made if $f$ were defined only in some open set containing $a.$

Here is the problem I am facing. Why the domain of $f$ needs to be open for the definition to work?

We taking a limit so we need to be able to approach our point, if we are possible on the boundary then there are direction from which we cannot approach our point. — Jürgen Sukumaran, May 29 '18 at 14:19

Jürgen Sukumaran · Accepted Answer · 2018-05-30T21:32:37.570

If $f$ is not defined in some neighborhood (open set) containing $a$ then we cannot make sense of an expression such as $f(a+h)$. This is what I mean by there are directions we cannot approach from. If $a$ is on the boundary of the domain of $f$ then for any neighborhood of $a$ we have point which are not in the domain of $f$. These are valid points to use when taking the limit though, which means will have issues when we try to evaluate them in expressions like $f(a+h)$.

score 1 · Answer 2 · answered May 29 '18 at 14:26

The definition of a limit is as follows:

Given a metric spaces $X, Y$ and a function $f: X \to Y$, we say that $$\lim_{x \to a} f(x) = y$$

if for any $\epsilon > 0$, there exists an open ball $B$ around $a$ such that for $b \in B$,

$$d(f(b),y) < \epsilon$$

As such, the definition of the derivative requires the domain contain open sets around $a$.

Why $Dom(f)$ needs to be open for $Df(a)$ to work?

2 Answers2