TITLE: Why is differential calculus built on open sets?
QUESTION [16 upvotes]: For example in W. Rudin: Principles of Mathematical Analysis in every theorem or definition regarding the derivatives of a function from $\mathbb{R}^n \to \mathbb{R}^m$ there it always says at the beginning: "Let $E$ be an open set in $\mathbb{R}^n$, $f$ maps $E$ to $\mathbb{R}^m$ ...".
Why is the differential calculus built on the open subsets of $\mathbb{R}^n$? What is the precise (maybe topological) explanation?

REPLY [20 votes]: The best way to understand this is to understand what an open set is. Recall that in metric spaces like $\mathbb{R}^n$ we define an open set as a set $U$ such that for every $p \in U$ there is some real number $\epsilon > 0$ such that the ball of radius $\epsilon$ and center $p$ is inside of $U$. Now, it says then that points less than a distance $\epsilon$ from $p$ are still in $U$ and we can understand this as saying "as long as we move $p$ arround without going a distance greater than $\epsilon$ we are still in $U$". So an open set is really a set on which for every point there is a small amount that the point can be moved around and will still be inside of the set.
Now let's take the derivative. First, imagine we have a function $f : \mathbb{R}^n \to \mathbb{R}^m$, so this function is defined inside all of $\mathbb{R}^n$. We say that $f$ is differentiable at $a$ if there exists a linear map called derivative $Df(a) \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$ such that
$$\lim_{h\to 0}\frac{|f(a+h)-f(a)-Df(a)(h)|}{|h|}=0$$
So, we start at $a$ and move along $h$. Compute $f$ when we begin, that's $f(a)$ and compute it again when we end moving, that's $f(a+h)$ then we take the difference $f(a+h)-f(a)$. So, $f$ is differentiable if we can approximate this with a linear map (this is what the definition says).
Now, what was important in that discussion, is that we could vary $f$ from $f(a)$ to $f(a+h)$. Now, this was possible because even if $h$ was imensely large in norm, the point $a+h$ would still be in the domain of $f$. If you have an arbitrary set of $\mathbb{R}^n$ it could happen that $a+h$ would end up going away from the set. Since we shrink $h$ to zero, it's not needed that for every $h$ the point $a+h$ is in the set, but that for sufficiently small $h$ it is (smallness with respect to the norm of course).
Because of that we require $U$ to be open: it grants us that for every point there will be really an amount of distance that we can move the point $a$ to $a+h$ without going away from the domain of definition.
If you want an example, consider $A = \{(x,y)\in \mathbb{R}^2 : x^2+y^2 \leq 1\}$. This set is not open. Now consider the point $a=(0,1)$, if we take $\epsilon >0$ and consider the vector $h=(0,\epsilon)$, then $a+h=(0,1+\epsilon)$ and this point is not inside $A$ anymore. If $f : A \to \mathbb{R}^m$, then what should be $f(a+h)$? This is undefined, so the derivative doesn't make sense.