About the statement of Lemma 2-10 on p.35 in "Calculus on Manifolds" by Michael Spivak. Is $A$ open or closed?

Question

I am reading "Calculus on Manifolds" by Michael Spivak.

The following lemma is on p.35 in this book.

Lemma 2-10 : Let $A \subset \mathbb{R}^n$ be a rectangle and let $f : A \to \mathbb{R}^n$ be continuously differentiable. If there is a number $M$ such that $| D_j f^i (x) | \leq M$ for all $x$ in the interior of $A$, then $$ |f(x)-f(y)| \leq n^2 M |x-y| $$ for all $x,y \in A$.

---

If $A$ is an open rectangle, then "$| D_j f^i (x) | \leq M$ for all $x$ in the interior of $A$" is strange because the interior of $A$ is equal to $A$.
So, if $A$ is an open rectangle, "$| D_j f^i (x) | \leq M$ for all $x$ in $A$" is natural.
So, I think $A$ is a closed rectangle.

---

But if $A$ is a closed rectangle, then the author didn't need to write the following since this automatically holds:

If there is a number $M$ such that $| D_j f^i (x) | \leq M$ for all $x$ in the interior of $A$

---

Is $A$ open or closed?
Since this lemma is used in the proof of the inverse function theorem, so I think it is best to read the proof of the inverse function theorem first.
I think I can decide $A$ is open or closed.

Answer 1

The Lemma, in general, only assumes the inequality and not the openness or closedness of the rectangle, just that it is some rectangle. As you've pointed out, if $A$ is a closed rectangle, then the inequality he assumes is automatically satisfied. When the rectangle is open, the assumption is not redundant (e.g. $f(x) = \sqrt{x}$ and $A = (0,1)$).

Answer 2

I read the proof of the inverse function theorem.

I wonder why the author didn't write as follows:

Lemma 2-10' : Let $A \subset \mathbb{R}^n$ be a closed rectangle and let $f : A \to \mathbb{R}^n$ be continuously differentiable. Let $M$ be a number such that $| D_j f^i (x) | \leq M$ for all $x$ in $A$, then $$ |f(x)-f(y)| \leq n^2 M |x-y| $$ for all $x,y \in A$.

This lemma (Lemma 2-10') is sufficient for the proof of the inverse function theorem.

Answer 3

It is irrelevant whether $A$ is open, closed or neither. Spivak could prove the slightly more general

Lemma: Let $A \subset \mathbb{R}^n$ be a rectangle and let $f : A \to \mathbb{R}^n$ be continuous and differentiable in the interior of $A$. If there is a number $M$ such that $| D_j f^i (x) | \leq M$ for all $x$ in the interior of $A$, then $$ |f(x)-f(y)| \leq n^2 M |x-y| $$ for all $x,y \in A$.

Note that Spivak actually only considers functions $f : \mathbb R^n \to \mathbb R^m$ because "considering only functions defined on $\mathbb R^n$ streamlines the statement of theorems and produces no real loss of generality. He defines such $f$ to be differentiable on $A \subset \mathbb R^n$ if $f$ is differentiable at all $a \in A$. A function $f : A \to \mathbb R^m$ is called differentiable if it has an extension $\tilde f : \mathbb R^n \to \mathbb R^m$ which is differentiable on some open $U \supset A$. Note that if $A$ is open, then we may take any extension of $f$, for example $\tilde f(x) = 0$ for $x \notin A$. This works because differentiability at $a \in A$ only depends only the values of $f$ close to $a$, thus on the values of $f$ contained in the open set $A$.

The proof of the above Lemma uses the mean value theorem for functions $\phi : [a,b] \to \mathbb R$. It says thatif $\phi$ is continuous and differentiable on $(a,b)$, then $\phi(b) - \phi(a) = (b-a)\phi'(z)$ with some $z \in (a,b)$. Now osberve that if $f$ is differentiable in the interior of $A$, then all $D_jf^i$ exist in the interior of $A$.