Proof of the rank theorem in Rudin's PMA book I am studying Rudin's proof of the rank theorem (theorem 9.32 in Principles of Mathematical Analysis.) We have an invertible function $H(x)$ defined on an open set. He claims we can "shrink" the open set on which the function $H(x)$ is defined, and thereby guarantee that $H'(x)$ is invertible everywhere on this subset. Can anyone explain this process to me?
 A: even though the question is old, here is an answer that refers directly to rudins book.
Rudin does several steps at once, which confused me at first, too. First we apply the inverse function theorem to find open sets $U$ and $V_1$, such that $G$ is 1-to-1 on $U$ and $ a \in U $. This is possible, since $ G'(a) = I $ is invertable. We define H as the inverse of $G$ on $V_1$. Let me denote $b \in V_1$ as the element for which $H(b)=a$. Since $V_1$ is open we can find a ball $V_2$ around $b$ with a radius, such that $V_2 \subset V_1$. Again $V_2$ is open and $H'(b)$ is invertable, so that we can apply theorem 9.8, which states, the set of invertable linear operators in $R^n$ is open. Again we can find a ball $V$ with radius so small, that $H(x)$ is invertable on $V$ and $V \subset V_2$. The restriction of $H$ to $V$ fullfills all desired properties.
tl;dr: See theorem 9.8 (a): The set of invertable linear operators on $R^n$ is open.
Cheers
A: $U$ is a neighbourhood of $a$, and $G'(a)=I$, so $H'(G(a))$ is invertible, that is, determinant of $H'(G(a))$ is nonzero. Since $G$ is continuous, $H$ is continuously differentiable, and determinant is continuous, there is a neighbourhood of $a$ such that determinant of $H'(G(x))$ is nonzero as well, hence $H'(G(x))$ is invertible in such neighbourhood. -- Marcin Łoś
