Let $f\in C^1(\mathbb{R}^n\to\mathbb{R}^m)$ and let $a \in \mathbb{R}^n$
Prove there exists an open ball $B(a,\epsilon) $ such that for every $x\in B(a,\epsilon)$:
$rankD_f(a)\le rankD_f(x)$
I only know the basic definitions and don't know where to start.
Any help would be appreciated.
Let $k =\text{rank } Df(a)$. Let $U$ be the set of all $m \times n$ matrices with rank $\geq k$. As mentioned in the comments, $U$ is an open subset of the vector space $M_{m \times n}(\mathbb{R})$ (I'll prove this below). By assumption, $f$ is $\mathcal{C^1}$, which means $Df: \mathbb{R^n} \to M_{m \times n}(\mathbb{R})$ is continuous. Hence the preimage $(Df)^{-1}(U)$ is an open subset of $\mathbb{R^n}$ containing $a$. Thus, there is an $\varepsilon > 0$ such that for all $x \in B(a, \varepsilon)$, $Df(x) \in U$. This is exactly what we wanted to show.
Proof:
Let $A \in U$. Then, since $\text{rank}(A) \geq k$, there is some $k \times k$ minor of $A$ which has non-zero determinant. For the sake of definiteness, let's say this minor is obtained by selecting rows $(i_1, \dots, i_k)$ and columns $(j_1, \dots, j_k)$ of $A$. Now, define a "projection function" $\pi: M_{m\times n}(\mathbb{R}) \to M_{k \times k}(\mathbb{R})$, by the rule that $\pi(X)$ is the $k \times k$ minor obtained by selecting rows $(i_1, \dots, i_k)$ and columns $(j_1, \dots, j_k)$ of $X$. This projection $\pi$ is continuous. Also, the determinant function $\det: M_{k \times k}(\mathbb{R}) \to \mathbb{R}$ is continuous. So, the composition $\det \circ \pi$ is continuous. Since $\mathbb{R} \setminus\{0\}$ is open, the set $V = (\det \circ \pi)^{-1}(\mathbb{R} \setminus\{0\})$ is an open subset of $M_{m\times n}(\mathbb{R})$. By construction, $A \in V$, and $V \subset U$. Thus, we have shown that for every $A \in U$, there is an open subset $V$ of $M_{m\times n}(\mathbb{R})$ such that $A \in V \subset U$. Hence, $U$ is a union of open sets, and hence open as well.