Prove that $U$ is an open subset of $\Bbb {R}^{n}$

Question

Let $m\gt n$ and $f:\Bbb {R}^{n}\to \Bbb {R}^{m}$ be a smooth function. Let $U$ be the subset of $\Bbb {R}^n$ consisting of points $p$ so that $Df(p)$ has rank n. Prove that $U$ is an open subset of $\Bbb {R}^{n}$.

I have no idea how to start this question. What does $Df(p)$ has rank n mean?

Answer 1

If $p\in U$, fix local coordinates near both $p$ and $f(p)$. With respect to those coordinates, $Df(p)$ has matrix, with $n$ rows and $m$ columns. By the definition of $U$, $\operatorname{rank}Df(p)=n$, which means that the determinant of at least on of the $n\times n$ submatrices of that matrix is not $0$. But then it must be different from $0$ in some neighborhood of $p$, by continuity. So, every point of that neighborhoodbelongs to $U$. SO, $U$ is a neighborhood of each of its points and therefore it is an open set.

Answer 2

Hint

It's a full answer, but I prefer to put it as a Hint, because depending on the level of the OP, there can be some justifications to add to complete the proof.

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Suppose $U$ is not open, i.e., there is $p_m\to p$ s.t. $Df(p_m)$ has rank at most $n-1$ and $Df(p_m)\to Df(p)$. Let $x_m\in \mathbb R^n\setminus \{0\}$ s.t. $$Df(p_m)x_m=0\tag{E}$$ (why such $x_m$ exist for all $m$ ?). We can suppose WLOG that $\|x_m\|=1$ for all $m$ (why ?). Therefore, there is a subsequence of $(x_m)$ (still denoted $(x_m)$) that converges. Denote $x$ it's limit and remark that $x\neq 0$ (why ?). Taking $m\to \infty $, in $(E)$, we get we get $Df(p)x=0$ (why ?). This contradict the fact that the rank of $Df(p)$ is $n$.