Lower-rank matrices form a closed set, higher-rank an open one

Idea. Rank is characterised by the vanishing of minors, which are polynomials in the entries.

Step 1: minors and rank. $\operatorname{rank}A$ is the largest $k$ such that some $k\times k$ minor (determinant of a $k\times k$ submatrix) is non-zero. Hence
$$\operatorname{rank}A\le r\iff\text{all }(r+1)\times(r+1)\text{ minors of }A\text{ vanish.}$$

Step 2: $\{\operatorname{rank}\le r\}$ is closed. It is a finite intersection of zero sets of the continuous functions "minor", hence closed.

Step 3: $\{\operatorname{rank}\ge r\}$ is open. If $\operatorname{rank}A\ge r$, some $r\times r$ minor $m(A)\ne0$ (for $r=0$ the set is everything). Since $m$ is continuous, $m(B)\neq0$ for all $B$ in a neighbourhood of $A$, so $\operatorname{rank}B\ge r$ there. (This is the *lower semicontinuity* of rank: it can jump up, never down, under small perturbation.)

(Note the complement of $\{\operatorname{rank}\ge r\}$ is $\{\operatorname{rank}\le r-1\}$, consistent with Step 2 for $r-1$.)

Step 4: empty interior when $r<n$. Let $\operatorname{rank}A\le r<n$. Then $A$ is singular and by the density of $GL_n$ (earlier problem) $A+\varepsilon I\in GL_n$ for small $\varepsilon\ne0$, so $\operatorname{rank}(A+\varepsilon I)=n>r$. Thus every neighbourhood of $A$ contains matrices not in $\{\operatorname{rank}\le r\}$.

Remark.
- For $r\ge n$ the set $\{\operatorname{rank}\le r\}$ is all of $M_n$, which has non-empty interior: the restriction $r<n$ is needed.
- In words: *rank can only go up in a neighbourhood*. The set of rank-exactly-$r$ matrices is neither open nor closed (next problem).