Rank is lower but not upper semicontinuous

Idea. Having rank $\ge r$ is an *open* condition (some $r\times r$ minor is nonzero), while having rank $\le r$ is *closed*.

Step 1: minors. $\operatorname{rank}A\ge r$ iff some $r\times r$ minor of $A$ is nonzero. (If some minor is nonzero, the corresponding $r$ rows are independent. Conversely, if the rank is $\ge r$, one finds $r$ independent rows, and among the columns $r$ independent ones of the resulting $r\times n$ matrix.)

Step 2: lower semicontinuity. Let $r=\operatorname{rank}A$ and choose an $r\times r$ minor $m(A)\ne0$. Minors are polynomials in the entries, hence continuous, so $m(B)\ne0$ for all $B$ in a neighbourhood $U$ of $A$. For $B\in U$: $\operatorname{rank}B\ge r=\operatorname{rank}A$. Equivalently, $\{\operatorname{rank}\ge r\}$ is open for every $r$.

Step 3: not upper semicontinuous. Upper semicontinuity at $A$ would mean that $\operatorname{rank}B\le\operatorname{rank}A$ for all $B$ near $A$. At $A=0$, $B=\frac1kI\to0$ has rank $n>0$. So it fails at $0$. More generally it fails at every singular $A$, since $A+\varepsilon I$ is invertible for small $\varepsilon\ne0$ (Part 2).

Consequences.
- The *maximal* rank $n$ is attained on an open dense set, which is why "generic" matrices are invertible.
- $\{\operatorname{rank}\le r\}$ is closed, and by Step 3 has empty interior for $r<n$.

Remark. Rank is continuous exactly on $GL_n$ (the set where it is constant on a neighbourhood, being maximal). More precisely, the points of continuity of $\operatorname{rank}$ are the matrices of rank $n$: at a matrix of rank $r<n$, perturbation by $\varepsilon I$ raises the rank.