Matrices of rank exactly r: neither open nor closed

Idea. By the previous problem $\{\operatorname{rank}\le r\}$ is closed and $\{\operatorname{rank}\ge r+1\}$ is open. The set $\{\operatorname{rank}=r\}$ is their "difference" and sits in between.

Let $D_r=\operatorname{diag}(1,\dots,1,0,\dots,0)$ with $r$ ones; it has rank $r$.

Step 1: not open. $D_r+\varepsilon E_{nn}$ has diagonal $(1,\dots,1,0,\dots,0,\varepsilon)$ with $r+1$ non-zero entries (this uses $r<n$ so that position $n$ is a zero of $D_r$), so rank $r+1$. It tends to $D_r$ as $\varepsilon\to0$. So $D_r$ has no neighbourhood inside $\{\operatorname{rank}=r\}$.

Step 2: not closed. $\frac1kD_r$ has rank $r$ for every $k\ge1$ and converges to $0$, which has rank $0\neq r$ (here $r>0$ is used).

Step 3: the closure is $\{\operatorname{rank}\le r\}$.
- ($\subset$) $\{\operatorname{rank}=r\}\subset\{\operatorname{rank}\le r\}$ and the latter is closed.
- ($\supset$) Let $\operatorname{rank}A=s\le r$. Write the SVD $A=P\Sigma Q^T$ with singular values $\sigma_1\ge\dots\ge\sigma_s>0=\sigma_{s+1}=\dots=\sigma_n$. Define $A_\varepsilon=P\Sigma_\varepsilon Q^T$ where $\Sigma_\varepsilon$ replaces $\sigma_{s+1},\dots,\sigma_r$ by $\varepsilon>0$ (and keeps the rest). Then $\operatorname{rank}A_\varepsilon=r$ and $\|A_\varepsilon-A\|=\varepsilon\to0$.

Remark.
- Matrices of rank exactly $r$ form a *manifold* of dimension $r(2n-r)$, but not a closed one: its boundary consists of the lower-rank matrices.
- For $r=n$ the set $\{\operatorname{rank}=n\}=GL_n$ is open; for $r=0$ it is $\{0\}$, which is closed. Both are excluded by $0<r<n$.