The closure of the real-diagonalisable matrices

Idea. The only obstruction to being a limit of diagonalisable matrices is having non-real eigenvalues. Real-rooted matrices can always be nudged to have distinct real eigenvalues.

Let $\mathcal T=\{A:\chi_A\text{ splits over }\mathbb{R}\}$ (all roots real, with multiplicity).

Step 1: $\overline{\mathcal D_{\mathbb{R}}}\subset\mathcal T$. If $A$ is diagonalisable over $\mathbb{R}$ then $A\in\mathcal T$, so $\mathcal D_{\mathbb{R}}\subset\mathcal T$. It suffices to show that $\mathcal T$ is closed. Let $A_k\in\mathcal T$, $A_k\to A$, and let $z$ be a root of $\chi_A$. By continuity of roots (Part 3), for $k$ large there is a root $z_k$ of $\chi_{A_k}$ with $|z-z_k|\to0$. The roots $z_k$ are real, and the limit of real numbers is real, so $z\in\mathbb{R}$. Hence all roots of $\chi_A$ are real: $A\in\mathcal T$.

Step 2: $\mathcal T\subset\overline{\mathcal D_{\mathbb{R}}}$. Let $A\in\mathcal T$. Since all eigenvalues are real, $A$ is triangularisable over $\mathbb{R}$ (Schur: $A=PTP^{-1}$ with $P$ invertible, $T$ upper triangular with the eigenvalues on the diagonal; one can take $P$ orthogonal). Let $\Delta_\varepsilon=\varepsilon\operatorname{diag}(1,2,\dots,n)$. The matrix $T_\varepsilon=T+\Delta_\varepsilon$ is upper triangular with diagonal entries $t_{ii}+i\varepsilon$. These are pairwise distinct for all small $\varepsilon\ne0$: $t_{ii}+i\varepsilon=t_{jj}+j\varepsilon$ forces $\varepsilon=\frac{t_{ii}-t_{jj}}{j-i}$ for some pair $i\neq j$, which excludes only finitely many values of $\varepsilon$. For such $\varepsilon$ the matrix $T_\varepsilon$ has $n$ distinct real eigenvalues, so is diagonalisable over $\mathbb{R}$, and so is $A_\varepsilon=PT_\varepsilon P^{-1}$. Finally $\|A_\varepsilon-A\|=\|P\Delta_\varepsilon P^{-1}\|\le\varepsilon n\|P\|\|P^{-1}\|\to0$.

Remark.
- Example: $\begin{pmatrix}1&1\\0&1\end{pmatrix}\in\mathcal T$ is the limit of $\begin{pmatrix}1&1\\0&1+\varepsilon\end{pmatrix}$; but the rotation $\begin{pmatrix}0&-1\\1&0\end{pmatrix}\notin\mathcal T$ and is at positive distance from $\mathcal D_{\mathbb{R}}$.
- Thus the three sets satisfy $\mathcal S=\operatorname{int}\mathcal D_{\mathbb{R}}\subset\mathcal D_{\mathbb{R}}\subset\overline{\mathcal D_{\mathbb{R}}}=\mathcal T$, with both inclusions strict for $n\ge2$.