Orbits of non-diagonalisable matrices are not closed

Real Jordan form. Over $\mathbb{R}$, $A$ is similar to a block-diagonal matrix whose blocks are:
- for a real eigenvalue $\lambda$: the Jordan block $J_k(\lambda)=\lambda I_k+N_k$, with $N_k$ having ones on the superdiagonal;
- for a pair of complex eigenvalues $a\pm ib$: the block built from $C=\begin{pmatrix}a&-b\\b&a\end{pmatrix}$ on the diagonal and $I_2$ on the block superdiagonal.

So $A\sim S+N$ with $S$ block diagonal (the diagonal parts), $N$ the "superdiagonal-of-blocks" part, and $SN=NS$. Since $A$ is not diagonalisable over $\mathbb{C}$, some Jordan block has size $\ge2$, so $N\ne0$.

Rescaling. Index the diagonal positions of each Jordan block by $i=1,2,\dots$ (for a complex pair, each index stands for a $2\times2$ block) and let $D_t$ be the block diagonal matrix with scalar block $t^{-i}$ in position $i$ of each Jordan block. Then
- $D_tSD_t^{-1}=S$, because inside each position $S$ is a scalar $\lambda I$ or a $2\times2$ block $C$, and $D_t$ acts there as a scalar;
- an entry of $N$ linking position $i$ to position $i+1$ is multiplied by $t^{-i}/t^{-(i+1)}=t$: $D_tND_t^{-1}=tN$.

Hence $D_t(S+N)D_t^{-1}=S+tN\to S$ as $t\to0$.

Conclusion. $S\in\overline{\mathcal{O}(A)}$. But $S$ is diagonalisable over $\mathbb{C}$ while $A$ is not, so $S$ is not similar to $A$: $S\notin\mathcal{O}(A)$. Thus $\mathcal{O}(A)$ is not closed.

Check. For $J_2(1)=\begin{pmatrix}1&1\\0&1\end{pmatrix}$: $S=I$, $N=E_{12}$, and the limit is $I$, matching the previous problem.

Remark. The converse of this result is true: orbits of matrices diagonalisable over $\mathbb{C}$ are closed (the characterisation $\{B:p(B)=0,\chi_B=\chi_A\}$ with $p$ the squarefree polynomial vanishing at the distinct eigenvalues, now over $\mathbb{R}$). Lesson. Jordan blocks are "unstable": the nilpotent part can be shrunk to zero by conjugation.