Power-bounded matrices do not form a closed set

The sequence. For $0<\varepsilon<1$ let
$$A_\varepsilon=\begin{pmatrix}1-\varepsilon&1\\0&1-\varepsilon\end{pmatrix}\oplus0_{n-2}.$$
Its eigenvalues are $1-\varepsilon$ (and $0$), so $\rho(A_\varepsilon)<1$ and $A_\varepsilon^k\to0$ by the previous results. A convergent sequence is bounded, so $A_\varepsilon^k$ is bounded in $k$: $A_\varepsilon\in\mathcal{P}$.

The limit. As $\varepsilon\to0$, $A_\varepsilon\to A_0=\begin{pmatrix}1&1\\0&1\end{pmatrix}\oplus0$. By induction $\begin{pmatrix}1&1\\0&1\end{pmatrix}^k=\begin{pmatrix}1&k\\0&1\end{pmatrix}$, whose norm is at least $k$. Hence $\sup_k\|A_0^k\|=\infty$: $A_0\notin\mathcal{P}$.

Conclusion. $A_\varepsilon\in\mathcal{P}$, $A_\varepsilon\to A_0\notin\mathcal{P}$, so $\mathcal{P}$ is not closed.

Remark. What is true: $\mathcal{P}$ is contained in $\{\rho\le1\}$ and contains $\{\rho<1\}$; and the set $\{A:\|A^k\|\le1\ \forall k\}$ for a given norm is closed. A matrix is power bounded iff $\rho\le1$ and the eigenvalues of modulus $1$ are semisimple (no nontrivial Jordan block). The growth of $\|A_\varepsilon^k\|$ before decay is of order $\frac1{e\varepsilon}$, which blows up as $\varepsilon\to0$: power-boundedness holds but not *uniformly*.