I am not sure whether this question fits this forum (I will delete it if not appropriate here). But I asked this on MSE over a week ago with no answer and then put a bounty still got no answer. Here is the link on MSE and a copy the question:

Suppose $M \in \mathcal M(n \times n; \mathbb R)$ and $N \in \mathcal M(n \times m; \mathbb R)$ are fixed with $N\neq 0$. Let
\begin{align*}
E = \{X \in \mathcal{M}(m \times n; \mathbb R) : \rho(M-NX) < 1\},
\end{align*}
where $\rho(\cdot)$ denotes the spectral radius of a matrix.
I want to know whether the closure $\bar{E}$ of $E$ is equal to

$$F = \{X \in \mathcal{M}(m \times n; \mathbb R) : \rho(M-NX) \le 1\}.$$
**We will assume $E$ is not empty.**

If we define following composition of continuous maps \begin{align*} f: X \mapsto M-NX \mapsto (\lambda_1(M-NX), \dots, \lambda_n(M-NX)) \mapsto (|\lambda_1(M-NX)|, \dots, |\lambda_n(M-NX)|) \\ \mapsto \max( |\lambda_1(M-NX)|, \dots, |\lambda_n(M-NX)|). \end{align*} Then $f$ is continuous into $[0, \infty)$. We note $E = \{X: f^{-1}([0,1)\}$ and $F = \{X: f^{-1}([0,1])\}$. So $E$ is open and $F$ is closed. Clearly $\bar{E} \subset F$. But I could not determine the other direction: is it true or there is a counterexample to show not true. I tried to construct a sequence $\{X_n\}$ converging to $X \in F \setminus E$ by multiplying a factor $1-\varepsilon$ to $X$ but apparently to conclude $\bar{E} = F$ we need some kind of sublinearatily of spectral radius which is not true in general.