A question on subordinate matrix norm Suppose: 


*

*$B_i \in \mathbb{C}^{n \times n}$, $0<w_i\in \mathbb{R}$ $(i = 0,1,2,\ldots,m)$

*${\rm P}(x) ={\rm{B}_m} x ^m + \cdots + B_1 x  + A_0$ is a matrix polynomial,  and $x $ is a complex variable.

*${\rm q}(x) =\rm{w}_m x^m + \cdots +w_1 x + w_0$

*$U= \left\{ {x  \in \mathbb{C}:\left\| P(x)^{-1} \right\| \ge (\varepsilon q(|x|))}^{-1} \right\}$, where $\left\| \cdot \right\|$ is any subordinate matrix norm.


Why can the boundary of $U$ be written in the form $S = \left\{ x  \in \mathbb{C}:\left\| P(x )^{-1} \right\| = (\varepsilon q(|x|))^{-1} \right\}$?

$h(x)=\Vert P(x)^{-1} \Vert - (\varepsilon q(|x|))^{-1}$ then $U=h^{-1}[0, \infty)\Rightarrow U $ is closed. So $\partial U=\bar{U}-\mathring{U}\subseteq {U}-({h^{-1}(0,\infty)})^o={U}-{h^{-1}(0,\infty)}=h^{-1}(0)=S$
in last line we have used: $\color{red}{\mathring{A}\subseteq (A\cup B)^{o}}$ and $U=h^{-1}(0,\infty)\cup h^{-1}(0).$
