A question about the properties of the pseudospectrum Assume that $A\in \mathbb{C}^{n\times n}$. The $\epsilon-$pseudospectrum of $A$ is defined by
$$\sigma_{\epsilon}(A)=\{z\in C \quad | \quad \Arrowvert (zI-A)^{-1} \Arrowvert>\frac{1}{\epsilon}\}.$$
Why $\sigma_{\epsilon}(A)$ is a bounded open subset of $\mathbb{C}$?

Take $x\in \mathbb C^n$, $z\in \mathbb C$. Then it holds
$$
|z|\cdot \|x\| = \|zx\| \le \|(zI-A)x\| + \|Ax\| \le  \|(zI-A)x\| + \|A\|\cdot \|x\|,
$$
which implies
$$
\|(zI-A)x\| \ge (|z|-\|A\|)\|x\|.
$$
If $|z| \ge \|A\| + \epsilon$, then this implies
$$
\|(zI-A)^{-1}\| \le \frac1{|z|-\|A\|} \le \frac1\epsilon.
$$
Hence $\sigma_\epsilon(A)$ is bounded.
