Spectrum in Banach Algebra Let $A$ be a unital Banach algebra and $a\in A$. Let $U$ be an open subset of $\mathbb C$ containing $\sigma (a)$. Prove that there is $\delta>0$ such that for every $b\in A$, if $\|b-a\|<\delta$ then $\sigma(b)\subseteq U.$What I've tried, Since $U$ is open we can find an open ball $B$ with radius $\|a\|+\epsilon$ such that $\sigma(a)\subseteq B\subseteq U$. Let $\lambda \ge \|a\|+\epsilon$, this implies $\lambda-a$ is invertible with inverse $c$, let $\delta=\frac{1}{\|c\|}$ then if $\|b-a\|=\|\lambda-b-(\lambda-a)\|<\delta$ then $\lambda-b$ is inverible which implies $r(b)\le \|a\|+\epsilon$, Thus, $\sigma(b)\subseteq B\subseteq U$ as required.I have a feeling that I'm missing something or did something worng, can you help me?
 A: Such a $B$ need not exist (draw pictures), and I don't see how you have fixed a $\delta>0$ (your $\delta$ depends on which $\lambda$ you chose), but otherwise there are similarities to your work and the approach I'll share here.
Let $\rho(a)=\mathbb C\setminus \sigma(a)$ denote the resolvent set of $a$.  Define 
$f:\rho(a)\to (0,\infty)$ by $f(z)=\|(a-z)^{-1}\|$.  Then $f$ is continuous and $\lim\limits_{|z|\to\infty}f(z)=0$.  Because each set $\overline{B(0,R)}\setminus U$ is compact, it follows that $f$ takes on a maximum value $C>0$ on $\mathbb C\setminus U\subset \rho(a)$.  We will see that $\delta=\dfrac{1}{C}$ suffices.
Suppose that $\|a-b\|<\dfrac{1}{C}$ and $\lambda\in\mathbb C\setminus U$.  Then $\|(a-\lambda)^{-1}\|\leq C$, so $$\|(a-\lambda)-(b-\lambda)\|=\|a-b\|<\dfrac{1}{C}\leq\dfrac{1}{\|(a-\lambda)^{-1}\|}.$$
This implies that $b-\lambda$ is invertible.  Thus, $\lambda \in \rho(b)$.  Since $\lambda\in\mathbb C\setminus U$ was arbitrary, this shows that $\sigma(b)\subseteq U$.
