Spectral continuity: $W = \{x \in A : \sigma_A(x)\subset U\}$ is open in $A$. Let $A$ be a unital Banach algebra and $U \subset \Bbb C$ be an open set.
I need to prove that $W = \{x \in A : \sigma_A(x)\subset U\}$ is open in $A$.
Here $\sigma_A(x) = \{\lambda \in \Bbb C : x-\lambda I \in A$ is not invertible$\}$.
So, I know that $\sigma_A(x)$ is compact and not empty for every $x\in A$, so by compactness and the fact that $U$ is open we can take $\lambda_1,\dots ,\lambda_k \in \Bbb C$ s.t. $\sigma(x)\subset \bigcup_{i=1}^k B_{r_i}(\lambda_i) \subset U$. 
I thought this could help, but I can't see how to continue with this direction. 
Thanks for helping.
 A: Instead of showing that $W$ is open, we show that $A\setminus W$ is closed! Let $(a_n)_{n \geq 0} \subseteq A\setminus W$ and assume that $a_n \to a$. Our goal is to show that $a \in A \setminus W$. This means we have to exhibit an element $\lambda \in \sigma_A(a)$ with $\lambda \not\in U$.
Since for each $n$ we have that $a_n \in A\setminus W$, this means that $\sigma_A(a_n)$ is not contained in $U$, meaning that there is $\lambda_n \in \sigma_A(a_n)$ with $\lambda_n \not\in U$.  Thus, consider the sequence $(\lambda_n)_{n \geq 0}$.

Key step: since $a_n \to a$, there is $n_0$ large enough such that for $n \geq n_0$ we have that $\lambda_n \in \sigma_A(a)$.
Proof: by contradiction, assume not and extract a subsequence $(\lambda_{n_k})_{k \geq 0}$ such that each $\lambda_{n_k}-a$ is invertible in $A$. Then $$\|(\lambda_{n_k}-a_{n_k})-(\lambda_{n_k}-a)\| = \|a_{n_k}-a\| \to 0,$$contradicting that $\lambda_{n_k} \in \sigma_A(a_{n_k})$ for some large $k$, since the set of invertible elements in $A$ is open.

Now, since $\sigma_A(a)$ is compact, passing to a subsequence we may assume that we have a limit $\lambda_n \to \lambda \in \sigma_A(a)$. And this $\lambda$ cannot be in $U$, or else for large $n$ we would have $\lambda_n \in U$ (this is the step where the fact that $U$ is open finally kicks in), another contradiction.

Caveat lector: I'm slightly abusing notation and sweeping the embedding $\Bbb C \ni \lambda \mapsto \lambda 1 \in A$ under the rug.
