Suppose $a\_n \to a$ in a unital C\*-algebra $A$. If $\lambda\_n \in \sigma(a\_n)$ and $\lambda\_n \to \lambda$, then $\lambda \in \sigma(a)$. Does the converse hold?


So if $\lambda \in \sigma(a)$, does there exist a sequence $\lambda\_n \in \sigma(a\_n)$ with $\lambda\_n \to \lambda$?


It is true if $A$ is commutative or $A = M\_n(\mathbb{C})$, but in the general case I don't see a proof. It feels like such a basic thing should be known, but I cannot find it anywhere - and my StackExchange question was unanswered.



Continuous depedence of the spectrum on elements
The result is not true. 


The sequence of operators $U\_k$ on $\ell\_2(\mathbb{Z})$ defined by $U\_ke\_n = e\_{n-1}$ for $n\in\mathbb{Z}\setminus 0$ and $U\_ke\_0=k^{-1}e\_{-1}$ converges to $U\_\infty$ defined by $U\_\infty e\_n = e\_{n-1}$ for $n\in\mathbb{Z}\setminus 0$ and 
$U\_\infty e\_0=0$. 


The spectrum of $U\_\infty$ is the closed unit disc and the spectrum of $U\_k$ is just the boundary of the unit disc for all $k$. 


