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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.

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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 $k$.

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  • $\begingroup$ So it does not hold in general. Is there a counterexample if all the $a_n$ are normal? $\endgroup$
    – Marten Wortel
    Mar 14, 2014 at 12:32
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    $\begingroup$ @Martin Wortel: You can find some information in [J.B. Conway and Morrel. Operators that are points of spectral continuity. Integral Equations operator Theory 2 (1979), 174-198]. Introduction: A result of Newburgh implies that if each $A_k$ is a normal operator and $A_k$ converges to $A$ then $\sigma(A_k)$ converges to $\sigma(A)$ with respect to the Hausdorff metric for compact sets. See 3.1 Theorem for a characterization of the operators that are points of spectral continuity. $\endgroup$
    – M.González
    Mar 14, 2014 at 16:59

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