Convergence of the spectrum under norm resolvent convergence Suppose $\{A_n\}$ is a sequence of self-adjoint operators in a Hilbert space $\mathcal H$, and $A$ is a self-adjoint operator, with $A_n \to A$ in norm resolvent sense.

Since $A_n \to A$ in strong resolvent sense also, if $\lambda \in \sigma(A)$ then there exists $\lambda_n \in \sigma(A_n)$ for all $n$, such that $\lambda_n \to \lambda$.

Since $A_n \to A$ in norm resolvent sense, if $\rho \not \in \sigma(A)$ then there exists $N$ such that $\rho \not \in \sigma(A_n)$ for all $n &gt; N$.

Suppose now that we have $\lambda_n \in \sigma(A_n)$ for all $n$, and $\lambda_n \to \lambda$ for some $\lambda$. How can we see that $\lambda \in \sigma(A)$?
Assuming you mean that $\|A_n - A\| \to 0$, suppose that $\lambda \notin \sigma(A)$, then $(A-\lambda I)$ is invertible. The set of invertible elements is open, so $\exists \epsilon &gt; 0$ such that if 
$$
\|B - (A-\lambda I)\| &lt; \epsilon
$$
then $B$ must also be invertible. Now choose $B$ to be of the form $A_n - \lambda_n I$ for large enough $n$. This would contradict the fact that $\lambda_n \in \sigma(A_n)$.
Assuming you mean that $\|A_n - A\| \to 0$, suppose that $\lambda \notin \sigma(A)$, then $(A-\lambda I)$ is invertible. The set of invertible elements is open, so $\exists \epsilon &gt; 0$ such that if 
$$
\|B - (A-\lambda I)\| &lt; \epsilon
$$
then $B$ must also be invertible. Now choose $B$ to be of the form $A_n - \lambda_n I$ for large enough $n$. This would contradict the fact that $\lambda_n \in \sigma(A_n)$.