I'm working in the following excercise:
Let $(T_n)_{n\in\mathbb{N}} \subseteq\mathcal{L}(E)$ and $T\in \mathcal{L}(E)$ such that $T_n\to T$. Show that for every compac set $K\subseteq R(T)$ (resolvent set of T) exists $n_0\in\mathbb{N}$ such that $$ K \subseteq R(T_n) \quad \forall n \geq n_0. $$
My attempt:
Let $K$ a subset of $R(T)$, then $$ K = \bigcup_{i=1}^m B(\lambda_i,\varepsilon) $$ On the other hand, since $T_n \to T$, then $$ (T_n - \lambda Id)^{-1} \to (T - \lambda Id)^{-1} $$ even more, the last statement is true for every $\lambda$ in $K$.
Take $n_0 = \max\{m,s\}$ where $s\in\mathbb{N}$ came from convergence of $T_n\to T$.
Let $\lambda \in K$. We want to prove that $\lambda \in R(T_n)$ where $n\geq n_0$, i.e., we want to prove that $T_n-\lambda Id$ is invertible. Note that $$ T_n-\lambda Id = (T-\lambda_j Id)(Id - (T-\lambda_j Id)^{-1}((T-\lambda_j Id)-(T-\lambda Id))) $$ The previous equality comes from $$ T = A(Id-A^{-1}(A-T)) $$ where $\lambda_j$ satisfies that $|\lambda - \lambda_j|<\varepsilon$ (since $K$ is compact). Then, if $$\|(T-\lambda_j Id)^{-1}((T-\lambda_j Id)-(T-\lambda Id))\|<1$$ following the desire result. In fact, \begin{align}\|(T-\lambda_j Id)^{-1}((T-\lambda_j Id)-(T-\lambda Id))\| &\leq \|(T-\lambda_j Id)^{-1}\| \|(T-\lambda_j Id)-(T-\lambda Id)\| \\\\ & \leq M \|(T-\lambda_j Id)-(T-\lambda Id)\| \\\\ &\leq M (\|T_n-T\| + |\lambda-\lambda_j|) \\\\ & < 1 \end{align}
Is correct my proof? Thanks
In my opinion the simplest proof is by contradiction, based on the fact that the set of noninvertible operators is closed.
Assume by contradiction that there exist: an increasing sequence $n_k\in \mathbb{N}$ and $\lambda_k\in K,$ such that $T_{n_k}-\lambda_kI$ are not invertible. As $K$ is compact we may assume (by restricting to a subsequence) that $\lambda_k$ is convergent to $\lambda\in K.$ As $T_{n_k}-\lambda_kI\to T-\lambda I$ the operator $T-\lambda I$ is noninvertible, a contradiction.