We are reading a proof about the following limit
\begin{equation}\tag{1}
\lim\_{n \to \infty} \sigma\_1(T\_n)= \sigma\_1(T),
\end{equation}
where $T:D(T) \subseteq H \to H$ and $T\_n:D(T\_n) \subseteq H \to H$ are linear operators on a Hilbert space $H$ and
$$\sigma\_1(A)= \sigma(A) \cup \{ z \in \mathbb{C}: \|(z-A)^{-1} \|>1\}$$
for a linear operator $A$. In the proof it is said that it is enough to show the following:


(i) If $K \subseteq \sigma\_1(T)$ is compact then there is $N \in \mathbb{N}$ such that $K \subseteq \sigma\_1(T\_n)$ for all $n \geq N$.


(ii) If $K$ is compact and $K \cap \overline{\sigma\_1(T)}=\emptyset$ then there is $N \in \mathbb{N}$ such that $K \cap \overline{\sigma\_1(T\_n)}=\emptyset$ for all $n \geq N$.


But we don't know why (i) and (ii) implies (1)?.


The definition of convergence of sets which is used in (1) is the following:
Let $\{X\_n\}$ a sequence of subsets of $\mathbb{C}$. Then $x \in \limsup X\_n$ iff there exists a subsequence $\{ X\_{n\_k}\}$ and a sequence $\{ x\_k \}$ such that $x\_k \in X\_{n\_k}$ and $\lim\_{k \to \infty} x\_k=x$. On the other hand, $x \in \liminf X\_n$ iff there is sequence $\{x\_n \}$ with $x\_n \in X\_n$ suc that $\lim\_{n \to \infty} x\_n=x$. The limit $\lim\_{n \to \infty} X\_n$ exists if $\limsup X\_n=\liminf X\_n$ and we define $\lim\_{n \to \infty} X\_n=\limsup X\_n$.


**Our attempt:**


We think that (i) implies that $\sigma\_1(T) \subseteq \liminf \sigma\_1(T\_n)$ because if $z \in \sigma\_1(T)$ then $\{z \}$ is compact.


From (ii) we can get that $\limsup \overline{\sigma\_1(T\_n)} \subseteq \overline{\sigma\_1(T)}$. Indeed if $z \notin \overline{\sigma\_1(T)}$, then there exists $\varepsilon>0$ such that $\overline{B(z;\varepsilon )} \cap \overline{\sigma\_1(T)}= \emptyset$ and from (ii) we get $\overline{B(z;\varepsilon )} \cap \overline{\sigma\_1(T\_n)}= \emptyset$ for all large enough $n$, so $z \notin \limsup \overline{\sigma\_1(T\_n)}$.


But how can we show that $\limsup \sigma\_1(T\_n) \subseteq \sigma\_1(T)$?


Thank you for any help you can provide us.



Understanding a proof about limit of a sequence of open sets
You have hit the nail on the head and indeed there is a mistake with the proof you are reading. The argument already fails for matrices. If $T\_n \equiv T \in \mathbb{C}^{n \times n}$ then $\sigma\_1(T)$ is open by definition, but 
$$ \lim \sigma\_1(T\_n) = \overline{\sigma\_1(T)}, $$
as any element in the closure of $\sigma\_1(T)$ is the limit of a sequence lying in $\sigma\_1(T)$. This sequence we can interpret as a sequence lying in the $\sigma\_1(T\_n)$.


The conditions (i) and (ii) provided are trivially satisfied as we have $\sigma\_1(T\_n) = \sigma\_1(T)$. So (i) and (ii) do not imply the claimed equality. Indeed your definition of the limit of a sequence of sets implies that the limit is closed, if it exists. So the claim has to be modified accordingly.


