# Understanding a proof about limit of a sequence of open sets

We are reading a proof about the following limit $$$$\tag{1} \lim_{n \to \infty} \sigma_1(T_n)= \sigma_1(T),$$$$ where $$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 $$\limsup \sigma_1(T_n) \subseteq \sigma_1(T)$$?

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.
• Thank you so much. I had been coffused about that proof for weeks. I think that the claim is true if we change $\sigma_1(T)$ by $\overline{ \sigma_1(T)}$. You agree with me? Commented Jul 24, 2019 at 23:32
• I do. This follows from your arguments if you note that from the definition already $\liminf$ and $\limsup$ are closed. Commented Jul 24, 2019 at 23:59