TITLE: Why approximative spectrum of operator is closed
QUESTION [6 upvotes]: Let $A$ be a linear operator on Hilbert space $H$. 
We say that $\lambda \in \mathbb{C}$ is in approximative spectrum of $A$ iff there exists a sequence $(x_n)$ of vectors such that $\|x_n\|=1$ and $\|Ax_n-\lambda x_n\|\rightarrow 0$.
Equivalently, $\lambda$ is in approximative spectrum of $A$ iff for every $\varepsilon >0$ there exists $x  \neq 0$ such that $\|Ax-\lambda x\| \leq \varepsilon \|x\|$.
How to prove that approximative spectrum is a closed subset of $\mathbb{C}$ ?
Thanks.

REPLY [5 votes]: Let $\lambda$ which is not in the approximative spectrum. We can find a positive integer $n$ such that for all $x$: $\lVert Ax-\lambda x\rVert\geq n^{-1}\lVert x\rVert$. For $\lambda_1\in\mathbb C$ such that $|\lambda_1-\lambda|<n^{-1}$ we have 
$$\lVert Ax-\lambda_1 x\rVert=\lVert Ax-(\lambda_1-\lambda) x-\lambda x\rVert\geq \lVert Ax-\lambda x\rVert-|\lambda_1-\lambda|\lVert x\rVert\geq (n^{-1}-|\lambda_1-\lambda|)\lVert x\rVert$$ 
so $\lambda_1$ is not in the approximative spectrum. This show that the approximative spectrum has an open complement, therefore it is closed.