# Banach Spaces, Convergence and Spectrum

Let $$V$$ be a Banach space and $$T_n → T$$ in $$B(V)$$. Assume $$λ_n ∈ σ(T_n)$$ and $$λ_n → λ$$, I want to show that $$λ ∈ σ(T)$$.

Okay, so if $$\lVert T_n-T\rVert_{\mathcal B(V)}\to 0$$ and $$\lambda_n\to \lambda$$, then $$\lVert (T_n-\lambda_nI)-(T-\lambda I)\rVert_{\mathcal B(V)}\to 0$$. Right? Hmm how do I continue to conclude $$λ ∈ σ(T)$$~

## 2 Answers

The set of invertible bounded operators is a neighborhood of $$I$$ (if $$\|H\| < 1$$, then $$\sum_{n \geq 0}{H^n}=(I-H)^{-1}$$).

Therefore (using left translation), the set of invertible bounded operators is open.

Thus the set of singular bounded operators is closed.

If $$\|I-T\| <1$$ the $$T$$ is invertible with inverse $$I+(I-T)+(I-T)^{2}+\cdots$$. Now, if $$A$$ is invertible and $$\|A-B\|<\frac 1 {\|A^{-1}\|}$$ then $$\|I-BA^{-1}\|< 1$$ which makes $$BA^{-1}$$ invertible. But then $$B$$ is itself invertible. Hence invertible operators form an open set. Can you complete the argument now?.