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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 $λ ∈ σ(T)$~

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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.

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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?.

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