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This is an old pre lim question that I'm doing for practice. So suppose we have a sequence $A_n$ of bounded linear operators on a Banach space which converges to a bounded linear operator $A$. If we assume the spectrum $\sigma_0=\sigma(A_i)$ are all equal. Prove that $\sigma_0\subseteq \sigma(A)$. Now if $\lambda\in \sigma_0$ then $A_n-\lambda I\rightarrow A-\lambda I$ since $A_n \rightarrow A$. By definition $A_n-\lambda I$ are not bijective, so I'm stuck on showing that $A-\lambda I$ isn't bijective. Does anyone have any hints for this?

Not sure if it matters, but recall from a long time ago about the the resolvent set of the limit of such operators is contained in the union of the resolvents of the sequence from which our result would be immediate, but I'm more sure if I'm even remembering this correctly.

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The set of all bijective bounded linear operators on a Banach space is open. Can you take it from her ?

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  • $\begingroup$ Ah, yes I got it. Thanks. $\endgroup$ – thegamer Sep 13 '17 at 3:55

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