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I'm trying to prove that the spectrum of an operator is closed.

I was thinking of using the fact that the set of invertible operators is open, and hence the set of non-invertibles operators must be closed. Is it ok?

Also, if I wanted to go by a more direct way, such as using a sequence of eigenvalues, how would I do it?

Edit: So, we know that $\{T-\lambda: \exists (T-\lambda)^{-1}\}$ is open in $B(X,Y)$.

Define $F_A(\lambda)=A-\lambda$. We then know that $|F_A(\lambda)-F_A(\mu)|=|\lambda-\mu|$ making $F_A$ continuous.

$\sigma(A)^c=F_A^{-1}(\{T-\lambda: \exists (T-\lambda)^{-1}\})$ is open by continuity and our sentence above.

So, $\sigma(A)$ is closed.

Is this correct?

An old man in the sea.
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    Suggestion: Given an operator $T$, convince yourself that the set of complex numbers $\lambda$ such that $(\lambda I - T)^{-1}$ exists and is a bounded operator is an open set. It is well known that the spectrum is a nonempty compact set in $\mathbb{C}$ – Mittens Nov 06 '19 at 19:58
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    @OliverDiaz Thank you for your comment, but I don't see how it helps me. You're just restating my question in your first sentence, and in the second you're giving a stronger statement than my doubt, by using compactness... – An old man in the sea. Nov 06 '19 at 20:03
  • It is not too difficult to show that the residual set, the set of numbers $\lambda$ for which $(\lambda I -T)$ is invertible and has continuous inverse is open. you can start by looking at the geometric series $\sum_n \lambda^{-n}T^n$ for $\lambda$ large enough (say $|\lambda| >|T|$) – Mittens Nov 06 '19 at 20:44
  • @OliverDiaz I've added a possible answer to my doubt. Do you think it's correct? – An old man in the sea. Nov 06 '19 at 20:46
  • It's not quite correct, $F_A$ isn't linear if $A \neq 0$. It's affine, but the only property you need is that it's continuous (and that it is, it's even isometric). Also, you write "open in $B(X,Y)$", but for $T - \lambda$ to make sense you need $Y = X$. But the argument as such is correct. – Daniel Fischer Nov 06 '19 at 21:06
  • @DanielFischer Many thanks! I don't know what happened, I was thinking continuous, but I wrote linear... – An old man in the sea. Nov 06 '19 at 21:23

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Using Daniel Fischer's comment:

we know that $\{T-\lambda: \exists (T-\lambda)^{-1}\}$ is open in $B(X)$.

Define $F_A(\lambda)=A-\lambda$. We then know that $|F_A(\lambda)-F_A(\mu)|=|\lambda-\mu|$ making $F_A$ continuous.

$\sigma(A)^c=F_A^{-1}(\{T-\lambda: \exists (T-\lambda)^{-1}\})$ is open by continuity and our sentence above.

So, $\sigma(A)$ is closed.

An old man in the sea.
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The map $\lambda\mapsto L(X)$ is continuous and $\{T\in L(X): T^{-1} \text{exists and is bounded}\}$ is open. Thus $\operatorname{res}_T:=\{\lambda: (\lambda I-T)^{-1}\text{ exists and is bounded}\}$ is open in $\mathbb{C}$. From this, it we get that the spectrum, being the complement of $\operatorname{res}_T$, is closed.

Mittens
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