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Let $T$ be an bounded linear operator on a Banach space $X$. Suppose the spectrum of $T$, $\sigma(T)$ has infinitely many connected components, then $\sigma(T)$ must contain infinitely many connected components that are relatively open in $\sigma(T)$.

I come across the above statement in operator algebra paper but I have no idea why this is true. Since obviously this was not necessarily true if $\sigma(T)$ was replaced by a general set. Thus the answer must involve some structure property of spectrum of operators. But I am not sure what kind of property we need.

Anyone has a suggestion?

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Compactness. As it is the only property that may distinguish $\sigma(T)$ from a general subset of $\mathbb C$. –  martini May 23 '12 at 14:26

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up vote 4 down vote accepted

The statement is false. It is a standard exercise in Hilbert space theory that every non-empty compact subset of $\mathbb{C}$ arises as the spectrum of some (diagonal) operator.

Now let $T$ be an operator whose spectrum $\sigma(T)$ is the Cantor set $C \cong \{0,1\}^\mathbb{N}$. Its connected components are the points of the Cantor set, but no point of the Cantor set is relatively open.

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So if the statement was false then the paper would need some improvement. –  Hui Yu May 24 '12 at 0:45
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@HuiYu: As t.b. said, you can see this using diagonal operators. See these questions for reference: 1, 2, 3, 4. (For the dense sequence in $C$, you could take e.g. the endpoints of the removed middle thirds.) You could also let $\mu$ be the Lebesgue-Stieltjes measure of the Cantor function, and let $T$ be multiplication by $x$ on $L^2(\mu)$. –  Jonas Meyer May 24 '12 at 1:18
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@HuiYu: I looked at the paper and it looks more like a slip than a serious error to me, but I'm not too acquainted with the techniques used there. What I think they really need for the argument to work is that they can find a countable set of disjoint rectifiable curves in the resolvent set, separating countably many pieces of the spectrum (so that they can decompose parts of the operator using Riesz projections) and I'm pretty sure that this is possible for compact planar sets with infinitely many connected components, although I haven't carefully written down an argument. –  t.b. May 24 '12 at 10:26
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@JonasMeyer:thanks very much for your links to those related questions! They are really helpful! –  Hui Yu May 24 '12 at 15:27
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@t.b.:I agree with you now. I might be talking to the authors of that paper later about this. Thanks so much! –  Hui Yu May 24 '12 at 15:29

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