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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 $\mathbb C$.2012-05-23

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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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    t.b. and @HuiYu: You're welcome. (t.b.: Thanks, I'm not sure if I was either, but my inbox says that I have 3 comments on this post, which might mean that I was. If so, maybe they improved the ping system.)2012-05-24