# Connected components that are relatively open in $\sigma(T)$

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

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.
@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