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Let $T\in B(l^2)$ be s.t. $Tx=(\alpha_1 x_1, \alpha_2 x_2, \cdots )$, where the set of all $\alpha$ is dense in $[0,1]$.

I've shown that the set of all eigenvalues is $A=(\alpha_j)_1^\infty$. The resolvend operator, where it exists, is bounded. Therefore, the continuous spectrum is empty. The range of $T-\lambda I$ is the entire $l^2$. Therefore, the residual spectrum is empty.

But is this true? I never used the fact that $\alpha$ is dense in $[0,1]$, which leads me to believe that at least one of my conclusions above is false.

(If possible, I prefer hints over solutions.)

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    The differense that "A is dense in [0,1]" makes is that I know $||T|| < 1$.2012-12-17
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    Haven't you learned yet that the spectrum is always closed in $\Bbb{C}$?2012-12-17
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    @ChrisEagle Yes, and on a Banach space, compact.2012-12-17
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    Then you know your answer is incorrect. So you should examine your argument to find the errors.2012-12-17
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    Thanks, I think I've found the flaw. (Can't delete my question, though?)2012-12-17
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    You shouldn't delete your question: you should post an answer yourself. You may need to wait a while since you're a new user.2012-12-17
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    Ok, I'll write an answer in 8h then :)2012-12-17

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