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Let $H$ be a complex, infinite-dimensional, separable Hilbert space. Let $T \in B(H)$ be an operator with disconnected spectrum. In the introduction of the paper:

Jiang, C.; Sun, S.; Wang, Z. (1997). Essentially normal operator +compact operator = strongly irreducible operator.Transactions of the American Mathematical Society, vol 349, 1, 217-233

it is stated that $T+K$ is strongly reducible for every compact operator $K$. I know that any operator $A$ having disconnected spectrum cannot be strongly irreducible, but how can we prove the the above sum is never strongly irreducible, whichever is $K$ compact? Any suggestion is greatly appreciated.

EDIT: to be more precise, on the first page of the link in the comments, it is stated that:

... P. R. Halmos proved that for each $T \in \mathcal{L}(H)$ and $\epsilon > 0$, there exists a $K$ compact with $\| K \| < \epsilon$ such that $T+K$ is irreducible [11]. It is obvious that if $\sigma(T)$ is not connected, then for each $K$ compact, $T+K$ is still strongly reducible. Thus D. A. Herrero asked the following questions ...

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  • $\begingroup$ It appears you misquoted the statement, if you are referring to what the authors call the Main Theorem (and I don't find anything else on the topic in the introduction). See here: ams.org/journals/tran/1997-349-01/S0002-9947-97-01754-6/… $\endgroup$ Jul 25, 2020 at 19:08
  • $\begingroup$ Actually, I'm not referring to the main Theorem. On the first page, before introducing the two conjectures by Herrero, it is explained why such questions are only considered for operators with connected spectrum, namely because the conjecture is said to be obviously false if the spectrum is disconnected $\endgroup$ Jul 25, 2020 at 19:34
  • $\begingroup$ Thanks for the clarification. And I suppose by "strongly reducible" they mean not strongly irreducible (I don't see any other interpretation, though it's perhaps not exactly what the grammar would suggest). $\endgroup$ Jul 26, 2020 at 14:42
  • $\begingroup$ Yes, that's precisely the definition given in other papers on Herrero's conjectures by the same authors. $\endgroup$ Jul 26, 2020 at 14:47
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    $\begingroup$ I have looked at some other papers on this topic, and I have found out that the reason why $T+K$ is not strongly irreducible (for all compact $K$) follows from Riesz decomposition Theorem: thanks to disconnectedness, this result implies the existence of nontrivial (Riesz) projections commuting with the operators considered. Consequently, the above operator is not strongly irreducible (for every compact $K$). $\endgroup$ Jul 28, 2020 at 12:11

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