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I want to show that the set of all compact operators $K(H)$ is the unique ideal in $B(H)$. Is there any relation between invertibility and compactness of an operator?

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1. Assuming $H$ is separable, it's the unique proper closed ideal. (the operators of finite rank are also an ideal, as well as the Schatten $p$-ideals, for instance) 2. An invertible operator is compact if and only if $H$ is finite-dimensional. – t.b. Jun 25 '11 at 6:28
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Since this is a standard result explained in textbooks on operator algebras, is it homework? What have you tried, where are you stuck, what hints did you get, what auxilliary results are you allowed to use? – Tim van Beek Jun 25 '11 at 9:14

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