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  "body": "\n## Definition\n\nLet $H$ be a [[Hilbert space|HM2HHE3J51]] and let $\\mathcal{B}(H)$ be the [[algebra of bounded operators|BMDG4JQPEH]] on $H$ with the operator norm.\nThe *ideal of compact operators* on $H$ is the set\n$$\\mathcal{K}(H) = \\set{T \\in \\mathcal{B}(H) : T \\text{ is a compact operator}},$$\nwith compactness in the sense of [[compact operators|M2NEP1K5AE]].\nIt is a norm-closed [[two-sided ideal|0CD87SFM2N]] of $\\mathcal{B}(H)$ and is closed under taking adjoints.\n\n## Remarks\n\nExample: on $H = \\ell^2$, the diagonal operator with entries $1, 1/2, 1/3, \\ldots$ belongs to $\\mathcal{K}(H)$, since it is the norm limit of its finite-rank truncations.\nFor any bounded $A$, the products $AT$ and $TA$ are again in $\\mathcal{K}(H)$.\nIndeed, a bounded operator maps bounded sets to bounded sets and relatively compact sets to relatively compact sets.\n\nNon-example: if $H$ is infinite-dimensional, the identity $I$ is not in $\\mathcal{K}(H)$, since an orthonormal sequence has no convergent subsequence; so $\\mathcal{K}(H)$ is a proper ideal.\nIf $H$ is finite-dimensional, every operator is compact and $\\mathcal{K}(H) = \\mathcal{B}(H)$.\n\nOther notations are $\\mathcal{K}$ and $\\mathcal{B}_0(H)$.\nOn a Hilbert space, $\\mathcal{K}(H)$ equals the norm closure of the finite-rank operators.\n",
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