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  "body": "\n## Definition\n\nLet $X$ and $Y$ be [[normed spaces|03J5RAKB30]] over $\\R$ or $\\C$ and let $T \\colon X \\to Y$ be a [[linear map|0RJF668P1T]].\nLet $B_X = \\set{x \\in X : \\norm{x} \\le 1}$ be the closed unit ball of $X$.\nThe map $T$ is *compact* when the [[closure|0QEAJB91GZ]] of $T(B_X)$ is a [[compact subset|0R6EWFKRCF]] of $Y$.\n\n## Remarks\n\nEquivalent form: $T$ is compact if and only if for every bounded sequence $(x_n)$ in $X$, the sequence $(Tx_n)$ has a convergent subsequence in $Y$.\nA compact operator is bounded, since a compact set is bounded.\n\nExample: on $\\ell^2$ let $T$ be the diagonal operator $T(x_1, x_2, \\ldots) = (x_1, x_2/2, x_3/3, \\ldots)$.\nThe finite-rank operators $T_N$ that keep the first $N$ entries of $Tx$ and set the rest to $0$ satisfy $\\norm{T - T_N} \\le 1/(N+1)$.\nA norm limit of finite-rank operators into a Banach space is compact, because the image of the unit ball is then totally bounded; so $T$ is compact.\n\nNon-example: the identity of $\\ell^2$ is not compact.\nThe basis vectors $e_n$ lie in the unit ball, and $\\norm{e_n - e_m} = \\sqrt{2}$ for $n \\ne m$, so $(e_n)$ has no convergent subsequence.\n\nThe set of compact operators on a Banach space is a closed two-sided ideal of the algebra of bounded operators, the [[ideal of compact operators|2WAZF1RV2Y]] in the Hilbert space case.\n",
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