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  "body": "\n## Example\n\nLet $\\ell^\\infty$ be the set of [[bounded sequences|0A8JNBGT2Z]] $x = (x_n)_{n \\ge 1}$ of real (or complex) numbers, with the norm $\\norm{x}_\\infty = \\sup_n \\abs{x_n}$.\nThen $\\ell^\\infty$ is a [[sequence space|XYTN19EPF2]] and a [[Banach space|0BDTTM24DP]]; this is the case $p = \\infty$ of the [[spaces ℓ^p|08P6T2WH6R]].\n\nCheck of completeness: let $(x^{(j)})$ be a Cauchy sequence in $\\ell^\\infty$.\nFor each $n$, $\\abs{x^{(j)}_n - x^{(i)}_n} \\le \\norm{x^{(j)} - x^{(i)}}_\\infty$, so $(x^{(j)}_n)_j$ is a Cauchy sequence of numbers with a limit $x_n$.\nGiven $\\varepsilon \u003e 0$, choose $J$ with $\\norm{x^{(j)} - x^{(i)}}_\\infty \\le \\varepsilon$ for $i, j \\ge J$.\nLetting $i \\to \\infty$ gives $\\abs{x^{(j)}_n - x_n} \\le \\varepsilon$ for all $n$ and all $j \\ge J$.\nSo $x$ is bounded and $x^{(j)} \\to x$ in norm.\n\n## Remarks\n\nThe space $\\ell^\\infty$ is not [[separable|0Z2P41SVDP]].\nThe sequences with all terms in $\\set{0, 1}$ form an uncountable set, and any two of them are at distance $1$.\nSo the open balls of radius $1/2$ around them are disjoint, and no countable set meets all of them.\nThe spaces of convergent sequences and of null sequences are closed subspaces of $\\ell^\\infty$.\nThe space $\\ell^\\infty$ is isometrically isomorphic to the dual of $\\ell^1$.\n\n## Explanations\n\n### L1 human\n\nA sequence is bounded when all its terms stay within some fixed distance of $0$.\nThe bounded sequences form a vector space, and the size of a sequence is the least bound on its terms.\nThis space is so large that no countable family of sequences can approximate all of its elements.\n",
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