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  "body": "\n## Example\n\nLet $c$ be the set of [[convergent sequences|0W4CN9E2W5]] $x = (x_n)_{n \\ge 1}$ of real (or complex) numbers, with the norm $\\norm{x}_\\infty = \\sup_{n} \\abs{x_n}$.\nThen $c$ is a [[sequence space|XYTN19EPF2]] and a [[Banach space|0BDTTM24DP]].\n\nCheck that $c$ is a sequence space: the sum of two convergent sequences converges, and so does a scalar multiple of a convergent sequence.\n\nCheck that $c$ is complete: convergent sequences are bounded, so $c$ is a subspace of the Banach space $\\ell^\\infty$ of [[bounded sequences|G3F2CWQSQH]].\nSo it suffices to show that $c$ is closed in $\\ell^\\infty$.\nLet $x^{(j)} \\in c$ with $\\norm{x^{(j)} - x}_\\infty \\to 0$ for some $x \\in \\ell^\\infty$.\nFor all $m, n$ and $j$,\n$$\\abs{x_m - x_n} \\le 2 \\norm{x - x^{(j)}}_\\infty + \\abs{x^{(j)}_m - x^{(j)}_n}.$$\nChoose $j$ to make the first term small, then use that $x^{(j)}$ is a [[Cauchy sequence|0QZMTN5AXX]].\nSo $x$ is a Cauchy sequence of numbers, hence convergent, and $x \\in c$.\n\n## Remarks\n\nThe map $x \\mapsto \\lim_n x_n$ is a linear functional on $c$ with $\\abs{\\lim_n x_n} \\le \\norm{x}_\\infty$, so it is continuous.\nIts kernel is the [[space of null sequences|WDH6QSH1Z6]] $c_0$.\n\n## Explanations\n\n### L1 human\n\nThe convergent sequences form a vector space: you can add them and scale them, and they still converge.\nMeasuring the size of a sequence by its largest term in absolute value makes this space complete, so limits of convergent sequences, in this size, are again convergent sequences.\n",
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