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  "body": "\n## Example\n\nLet $c$ be the set of real [[convergent sequences|0W4CN9E2W5]] $x = (x_n)_{n \\ge 1}$, with the norm $\\norm{x}_\\infty = \\sup_n \\abs{x_n}$.\nThen $c$ is a [[Banach space|06GZ956657]].\n\nCheck.\nEvery convergent sequence is [[bounded|0A8JNBGT2Z]], so $c$ is a linear subspace of the Banach space $\\ell^\\infty$ of bounded sequences.\nIt suffices to show that $c$ is closed in $\\ell^\\infty$.\nLet $x^{(k)} \\in c$ tend to $x$ in $\\ell^\\infty$.\nFor all $k$, $m$ and $n$,\n$$\\abs{x_m - x_n} \\le 2 \\norm{x - x^{(k)}}_\\infty + \\abs{x^{(k)}_m - x^{(k)}_n}.$$\nChoose $k$ to make the first term small, then $m, n$ large to make the second term small.\nSo $x$ is a Cauchy sequence of real numbers, and it converges.\n\n## Remarks\n\nThe sequences with limit $0$ form the closed subspace $c_0$ of codimension $1$ in $c$.\nThe map $x \\mapsto (\\ell, x_1 - \\ell, x_2 - \\ell, \\dots)$, with $\\ell = \\lim_n x_n$, is a linear isomorphism of $c$ onto $c_0$.\nBut $c$ and $c_0$ are not isometric.\nThe constant sequence $(1, 1, \\dots)$ is an extreme point of the unit ball of $c$, while the unit ball of $c_0$ has no extreme points.\nThe dual of $c$ is isometric to $\\ell^1$.\n\n## Explanations\n\n### L1 human\n\nThe space $c$ collects all sequences of real numbers that settle down to a limit.\nTwo sequences are close when they are close at every index.\nA uniform limit of sequences that settle down also settles down, so the space has no holes.\n",
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