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  "body": "\n## Statement\n\nLet $(X, d)$ be a nonempty [[metric space|0VKD1TP7F3]].\nLet $\\ell^\\infty(X)$ be the [[Banach space|06GZ956657]] of bounded functions $X \\to \\R$ with the norm $\\norm{f}_\\infty = \\sup_{y \\in X} \\abs{f(y)}$.\nFix a point $x_0 \\in X$.\nFor $x \\in X$, define $\\Phi(x) \\colon X \\to \\R$ by\n$$\\Phi(x)(y) = d(x, y) - d(x_0, y).$$\nThen each $\\Phi(x)$ lies in $\\ell^\\infty(X)$, and $\\Phi \\colon X \\to \\ell^\\infty(X)$ is an [[isometry|02HPP5VCZH]] onto its image.\nThat is, $\\norm{\\Phi(x) - \\Phi(x')}_\\infty = d(x, x')$ for all $x, x' \\in X$.\n\n## Remarks\n\nBy the triangle inequality, $\\abs{d(x, y) - d(x', y)} \\le d(x, x')$ for every $y$.\nWith $x' = x_0$, this shows that $\\Phi(x)$ is bounded by $d(x, x_0)$.\nIn general, it gives $\\norm{\\Phi(x) - \\Phi(x')}_\\infty \\le d(x, x')$.\nTaking $y = x'$ gives the value $d(x, x')$, so equality holds.\n\nIf $X$ is bounded, the simpler map $x \\mapsto d(x, \\cdot)$ already works.\nIf $X$ is [[separable|0Z2P41SVDP]], the same idea with a countable dense subset embeds $X$ isometrically in the sequence space $\\ell^\\infty$.\n\n## Explanations\n\n### L1 human\n\nEvery metric space, however abstract, can be placed inside a space of real functions without changing any distance.\nEach point $x$ is recorded by the list of its distances to all other points.\nThis turns questions about metric spaces into questions about normed vector spaces.\n\n### L2 human\n\nSince $\\ell^\\infty(X)$ is complete, the closure of $\\Phi(X)$ is a complete metric space containing $X$ densely.\nThis gives a quick construction of the [[completion of a metric space|06RW4RKWH7]].\nThe embedding also lets one take convex hulls and straight segments of points of $X$ inside a Banach space.\nFor compact $X$ and $Y$, it shows that the distance on the [[Gromov–Hausdorff space|ETRF2FHAEE]] can be computed inside the single space $Z = \\ell^\\infty$.\n",
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