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  "body": "\n## Statement\n\nLet $(X, d)$ be a [[metric space|0VKD1TP7F3]] with the topology induced by $d$.\nThen $X$ is [[paracompact|8R29G6NG9Q]].\nThat is, every [[open cover|049PXYHW1T]] of $X$ has an open refinement that is a [[locally finite family|DZ297MMGV8]].\n\n## Remarks\n\nThe theorem is due to A. H. Stone.\nA short proof by Mary Ellen Rudin well-orders the members of the cover.\nFor each $n$ and each member $U$, it takes the union of balls of radius $2^{-n}$ around suitable points whose first containing member is $U$.\nThe radii are chosen so that each point has a ball meeting only finitely many of the new sets.\nThe proof uses the axiom of choice through the well-ordering.\n\nExample at work: every metric space is therefore [[normal|0NSR9V8TED]] and admits partitions of unity subordinate to any open cover.\n\nThe metric is essential only through the topology it induces.\nSo every metrizable space is paracompact.\n\n## Explanations\n\n### L1 human\n\nMetric spaces can be very large and far from compact.\nStill, every open cover of a metric space can be replaced by a finer cover that is finite near each point.\nThis makes metric spaces good spaces for gluing local objects.\n\n### L2 human\n\nStone's theorem shows that paracompactness is a necessary condition for metrizability.\nMetrization theorems such as the [[Nagata–Smirnov theorem|KB11T6389G]] go in the other direction.\nCombined with [[Dieudonné's theorem|B6Y8PEY72W]], it gives another proof that metric spaces are normal.\nIt is also why paracompactness is automatic for metrizable manifolds.\n",
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