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  "body": "\n## Definition\n\nLet $(X, d)$ be a [[compact|0TVJZYB6SH]] [[metric space|0VKD1TP7F3]] and let $f \\colon X \\to X$ be a [[continuous map|0VGFHTM75Y]].\nThe system $(X, f)$ is *equicontinuous* if the family $\\set{f^n : n \\ge 0}$ is [[equicontinuous|0XN66MZN0Y]].\nSince $X$ is compact, this means uniform equicontinuity: for every $\\varepsilon \u003e 0$ there is $\\delta \u003e 0$ with $d(f^n(x), f^n(y)) \u003c \\varepsilon$ for all $n \\ge 0$ and all $x, y \\in X$ with $d(x, y) \u003c \\delta$.\n\n## Remarks\n\nFor a homeomorphism one often asks the same of the family $\\set{f^n : n \\in \\Z}$.\nExample: every [[isometry|02HPP5VCZH]] $f$ is equicontinuous with $\\delta = \\varepsilon$, since $d(f^n(x), f^n(y)) = d(x, y)$; for instance the rotation $x \\mapsto x + \\alpha \\bmod 1$ of the circle.\nNon-example: the doubling map $f(x) = 2x \\bmod 1$ on the circle $\\R / \\Z$.\nGiven $\\delta \u003e 0$, take $x = 0$ and $y$ with $0 \u003c y \u003c \\min(\\delta, 1/4)$, and let $n$ be the first integer with $2^n y \\ge 1/4$.\nThen $f^n(y) = 2^n y$ lies in $[1/4, 1/2)$, at distance at least $1/4$ from $f^n(0) = 0$, so the condition fails for $\\varepsilon = 1/4$.\n\n## Explanations\n\n### L1 human\n\nA system is equicontinuous when nearby points stay nearby under all iterates, with one closeness threshold that works for all times.\n",
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