{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Statement\n\nLet $(X, d)$ be a [[complete metric space|08EWT7J0SC]], for example $\\R$ or $\\C$, and let $(a_{m,n})_{m, n \\ge 1}$ be a double sequence in $X$.\nAssume:\n\n1. for each $m$, the limit $b_m = \\lim_{n \\to \\infty} a_{m,n}$ exists;\n2. there is a sequence $(c_n)$ such that $a_{m,n} \\to c_n$ as $m \\to \\infty$, [[uniformly|08FPABDQ4C]] in $n$, that is, $\\sup_n d(a_{m,n}, c_n) \\to 0$.\n\nThen the limits $\\lim_{m \\to \\infty} b_m$ and $\\lim_{n \\to \\infty} c_n$ exist, the [[Pringsheim double limit|4FK55CYMVZ]] $\\lim_{m,n \\to \\infty} a_{m,n}$ exists, and all three are equal.\n\n## Remarks\n\nProof idea: uniform convergence gives an $M$ such that $d(a_{m,n}, a_{m',n}) \u003c 2\\varepsilon$ for all $n$ when $m, m' \\ge M$.\nLetting $n \\to \\infty$ gives $d(b_m, b_{m'}) \\le 2\\varepsilon$, so $(b_m)$ is a Cauchy sequence and has a limit $L$ by completeness.\nThe triangle inequality $d(c_n, L) \\le d(c_n, a_{M,n}) + d(a_{M,n}, b_M) + d(b_M, L)$ then shows $c_n \\to L$, and the same splitting gives the double limit.\nExample: for $a_{m,n} = \\frac{1}{m} + \\frac{1}{n}$ the convergence in $m$ is uniform, and all limits equal $0$.\nWithout uniformity the conclusion fails: $a_{m,n} = \\frac{m}{m+n}$ gives $\\lim_m \\lim_n a_{m,n} = 0$ but $\\lim_n \\lim_m a_{m,n} = 1$.\nWithout completeness it also fails: in $\\mathbb{Q}$, take rationals $x_k \\to \\sqrt{2}$ and $a_{m,n} = x_{\\min(m,n)}$; both hypotheses hold, but $b_m = x_m$ has no limit in $\\mathbb{Q}$.\n\n## Explanations\n\n### L1 human\n\nTwo limits taken one after the other may give different answers, depending on the order.\nThe theorem says that if one of the inner limits is reached uniformly, the order does not matter, and the double limit exists too.\n\n### L2 human\n\nThis is the abstract principle behind many interchange results in analysis.\nThe theorem that a uniform limit of continuous functions is continuous, and the theorem that uniform limits commute with integration, follow the same pattern.\nFor [[double series|PAXCH4FFSV]], it gives conditions under which summing by rows, by columns, or over squares gives the same value.\n",
  "commit": "547b6eddb8f2feb8d579f2d633d22a27b7e132fc",
  "front": {
    "about": [
      "QYTGCW1GTK"
    ],
    "kind": "statement",
    "msc": [
      "40B05",
      "40A30"
    ],
    "name": "Moore–Osgood theorem",
    "role": "theorem",
    "sources": [
      {
        "url": "https://en.wikipedia.org/wiki/Iterated_limit"
      }
    ],
    "tag": "QN2210XCST",
    "uses": [
      "4FK55CYMVZ",
      "08FPABDQ4C",
      "08EWT7J0SC"
    ]
  },
  "kind": "statement",
  "link": "https://math2.tamnd.com/node/QN2210XCST/",
  "links": [
    {
      "field": "about",
      "target": "QYTGCW1GTK"
    },
    {
      "field": "uses",
      "target": "4FK55CYMVZ"
    },
    {
      "field": "uses",
      "target": "08FPABDQ4C"
    },
    {
      "field": "uses",
      "target": "08EWT7J0SC"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/QN2210XCST.md",
  "msc": [
    "40B05",
    "40A30"
  ],
  "name": "Moore–Osgood theorem",
  "prov": null,
  "role": "theorem",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/QN2/QN2210XCST.md",
  "status": null,
  "tag": "QN2210XCST",
  "uri": "m2://QN2210XCST",
  "url": "https://math2.tamnd.com/wiki/moore-osgood-theorem/",
  "writing": {
    "tag": "QN2210XCST",
    "kind": "statement",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      },
      {
        "what": "a source with a locator",
        "have": 0,
        "want": 1
      }
    ]
  }
}