{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe take a shortest representation of $x$ as a convex combination and show that it has at most $d + 1$ points.\n\n1. Let $K$ be the set of all [[convex combinations|21SXYJYC59]] $\\sum_{i=1}^k \\lambda_i y_i$ with $k \\geq 1$, $y_i \\in P$, $\\lambda_i \\geq 0$ and $\\sum_i \\lambda_i = 1$. Each $p \\in P$ is in $K$, with $k = 1$.\n2. The set $K$ is [[convex|02AS45T3H9]]. Let $x = \\sum_i \\lambda_i y_i$ and $x' = \\sum_j \\mu_j z_j$ be in $K$, and let $0 \\leq t \\leq 1$. Then $(1 - t)x + tx'$ is the combination of all $y_i$ and $z_j$ with the weights $(1 - t)\\lambda_i$ and $t\\mu_j$.\n3. These weights are nonnegative, and they add up to $(1 - t) + t = 1$. So $(1 - t)x + tx' \\in K$.\n4. The [[convex hull|0FQSDC2EQ2]] $\\operatorname{conv}(P)$ is the intersection of all convex sets that contain $P$, and $K$ is one of them. So $x \\in \\operatorname{conv}(P)$ implies $x \\in K$.\n5. By the [[well-ordering principle|01BX15HSRR]], there is a least $k$ such that $x = \\sum_{i=1}^k \\lambda_i y_i$ is a convex combination of $k$ points $y_i \\in P$. Fix such a representation.\n6. Suppose that $k \\geq d + 2$. The $k - 1 \\geq d + 1$ vectors $y_i - y_1$, for $2 \\leq i \\leq k$, lie in $\\mathbb{R}^d$, which is spanned by $d$ vectors.\n7. By the [[Steinitz exchange lemma|08ZDRN6H41]], these vectors are not [[linearly independent|0CRMGPAH83]]. So there are real numbers $\\alpha_2, \\dots, \\alpha_k$, not all zero, with $\\sum_{i=2}^k \\alpha_i (y_i - y_1) = 0$.\n8. Put $\\alpha_1 = -\\sum_{i=2}^k \\alpha_i$. Then $\\sum_{i=1}^k \\alpha_i y_i = 0$ and $\\sum_{i=1}^k \\alpha_i = 0$, and not all $\\alpha_i$ are zero.\n9. Since the $\\alpha_i$ add up to $0$ and are not all zero, some $\\alpha_i$ is positive. Let $J = \\set{i : \\alpha_i \u003e 0}$, and let $t = \\min_{i \\in J} \\lambda_i / \\alpha_i$. Then $t \\geq 0$, and the minimum is attained at some $j \\in J$.\n10. Put $\\lambda_i' = \\lambda_i - t\\alpha_i$. For $i \\in J$, $t \\leq \\lambda_i/\\alpha_i$ gives $\\lambda_i' \\geq 0$. For $i \\notin J$, $\\alpha_i \\leq 0$ gives $\\lambda_i' \\geq \\lambda_i \\geq 0$. Also $\\lambda_j' = 0$.\n11. By step 8, $\\sum_i \\lambda_i' = \\sum_i \\lambda_i - t\\sum_i \\alpha_i = 1$ and $\\sum_i \\lambda_i' y_i = x - t \\cdot 0 = x$.\n12. Removing the term with $\\lambda_j' = 0$, we get $x$ as a convex combination of $k - 1$ points of $P$. This contradicts the choice of $k$ in step 5. Hence $k \\leq d + 1$.\n13. Write $p_i = y_i$. It remains to show that $x \\in \\operatorname{conv}\\set{p_1, \\dots, p_k}$. We show that each convex set $S$ that contains $p_1, \\dots, p_k$ contains each convex combination $\\sum_{i=1}^k \\mu_i p_i$, by induction on $k$.\n14. For $k = 1$, the combination is $p_1 \\in S$. Let $k \\geq 2$. If $\\mu_k = 1$, then all other $\\mu_i$ are $0$, and the combination is $p_k \\in S$.\n15. If $\\mu_k \u003c 1$, put $z = \\sum_{i \u003c k} \\frac{\\mu_i}{1 - \\mu_k} p_i$. This is a convex combination of $k - 1$ points, so $z \\in S$ by induction.\n16. Then $\\sum_i \\mu_i p_i = (1 - \\mu_k)z + \\mu_k p_k$, which lies in $S$ because $S$ is convex.\n17. So $x$ lies in every convex set that contains $p_1, \\dots, p_k$, that is, $x \\in \\operatorname{conv}\\set{p_1, \\dots, p_k}$, with $k \\leq d + 1$. Step 5 also gives the equivalent form: $x$ is a convex combination of at most $d + 1$ points of $P$.\n\n## Idea\n\nIf more than $d + 1$ points are used, they satisfy a linear relation whose coefficients add up to zero. Sliding the weights along this relation keeps the point $x$ and the total weight $1$ fixed. Stop at the first moment when a weight reaches zero, and one point drops out.\n",
  "commit": "68a8c3dc73fd55004b2796431c607c30960d1a85",
  "front": {
    "kind": "proof",
    "msc": [
      "52A35"
    ],
    "name": "Proof of Carathéodory's theorem by removing points",
    "proves": "0201Q4P2B5",
    "sources": [
      {
        "locator": "Proof",
        "url": "https://proofwiki.org/wiki/Carath%C3%A9odory%27s_Theorem_(Convex_Analysis)"
      }
    ],
    "tag": "WYD7QF0XEH",
    "uses": [
      "0FQSDC2EQ2",
      "21SXYJYC59",
      "02AS45T3H9",
      "01BX15HSRR",
      "0CRMGPAH83",
      "08ZDRN6H41"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/WYD7QF0XEH/",
  "links": [
    {
      "field": "proves",
      "target": "0201Q4P2B5"
    },
    {
      "field": "uses",
      "target": "0FQSDC2EQ2"
    },
    {
      "field": "uses",
      "target": "21SXYJYC59"
    },
    {
      "field": "uses",
      "target": "02AS45T3H9"
    },
    {
      "field": "uses",
      "target": "01BX15HSRR"
    },
    {
      "field": "uses",
      "target": "0CRMGPAH83"
    },
    {
      "field": "uses",
      "target": "08ZDRN6H41"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/WYD7QF0XEH.md",
  "msc": [
    "52A35"
  ],
  "name": "Proof of Carathéodory's theorem by removing points",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/WYD/WYD7QF0XEH.md",
  "status": null,
  "tag": "WYD7QF0XEH",
  "uri": "m2://WYD7QF0XEH",
  "url": "https://math2.tamnd.com/wiki/proof-of-caratheodorys-theorem-by-removing-points/",
  "writing": {
    "tag": "WYD7QF0XEH",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}