{
  "index": "1954-B-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "3. Let \\( a \\) and \\( b \\) denote real numbers such that \\( a<b \\). The symbol \\( (a, b) \\) will denote the closed interval with the end points \\( a, b \\). Let there be given a collection of closed intervals \\( \\left(a_{1}, b_{1}\\right), \\ldots,\\left(a_{n}, b_{n}\\right) \\) such that any two of these closed intervals have at least one point in common. Prove that there exists then a point which is contained in every one of these intervals.",
  "solution": "Solution. Two closed intervals \\( (c, d) \\) and \\( (e, f) \\) overlap if and only it \\( c \\leq f \\) and \\( e \\leq d \\).\n\nSince any two of the given intervals overlap we have\n\\[\na_{n} \\leq b_{p}\n\\]\nfor all \\( n \\) and all \\( p \\). Hence \\( \\left\\{a_{n}\\right\\} \\) is bounded above (and non-empty) so we can let \\( \\xi \\) be the least upper bound of \\( \\left\\{a_{n}\\right\\} \\). Since (1) shows that each \\( b_{p} \\) is an upper bound for \\( \\left\\{a_{n}\\right\\} \\), we have \\( \\xi \\leq b_{p} \\) for all \\( p \\), while \\( a_{p} \\leq \\xi \\) because \\( \\xi \\) is an upper bound for \\( \\left\\{a_{n}\\right\\} \\). Thus \\( \\xi \\in\\left(a_{p}, b_{p}\\right) \\) for each \\( p \\).\n\nRemarks. We have worded the proof so that it remains valid for an infinite collection of closed intervals that intersect in pairs. This is Helly's theorem in dimension one. See L. Danzer, B. Grunbaum, V. Klee, \"Helly's Theorem and Its Relatives,\" Proc. Symposium in Pure Mathematics, 7: Convexity, American Mathematical Society, Providence, R.I., 1963. Also Yaglom and Boltyanskii, Convex Figures, Holt, Rinehart and Winston, New York, 1961.",
  "vars": [
    "a",
    "b",
    "a_1",
    "b_1",
    "a_n",
    "b_n",
    "c",
    "d",
    "e",
    "f",
    "a_p",
    "b_p",
    "\\\\xi"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "leftend",
        "b": "rightend",
        "a_1": "firstleft",
        "b_1": "firstright",
        "a_n": "indexleft",
        "b_n": "indexright",
        "c": "lefttemp",
        "d": "righttemp",
        "e": "secondleft",
        "f": "secondright",
        "a_p": "pleftend",
        "b_p": "prightend",
        "\\xi": "supremum"
      },
      "question": "3. Let \\( leftend \\) and \\( rightend \\) denote real numbers such that \\( leftend<rightend \\). The symbol \\( (leftend, rightend) \\) will denote the closed interval with the end points \\( leftend, rightend \\). Let there be given a collection of closed intervals \\( \\left(firstleft, firstright\\right), \\ldots,\\left(indexleft, indexright\\right) \\) such that any two of these closed intervals have at least one point in common. Prove that there exists then a point which is contained in every one of these intervals.",
      "solution": "Solution. Two closed intervals \\( (lefttemp, righttemp) \\) and \\( (secondleft, secondright) \\) overlap if and only if \\( lefttemp \\leq secondright \\) and \\( secondleft \\leq righttemp \\).\n\nSince any two of the given intervals overlap we have\n\\[\nindexleft \\leq prightend\n\\]\nfor all \\( n \\) and all \\( p \\). Hence \\( \\left\\{indexleft\\right\\} \\) is bounded above (and non-empty) so we can let \\( supremum \\) be the least upper bound of \\( \\left\\{indexleft\\right\\} \\). Since (1) shows that each \\( prightend \\) is an upper bound for \\( \\left\\{indexleft\\right\\} \\), we have \\( supremum \\leq prightend \\) for all \\( p \\), while \\( pleftend \\leq supremum \\) because \\( supremum \\) is an upper bound for \\( \\left\\{indexleft\\right\\} \\). Thus \\( supremum \\in\\left(pleftend, prightend\\right) \\) for each \\( p \\).\n\nRemarks. We have worded the proof so that it remains valid for an infinite collection of closed intervals that intersect in pairs. This is Helly's theorem in dimension one. See L. Danzer, B. Grunbaum, V. Klee, \\\"Helly's Theorem and Its Relatives,\\\" Proc. Symposium in Pure Mathematics, 7: Convexity, American Mathematical Society, Providence, R.I., 1963. Also Yaglom and Boltyanskii, Convex Figures, Holt, Rinehart and Winston, New York, 1961."
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "sandpaper",
        "b": "wheelhouse",
        "a_1": "blueberry",
        "b_1": "raincloud",
        "a_n": "doorstep",
        "b_n": "hummingbird",
        "c": "moonlight",
        "d": "printerink",
        "e": "sunflower",
        "f": "babysteps",
        "a_p": "skylights",
        "b_p": "paperclips",
        "\\xi": "cottonseed"
      },
      "question": "3. Let \\( sandpaper \\) and \\( wheelhouse \\) denote real numbers such that \\( sandpaper<wheelhouse \\). The symbol \\( (sandpaper, wheelhouse) \\) will denote the closed interval with the end points \\( sandpaper, wheelhouse \\). Let there be given a collection of closed intervals \\( \\left(blueberry, raincloud\\right), \\ldots,\\left(doorstep, hummingbird\\right) \\) such that any two of these closed intervals have at least one point in common. Prove that there exists then a point which is contained in every one of these intervals.",
      "solution": "Solution. Two closed intervals \\( (moonlight, printerink) \\) and \\( (sunflower, babysteps) \\) overlap if and only it \\( moonlight \\leq babysteps \\) and \\( sunflower \\leq printerink \\).\n\nSince any two of the given intervals overlap we have\n\\[\ndoorstep \\leq paperclips\n\\]\nfor all \\( n \\) and all \\( p \\). Hence \\( \\left\\{doorstep\\right\\} \\) is bounded above (and non-empty) so we can let \\( cottonseed \\) be the least upper bound of \\( \\left\\{doorstep\\right\\} \\). Since (1) shows that each \\( paperclips \\) is an upper bound for \\( \\left\\{doorstep\\right\\} \\), we have \\( cottonseed \\leq paperclips \\) for all \\( p \\), while \\( skylights \\leq cottonseed \\) because \\( cottonseed \\) is an upper bound for \\( \\left\\{doorstep\\right\\} \\). Thus \\( cottonseed \\in\\left(skylights, paperclips\\right) \\) for each \\( p \\).\n\nRemarks. We have worded the proof so that it remains valid for an infinite collection of closed intervals that intersect in pairs. This is Helly's theorem in dimension one. See L. Danzer, B. Grunbaum, V. Klee, \"Helly's Theorem and Its Relatives,\" Proc. Symposium in Pure Mathematics, 7: Convexity, American Mathematical Society, Providence, R.I., 1963. Also Yaglom and Boltyanskii, Convex Figures, Holt, Rinehart and Winston, New York, 1961."
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "uppermargin",
        "b": "lowermargin",
        "a_1": "upperfirst",
        "b_1": "lowerfirst",
        "a_n": "uppergeneral",
        "b_n": "lowergeneral",
        "c": "higherlimit",
        "d": "lowerlimit",
        "e": "topboundary",
        "f": "bottombound",
        "a_p": "upperparticular",
        "b_p": "lowerparticular",
        "\\xi": "inferior"
      },
      "question": "3. Let \\( uppermargin \\) and \\( lowermargin \\) denote real numbers such that \\( uppermargin<lowermargin \\). The symbol \\( (uppermargin, lowermargin) \\) will denote the closed interval with the end points \\( uppermargin, lowermargin \\). Let there be given a collection of closed intervals \\( \\left(upperfirst, lowerfirst\\right), \\ldots,\\left(uppergeneral, lowergeneral\\right) \\) such that any two of these closed intervals have at least one point in common. Prove that there exists then a point which is contained in every one of these intervals.",
      "solution": "Solution. Two closed intervals \\( (higherlimit, lowerlimit) \\) and \\( (topboundary, bottombound) \\) overlap if and only if \\( higherlimit \\leq bottombound \\) and \\( topboundary \\leq lowerlimit \\).\n\nSince any two of the given intervals overlap we have\n\\[\nuppergeneral \\leq lowerparticular\n\\]\nfor all \\( n \\) and all \\( p \\). Hence \\( \\left\\{uppergeneral\\right\\} \\) is bounded above (and non-empty) so we can let \\( inferior \\) be the least upper bound of \\( \\left\\{uppergeneral\\right\\} \\). Since (1) shows that each \\( lowerparticular \\) is an upper bound for \\( \\left\\{uppergeneral\\right\\} \\), we have \\( inferior \\leq lowerparticular \\) for all \\( p \\), while \\( upperparticular \\leq inferior \\) because \\( inferior \\) is an upper bound for \\( \\left\\{uppergeneral\\right\\} \\). Thus \\( inferior \\in\\left(upperparticular, lowerparticular\\right) \\) for each \\( p \\).\n\nRemarks. We have worded the proof so that it remains valid for an infinite collection of closed intervals that intersect in pairs. This is Helly's theorem in dimension one. See L. Danzer, B. Grunbaum, V. Klee, \"Helly's Theorem and Its Relatives,\" Proc. Symposium in Pure Mathematics, 7: Convexity, American Mathematical Society, Providence, R.I., 1963. Also Yaglom and Boltyanskii, Convex Figures, Holt, Rinehart and Winston, New York, 1961."
    },
    "garbled_string": {
      "map": {
        "a": "kjsnqpfm",
        "b": "zcnmwtrh",
        "a_1": "pqvmksal",
        "b_1": "ldfgrnce",
        "a_n": "cpfshjra",
        "b_n": "xzlhgnop",
        "c": "nbztrflk",
        "d": "hmjgrsca",
        "e": "qprnvldk",
        "f": "sntlwgav",
        "a_p": "brqltwex",
        "b_p": "vlsgcnat",
        "\\xi": "jdflqwer"
      },
      "question": "3. Let \\( kjsnqpfm \\) and \\( zcnmwtrh \\) denote real numbers such that \\( kjsnqpfm<zcnmwtrh \\). The symbol \\( (kjsnqpfm, zcnmwtrh) \\) will denote the closed interval with the end points \\( kjsnqpfm, zcnmwtrh \\). Let there be given a collection of closed intervals \\( \\left(pqvmksal, ldfgrnce\\right), \\ldots,\\left(cpfshjra, xzlhgnop\\right) \\) such that any two of these closed intervals have at least one point in common. Prove that there exists then a point which is contained in every one of these intervals.",
      "solution": "Solution. Two closed intervals \\( (nbztrflk, hmjgrsca) \\) and \\( (qprnvldk, sntlwgav) \\) overlap if and only it \\( nbztrflk \\leq sntlwgav \\) and \\( qprnvldk \\leq hmjgrsca \\).\n\nSince any two of the given intervals overlap we have\n\\[\ncpfshjra \\leq vlsgcnat\n\\]\nfor all \\( n \\) and all \\( p \\). Hence \\( \\left\\{cpfshjra\\right\\} \\) is bounded above (and non-empty) so we can let \\( jdflqwer \\) be the least upper bound of \\( \\left\\{cpfshjra\\right\\} \\). Since (1) shows that each \\( vlsgcnat \\) is an upper bound for \\( \\left\\{cpfshjra\\right\\} \\), we have \\( jdflqwer \\leq vlsgcnat \\) for all \\( p \\), while \\( brqltwex \\leq jdflqwer \\) because \\( jdflqwer \\) is an upper bound for \\( \\left\\{cpfshjra\\right\\} \\). Thus \\( jdflqwer \\in\\left(brqltwex, vlsgcnat\\right) \\) for each \\( p \\).\n\nRemarks. We have worded the proof so that it remains valid for an infinite collection of closed intervals that intersect in pairs. This is Helly's theorem in dimension one. See L. Danzer, B. Grunbaum, V. Klee, \"Helly's Theorem and Its Relatives,\" Proc. Symposium in Pure Mathematics, 7: Convexity, American Mathematical Society, Providence, R.I., 1963. Also Yaglom and Boltyanskii, Convex Figures, Holt, Rinehart and Winston, New York, 1961."
    },
    "kernel_variant": {
      "question": "(Sharp Quantitative Helly–type Theorem for Half–spaces)\n\nFix an integer $d\\ge 1$ and a real number $R>0$.  \nLet $A$ be an (arbitrary, possibly infinite) index set with cardinality  \n\\[\n|A|\\;\\ge\\;d+1 .\n\\tag{C}\n\\]  \nFor every $\\alpha\\in A$ put  \n\\[\nH_{\\alpha}\\;=\\;\\bigl\\{x\\in\\mathbb{R}^{d}\\;:\\;\\langle v_{\\alpha},x\\rangle\\le b_{\\alpha}\\bigr\\},\n\\qquad v_{\\alpha}\\in\\mathbb{R}^{d}\\setminus\\{0\\},\\; b_{\\alpha}\\in\\mathbb{R},\n\\]\na closed affine half–space.\n\nAssume the following \\emph{quantitative $(d+1)$–wise intersection property}:\n\n(Q)  \n(i)  For every finite subfamily $\\{\\alpha_{1},\\dots ,\\alpha_{k}\\}\\subset A$ with $k\\le d+1$ one has  \n\\[\nH_{\\alpha_{1}}\\cap\\dots\\cap H_{\\alpha_{k}}\\neq\\varnothing;\n\\tag{Q1}\n\\]\n\n(ii)  Whenever $k=d+1$ the whole intersection already lies in the Euclidean ball  \n\\[\nH_{\\alpha_{1}}\\cap\\dots\\cap H_{\\alpha_{d+1}}\\;\\subset\\;B(0,R),\n\\qquad \nB(0,R):=\\bigl\\{x\\in\\mathbb{R}^{d}: \\|x\\|_{2}\\le R\\bigr\\}.\n\\tag{Q2}\n\\]\n\nProve that\n\n1.\\; the total intersection $\\displaystyle\\bigcap_{\\alpha\\in A}H_{\\alpha}$ is non–empty;\n\n2.\\; every point of this intersection in fact lies in $B(0,R)$, i.e.  \n   $\\displaystyle\\bigcap_{\\alpha\\in A}H_{\\alpha}\\;\\subset\\;B(0,R)$;\n\n3.\\; the number $d+1$ occurring in (Q) is best possible:  \n   for every integer $d\\ge 1$ and every radius $R>0$ there exists a family\n   $\\{H_{\\alpha}\\}_{\\alpha\\in A}$ of closed half–spaces in $\\mathbb{R}^{d}$ satisfying\n\n   (a) property (Q1) for \\emph{every} subfamily of at most $d$ members,  \n\n   but such that\n\n   (b) $\\displaystyle\\bigcap_{\\alpha\\in A}H_{\\alpha}=\\varnothing$.\n\n   Consequently the quantitative conclusion in 2.\\ cannot, in general,\n   be guaranteed if one only knows the validity of (Q1) for subfamilies of\n   size $\\le d$.\n\n-------------------------------------------------------------",
      "solution": "We treat the three assertions in turn.\n\n--------------------------------------------------------------------\n1.\\; Non–emptiness of the global intersection.\n\nIntroduce the auxiliary convex sets  \n\\[\nK_{\\alpha}:=H_{\\alpha}\\cap B(0,R)\\qquad (\\alpha\\in A).\n\\]\nBecause $B(0,R)$ is compact and convex, each $K_{\\alpha}$ is a closed convex\nsubset of a compact set; hence every finite intersection of the $K_{\\alpha}$\nis compact.\n\n\\medskip\n\\emph{Step 1.1 – Finite subfamilies of size $\\le d+1$.}  \nLet $\\{\\alpha_{1},\\dots ,\\alpha_{k}\\}\\subset A$ with $k\\le d+1$.\n\n$\\bullet$ If $k=d+1$, then (Q2) gives\n\\[\nK_{\\alpha_{1}}\\cap\\dots\\cap K_{\\alpha_{d+1}}\n=\\bigl(H_{\\alpha_{1}}\\cap\\dots\\cap H_{\\alpha_{d+1}}\\bigr)\\cap B(0,R)\n=H_{\\alpha_{1}}\\cap\\dots\\cap H_{\\alpha_{d+1}}\\neq\\varnothing .\n\\]\n\n$\\bullet$ Suppose $k<d+1$.  Because of the cardinality condition (C) we can\nchoose additional, pairwise distinct indices\n$\\alpha_{k+1},\\dots ,\\alpha_{d+1}\\in A\\setminus\\{\\alpha_{1},\\dots ,\\alpha_{k}\\}$.\nApplying (Q2) to the enlarged $(d+1)$–tuple we get a point\n\\[\nx^{\\ast}\\in H_{\\alpha_{1}}\\cap\\dots\\cap H_{\\alpha_{d+1}}\\subset B(0,R),\n\\]\nhence $x^{\\ast}\\in K_{\\alpha_{j}}$ for every $1\\le j\\le k$.  Therefore  \n\\[\nK_{\\alpha_{1}}\\cap\\dots\\cap K_{\\alpha_{k}}\\neq\\varnothing .\n\\tag{1.1}\n\\]\n\n\\medskip\n\\emph{Step 1.2 – Arbitrary finite subfamilies.}  \nNow let $J\\subset A$ be \\emph{any} finite index set.  \nBecause of (1.1), every subfamily of\n$\\{K_{\\alpha}\\}_{\\alpha\\in J}$ containing at most $d+1$ members has\nnon–empty intersection.  Helly’s theorem for \\emph{finite} families of convex\nsets in $\\mathbb{R}^{d}$ therefore implies  \n\\[\n\\bigcap_{\\alpha\\in J}K_{\\alpha}\\neq\\varnothing.\n\\tag{1.2}\n\\]\n\n\\medskip\n\\emph{Step 1.3 – Passing to the whole family.}  \nRelation (1.2) shows that the collection $\\{K_{\\alpha}\\}_{\\alpha\\in A}$\nhas the \\emph{finite intersection property}.  \nSince every $K_{\\alpha}$ is a closed subset of the compact set $B(0,R)$,\ncompactness yields  \n\\[\n\\bigcap_{\\alpha\\in A}K_{\\alpha}\\neq\\varnothing .\n\\]\nChoose $x^{\\ast}$ in that intersection.  Then $x^{\\ast}\\in H_{\\alpha}$ for\nevery $\\alpha$ and $x^{\\ast}\\in B(0,R)$, proving  \n\\[\n\\bigcap_{\\alpha\\in A}H_{\\alpha}\\neq\\varnothing .\n\\]\n\n--------------------------------------------------------------------\n2.\\; A uniform radius bound for the whole intersection.\n\nLet $x\\in\\bigcap_{\\alpha\\in A}H_{\\alpha}$ and suppose,\nfor a contradiction, that $\\|x\\|_{2}>R$.  \nBecause $|A|\\ge d+1$, we can pick distinct indices\n$\\alpha_{1},\\dots ,\\alpha_{d+1}\\in A$.  \nThen $x$ lies in the intersection\n$H_{\\alpha_{1}}\\cap\\dots\\cap H_{\\alpha_{d+1}}$,\nyet by (Q2) this set is contained in $B(0,R)$,\ncontradicting $\\|x\\|_{2}>R$.  Therefore\n$\\|x\\|_{2}\\le R$ for every $x$ in the total intersection, and we have  \n\\[\n\\bigcap_{\\alpha\\in A}H_{\\alpha}\\;\\subset\\;B(0,R).\n\\]\n\n--------------------------------------------------------------------\n3.\\; Optimality of the number $d+1$.\n\nWe show that the quantitative Helly statement fails\nif one is only allowed to inspect subfamilies of \\emph{at most $d$} members.\n\n\\smallskip\n\\textbf{Step 3.1:  Any non–empty intersection of at most $d$ half–spaces is unbounded.}\n\nLet $1\\le k\\le d$ and let $G_{1},\\dots ,G_{k}$ be closed half–spaces in\n$\\mathbb{R}^{d}$ with non–empty intersection\n$G:=\\bigcap_{i=1}^{k}G_{i}$.  \nWrite $G_{i}=\\{x\\in\\mathbb{R}^{d}:\\langle a^{i},x\\rangle\\le b_{i}\\}$ with\n$a^{i}\\neq 0$ for $1\\le i\\le k$, and set  \n\\[\nA:=\\begin{pmatrix}(a^{1})^{\\top}\\\\\\vdots\\\\(a^{k})^{\\top}\\end{pmatrix}\\in\\mathbb{R}^{k\\times d}.\n\\]\nChoose $x_{0}\\in G$.\n\n\\emph{Case 1: $\\operatorname{rank}A\\le k-1$.}  \nThen $\\dim\\ker A\\ge 1$; pick a non–zero $w\\in\\ker A$.  \nFor every $t\\ge 0$,\n$\\langle a^{i},x_{0}+tw\\rangle=\\langle a^{i},x_{0}\\rangle\\le b_{i}$,\nso $x_{0}+tw\\in G$.  Hence $G$ contains the unbounded ray\n$\\{x_{0}+tw:t\\ge 0\\}$.\n\n\\emph{Case 2: $\\operatorname{rank}A=k=d$.}  \nThen $A$ is invertible.  Put\n$w:=-A^{-1}\\mathbf 1$, $\\mathbf 1:=(1,\\dots ,1)^{\\top}$.  \nBecause $Aw=-\\mathbf 1$, one has $\\langle a^{i},w\\rangle=-1$ for all $i$.\nFor $t>0$,\n$\\langle a^{i},x_{0}+tw\\rangle=\\langle a^{i},x_{0}\\rangle-t\\le b_{i}-t<b_{i}$,\nso $x_{0}+tw\\in G$ for every $t>0$.  \nAgain $G$ contains the entire ray $\\{x_{0}+tw:t\\ge 0\\}$.\n\n\\emph{Case 3: $k<d$ and $\\operatorname{rank}A=k$.}  \nHere $d-k\\ge 1$, so $\\ker A$ is non–trivial; choose $0\\neq w\\in\\ker A$\nand repeat the argument of Case~1.  \n\n\\noindent\nCombining the three cases we have established  \n\\[\n\\text{every non–empty intersection of at most $d$ half–spaces in $\\mathbb{R}^{d}$ is unbounded.}\n\\tag{3.1}\n\\]\n\n\\smallskip\n\\textbf{Step 3.2:  A concrete counterexample for the $d$–wise hypothesis.}\n\nFix $R>0$ and set $c:=R+1$.  \nDefine the following $d+1$ closed half–spaces in $\\mathbb{R}^{d}$:\n\\[\n\\begin{aligned}\nH_{0}&:=\\Bigl\\{x=(x_{1},\\dots ,x_{d})\\in\\mathbb{R}^{d}\\;:\\;\n           \\sum_{i=1}^{d}x_{i}\\le 0\\Bigr\\},\\\\[4pt]\nH_{i}&:=\\Bigl\\{x=(x_{1},\\dots ,x_{d})\\in\\mathbb{R}^{d}\\;:\\;\n           x_{i}\\ge c\\Bigr\\}, \\qquad 1\\le i\\le d.\n\\end{aligned}\n\\tag{3.2}\n\\]\n\n(a) \\emph{Every subfamily of at most $d$ members has non–empty intersection.}\n\n$\\bullet$ If the subfamily omits $H_{0}$, it is of the form\n$\\{H_{i}\\}_{i\\in I}$ with $I\\subseteq\\{1,\\dots ,d\\}$ and $|I|\\le d$.\nThe point $x^{(I)}$ defined by  \n\\[\nx^{(I)}_{i}=c \\text{ for } i\\in I,\\qquad \nx^{(I)}_{i}=c+1 \\text{ for } i\\notin I\n\\]\nlies in all selected half–spaces, so the intersection is non–empty (and unbounded).\n\n$\\bullet$ If the subfamily contains $H_{0}$ but omits, say, $H_{k}$ with $k\\ge 1$, take  \n\\[\nx_{k}:=-d(c+1),\\qquad\nx_{i}=c\\quad(i\\neq k).\n\\]\nThen $\\sum_{i=1}^{d}x_{i}= -d(c+1)+(d-1)c=-c<0$, so the point\n$x=(x_{1},\\dots ,x_{d})$ satisfies $H_{0}$ as well as every\n$H_{i}$ with $i\\neq k$.  Hence the intersection is again non–empty (and\nunbounded).\n\nThus property (Q1) holds for \\emph{every} subfamily of at most $d$ members.\n\n(b) \\emph{The total intersection is empty.}\n\nIf $x\\in\\bigcap_{i=0}^{d}H_{i}$ we would have\n$x_{i}\\ge c$ for all $1\\le i\\le d$, hence\n$\\sum_{i=1}^{d}x_{i}\\ge d\\,c>0$, contradicting the defining inequality of\n$H_{0}$.  Consequently  \n\\[\n\\bigcap_{i=0}^{d}H_{i}=\\varnothing .\n\\tag{3.3}\n\\]\n\n\\smallskip\n\\textbf{Step 3.3:  Consequences.}\n\nThe family $\\{H_{0},\\dots ,H_{d}\\}$ satisfies (a) and (b) above, establishing the optimality of the number $d+1$.  \nMoreover, by (3.1) every $d$–wise intersection is unbounded, so\n\\emph{no} ball $B(0,R)$ can cover all those intersections.  Therefore the radius\nbound asserted in part~2 cannot be guaranteed under a merely\n$d$–wise hypothesis, and the parameter $d+1$ is optimal.\n\n\\hfill$\\square$\n\n---------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.471249",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension:  the problem is set in ℝᵈ for arbitrary d, not in ℝ.  \n• Additional constraints:  the “radius-R” quantitative restriction forces the solver\n  to keep track of the size of every finite intersection, not merely its existence.  \n• Sophisticated structure:  half–spaces are unbounded; lifting to ℝ^{d+1} produces\n  cones and cylinders, so the argument mixes affine geometry with norm constraints.  \n• Deeper theory:  a successful solution needs the classical Helly theorem, but also\n  must recognise how to convert a norm bound into a linear inequality in one higher\n  dimension – a standard trick in convex analysis but far from obvious at\n  Olympiad level.  \n• Interacting concepts:  convexity, compactness of finite intersections,\n  quantitative vs. qualitative Helly, and a non-trivial minimal-example argument\n  all appear.\n\nThese layers of technicality and the necessity of invoking higher-dimensional\nHelly put the problem well beyond the scope of the original one-dimensional\ninterval exercise."
      }
    },
    "original_kernel_variant": {
      "question": "(Sharp Quantitative Helly–type Theorem for Half–spaces)\n\nFix an integer $d\\ge 1$ and a real number $R>0$.  \nLet $A$ be an (arbitrary, possibly infinite) index set with cardinality  \n\\[\n|A|\\;\\ge\\;d+1 .\n\\tag{C}\n\\]  \nFor every $\\alpha\\in A$ put  \n\\[\nH_{\\alpha}\\;=\\;\\bigl\\{x\\in\\mathbb{R}^{d}\\;:\\;\\langle v_{\\alpha},x\\rangle\\le b_{\\alpha}\\bigr\\},\n\\qquad v_{\\alpha}\\in\\mathbb{R}^{d}\\setminus\\{0\\},\\; b_{\\alpha}\\in\\mathbb{R},\n\\]\na closed affine half–space.\n\nAssume the following \\emph{quantitative $(d+1)$–wise intersection property}:\n\n(Q)  \n(i)  For every finite subfamily $\\{\\alpha_{1},\\dots ,\\alpha_{k}\\}\\subset A$ with $k\\le d+1$ one has  \n\\[\nH_{\\alpha_{1}}\\cap\\dots\\cap H_{\\alpha_{k}}\\neq\\varnothing;\n\\tag{Q1}\n\\]\n\n(ii)  Whenever $k=d+1$ the whole intersection already lies in the Euclidean ball  \n\\[\nH_{\\alpha_{1}}\\cap\\dots\\cap H_{\\alpha_{d+1}}\\;\\subset\\;B(0,R),\n\\qquad \nB(0,R):=\\bigl\\{x\\in\\mathbb{R}^{d}: \\|x\\|_{2}\\le R\\bigr\\}.\n\\tag{Q2}\n\\]\n\nProve that\n\n1.\\; the total intersection $\\displaystyle\\bigcap_{\\alpha\\in A}H_{\\alpha}$ is non–empty;\n\n2.\\; every point of this intersection in fact lies in $B(0,R)$, i.e.  \n   $\\displaystyle\\bigcap_{\\alpha\\in A}H_{\\alpha}\\;\\subset\\;B(0,R)$;\n\n3.\\; the number $d+1$ occurring in (Q) is best possible:  \n   for every integer $d\\ge 1$ and every radius $R>0$ there exists a family\n   $\\{H_{\\alpha}\\}_{\\alpha\\in A}$ of closed half–spaces in $\\mathbb{R}^{d}$ satisfying\n\n   (a) property (Q1) for \\emph{every} subfamily of at most $d$ members,  \n\n   but such that\n\n   (b) $\\displaystyle\\bigcap_{\\alpha\\in A}H_{\\alpha}=\\varnothing$.\n\n   Consequently the quantitative conclusion in 2.\\ cannot, in general,\n   be guaranteed if one only knows the validity of (Q1) for subfamilies of\n   size $\\le d$.\n\n-------------------------------------------------------------",
      "solution": "We treat the three assertions in turn.\n\n--------------------------------------------------------------------\n1.\\; Non–emptiness of the global intersection.\n\nIntroduce the auxiliary convex sets  \n\\[\nK_{\\alpha}:=H_{\\alpha}\\cap B(0,R)\\qquad (\\alpha\\in A).\n\\]\nBecause $B(0,R)$ is compact and convex, each $K_{\\alpha}$ is a closed convex\nsubset of a compact set; hence every finite intersection of the $K_{\\alpha}$\nis compact.\n\n\\medskip\n\\emph{Step 1.1 – Finite subfamilies of size $\\le d+1$.}  \nLet $\\{\\alpha_{1},\\dots ,\\alpha_{k}\\}\\subset A$ with $k\\le d+1$.\n\n$\\bullet$ If $k=d+1$, then (Q2) gives\n\\[\nK_{\\alpha_{1}}\\cap\\dots\\cap K_{\\alpha_{d+1}}\n=\\bigl(H_{\\alpha_{1}}\\cap\\dots\\cap H_{\\alpha_{d+1}}\\bigr)\\cap B(0,R)\n=H_{\\alpha_{1}}\\cap\\dots\\cap H_{\\alpha_{d+1}}\\neq\\varnothing .\n\\]\n\n$\\bullet$ Suppose $k<d+1$.  Because of the cardinality condition (C) we can\nchoose additional, pairwise distinct indices\n$\\alpha_{k+1},\\dots ,\\alpha_{d+1}\\in A\\setminus\\{\\alpha_{1},\\dots ,\\alpha_{k}\\}$.\nApplying (Q2) to the enlarged $(d+1)$–tuple we get a point\n\\[\nx^{\\ast}\\in H_{\\alpha_{1}}\\cap\\dots\\cap H_{\\alpha_{d+1}}\\subset B(0,R),\n\\]\nhence $x^{\\ast}\\in K_{\\alpha_{j}}$ for every $1\\le j\\le k$.  Therefore  \n\\[\nK_{\\alpha_{1}}\\cap\\dots\\cap K_{\\alpha_{k}}\\neq\\varnothing .\n\\tag{1.1}\n\\]\n\n\\medskip\n\\emph{Step 1.2 – Arbitrary finite subfamilies.}  \nNow let $J\\subset A$ be \\emph{any} finite index set.  \nBecause of (1.1), every subfamily of\n$\\{K_{\\alpha}\\}_{\\alpha\\in J}$ containing at most $d+1$ members has\nnon–empty intersection.  Helly’s theorem for \\emph{finite} families of convex\nsets in $\\mathbb{R}^{d}$ therefore implies  \n\\[\n\\bigcap_{\\alpha\\in J}K_{\\alpha}\\neq\\varnothing.\n\\tag{1.2}\n\\]\n\n\\medskip\n\\emph{Step 1.3 – Passing to the whole family.}  \nRelation (1.2) shows that the collection $\\{K_{\\alpha}\\}_{\\alpha\\in A}$\nhas the \\emph{finite intersection property}.  \nSince every $K_{\\alpha}$ is a closed subset of the compact set $B(0,R)$,\ncompactness yields  \n\\[\n\\bigcap_{\\alpha\\in A}K_{\\alpha}\\neq\\varnothing .\n\\]\nChoose $x^{\\ast}$ in that intersection.  Then $x^{\\ast}\\in H_{\\alpha}$ for\nevery $\\alpha$ and $x^{\\ast}\\in B(0,R)$, proving  \n\\[\n\\bigcap_{\\alpha\\in A}H_{\\alpha}\\neq\\varnothing .\n\\]\n\n--------------------------------------------------------------------\n2.\\; A uniform radius bound for the whole intersection.\n\nLet $x\\in\\bigcap_{\\alpha\\in A}H_{\\alpha}$ and suppose,\nfor a contradiction, that $\\|x\\|_{2}>R$.  \nBecause $|A|\\ge d+1$, we can pick distinct indices\n$\\alpha_{1},\\dots ,\\alpha_{d+1}\\in A$.  \nThen $x$ lies in the intersection\n$H_{\\alpha_{1}}\\cap\\dots\\cap H_{\\alpha_{d+1}}$,\nyet by (Q2) this set is contained in $B(0,R)$,\ncontradicting $\\|x\\|_{2}>R$.  Therefore\n$\\|x\\|_{2}\\le R$ for every $x$ in the total intersection, and we have  \n\\[\n\\bigcap_{\\alpha\\in A}H_{\\alpha}\\;\\subset\\;B(0,R).\n\\]\n\n--------------------------------------------------------------------\n3.\\; Optimality of the number $d+1$.\n\nWe show that the quantitative Helly statement fails\nif one is only allowed to inspect subfamilies of \\emph{at most $d$} members.\n\n\\smallskip\n\\textbf{Step 3.1:  Any non–empty intersection of at most $d$ half–spaces is unbounded.}\n\nLet $1\\le k\\le d$ and let $G_{1},\\dots ,G_{k}$ be closed half–spaces in\n$\\mathbb{R}^{d}$ with non–empty intersection\n$G:=\\bigcap_{i=1}^{k}G_{i}$.  \nWrite $G_{i}=\\{x\\in\\mathbb{R}^{d}:\\langle a^{i},x\\rangle\\le b_{i}\\}$ with\n$a^{i}\\neq 0$ for $1\\le i\\le k$, and set  \n\\[\nA:=\\begin{pmatrix}(a^{1})^{\\top}\\\\\\vdots\\\\(a^{k})^{\\top}\\end{pmatrix}\\in\\mathbb{R}^{k\\times d}.\n\\]\nChoose $x_{0}\\in G$.\n\n\\emph{Case 1: $\\operatorname{rank}A\\le k-1$.}  \nThen $\\dim\\ker A\\ge 1$; pick a non–zero $w\\in\\ker A$.  \nFor every $t\\ge 0$,\n$\\langle a^{i},x_{0}+tw\\rangle=\\langle a^{i},x_{0}\\rangle\\le b_{i}$,\nso $x_{0}+tw\\in G$.  Hence $G$ contains the unbounded ray\n$\\{x_{0}+tw:t\\ge 0\\}$.\n\n\\emph{Case 2: $\\operatorname{rank}A=k=d$.}  \nThen $A$ is invertible.  Put\n$w:=-A^{-1}\\mathbf 1$, $\\mathbf 1:=(1,\\dots ,1)^{\\top}$.  \nBecause $Aw=-\\mathbf 1$, one has $\\langle a^{i},w\\rangle=-1$ for all $i$.\nFor $t>0$,\n$\\langle a^{i},x_{0}+tw\\rangle=\\langle a^{i},x_{0}\\rangle-t\\le b_{i}-t<b_{i}$,\nso $x_{0}+tw\\in G$ for every $t>0$.  \nAgain $G$ contains the entire ray $\\{x_{0}+tw:t\\ge 0\\}$.\n\n\\emph{Case 3: $k<d$ and $\\operatorname{rank}A=k$.}  \nHere $d-k\\ge 1$, so $\\ker A$ is non–trivial; choose $0\\neq w\\in\\ker A$\nand repeat the argument of Case~1.  \n\n\\noindent\nCombining the three cases we have established  \n\\[\n\\text{every non–empty intersection of at most $d$ half–spaces in $\\mathbb{R}^{d}$ is unbounded.}\n\\tag{3.1}\n\\]\n\n\\smallskip\n\\textbf{Step 3.2:  A concrete counterexample for the $d$–wise hypothesis.}\n\nFix $R>0$ and set $c:=R+1$.  \nDefine the following $d+1$ closed half–spaces in $\\mathbb{R}^{d}$:\n\\[\n\\begin{aligned}\nH_{0}&:=\\Bigl\\{x=(x_{1},\\dots ,x_{d})\\in\\mathbb{R}^{d}\\;:\\;\n           \\sum_{i=1}^{d}x_{i}\\le 0\\Bigr\\},\\\\[4pt]\nH_{i}&:=\\Bigl\\{x=(x_{1},\\dots ,x_{d})\\in\\mathbb{R}^{d}\\;:\\;\n           x_{i}\\ge c\\Bigr\\}, \\qquad 1\\le i\\le d.\n\\end{aligned}\n\\tag{3.2}\n\\]\n\n(a) \\emph{Every subfamily of at most $d$ members has non–empty intersection.}\n\n$\\bullet$ If the subfamily omits $H_{0}$, it is of the form\n$\\{H_{i}\\}_{i\\in I}$ with $I\\subseteq\\{1,\\dots ,d\\}$ and $|I|\\le d$.\nThe point $x^{(I)}$ defined by  \n\\[\nx^{(I)}_{i}=c \\text{ for } i\\in I,\\qquad \nx^{(I)}_{i}=c+1 \\text{ for } i\\notin I\n\\]\nlies in all selected half–spaces, so the intersection is non–empty (and unbounded).\n\n$\\bullet$ If the subfamily contains $H_{0}$ but omits, say, $H_{k}$ with $k\\ge 1$, take  \n\\[\nx_{k}:=-d(c+1),\\qquad\nx_{i}=c\\quad(i\\neq k).\n\\]\nThen $\\sum_{i=1}^{d}x_{i}= -d(c+1)+(d-1)c=-c<0$, so the point\n$x=(x_{1},\\dots ,x_{d})$ satisfies $H_{0}$ as well as every\n$H_{i}$ with $i\\neq k$.  Hence the intersection is again non–empty (and\nunbounded).\n\nThus property (Q1) holds for \\emph{every} subfamily of at most $d$ members.\n\n(b) \\emph{The total intersection is empty.}\n\nIf $x\\in\\bigcap_{i=0}^{d}H_{i}$ we would have\n$x_{i}\\ge c$ for all $1\\le i\\le d$, hence\n$\\sum_{i=1}^{d}x_{i}\\ge d\\,c>0$, contradicting the defining inequality of\n$H_{0}$.  Consequently  \n\\[\n\\bigcap_{i=0}^{d}H_{i}=\\varnothing .\n\\tag{3.3}\n\\]\n\n\\smallskip\n\\textbf{Step 3.3:  Consequences.}\n\nThe family $\\{H_{0},\\dots ,H_{d}\\}$ satisfies (a) and (b) above, establishing the optimality of the number $d+1$.  \nMoreover, by (3.1) every $d$–wise intersection is unbounded, so\n\\emph{no} ball $B(0,R)$ can cover all those intersections.  Therefore the radius\nbound asserted in part~2 cannot be guaranteed under a merely\n$d$–wise hypothesis, and the parameter $d+1$ is optimal.\n\n\\hfill$\\square$\n\n---------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.395658",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension:  the problem is set in ℝᵈ for arbitrary d, not in ℝ.  \n• Additional constraints:  the “radius-R” quantitative restriction forces the solver\n  to keep track of the size of every finite intersection, not merely its existence.  \n• Sophisticated structure:  half–spaces are unbounded; lifting to ℝ^{d+1} produces\n  cones and cylinders, so the argument mixes affine geometry with norm constraints.  \n• Deeper theory:  a successful solution needs the classical Helly theorem, but also\n  must recognise how to convert a norm bound into a linear inequality in one higher\n  dimension – a standard trick in convex analysis but far from obvious at\n  Olympiad level.  \n• Interacting concepts:  convexity, compactness of finite intersections,\n  quantitative vs. qualitative Helly, and a non-trivial minimal-example argument\n  all appear.\n\nThese layers of technicality and the necessity of invoking higher-dimensional\nHelly put the problem well beyond the scope of the original one-dimensional\ninterval exercise."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}