{
  "index": "1989-B-4",
  "type": "COMB",
  "tag": [
    "COMB",
    "NT",
    "ANA"
  ],
  "difficulty": "",
  "question": "Can a countably infinite set have an uncountable collection of\nnon-empty subsets such that the intersection of any two of them is\nfinite?",
  "solution": "Solution 1. The set \\( \\mathbb{Q} \\) of rational numbers is countably infinite. For each real number \\( \\alpha \\), choose a sequence of distinct rational numbers tending to \\( \\alpha \\), and let \\( S_{\\alpha} \\) be the set of terms. If \\( \\alpha, \\beta \\) are distinct real numbers, then \\( S_{\\alpha} \\cap S_{\\beta} \\) is finite, since otherwise a sequence obtained by listing its elements would converge to both \\( \\alpha \\) and \\( \\beta \\). In particular, \\( S_{\\alpha} \\neq S_{\\beta} \\). Thus \\( \\left\\{S_{\\alpha}: \\alpha \\in \\mathbb{R}\\right\\} \\) is an uncountable collection of nonempty subsets of \\( \\mathbb{Q} \\) with the desired property.\n\nRemark. A minor variant on this solution would be to take the (countable) set of real numbers with terminating decimal expansions, and for each of the (uncountably many) irrational numbers \\( a \\), let \\( S_{a} \\) be the set of decimal approximations to \\( a \\) obtained by truncating the decimal expansion at some point.\n\nSolution 2. Let \\( A \\) denote the countably infinite set of finite strings of 0 's and 1's. For each infinite string \\( a=a_{1} a_{2} a_{3} \\ldots \\) of 0's and 1's, let \\( S_{a}=\\left\\{a_{1} a_{2} \\ldots a_{n}: n \\geq 1\\right\\} \\) denote the set of finite initial substrings. There are uncountably many \\( a \\), and if \\( a=a_{1} a_{2} \\ldots \\) and \\( b=b_{1} b_{2} \\ldots \\) are distinct infinite strings, say with \\( a_{m} \\neq b_{m} \\), then all strings in \\( S_{a} \\cap S_{b} \\) have length less than \\( m \\), so \\( S_{a} \\cap S_{b} \\) is finite.\n\nSolution 3. The set \\( \\mathbb{Z}^{2} \\) of lattice points in the plane is countably infinite. For each real \\( \\alpha \\), let \\( S_{\\alpha} \\) denote the set of points in \\( \\mathbb{Z}^{2} \\) whose distance to the line \\( y=\\alpha x \\) is at most 1. If \\( \\alpha, \\beta \\) are distinct real numbers, then \\( S_{\\alpha} \\cap S_{\\beta} \\) is a set of lattice points in a bounded region, so it is finite.\n\nSolution 4. Let \\( A \\) be a disjoint union of a countably infinite number of countably infinite sets, so \\( A \\) is countably infinite. Call a collection of subsets \\( \\mathcal{C} \\) of \\( A \\) good if it consists of an infinite number of countably infinite subsets of \\( A \\), and \\( S \\cap T \\) is finite for any distinct \\( S, T \\in \\mathcal{C} \\). By construction of \\( A \\), there exists a good collection (of disjoint subsets). Order the good collections by inclusion. For any chain of good collections, the union is also a good collection. Hence by Zorn's Lemma, there exists a maximal good collection \\( \\mathcal{C}_{\\text {max }} \\).\n\nSuppose \\( \\mathcal{C}_{\\text {max }} \\) were countable, say \\( \\mathcal{C}_{\\text {max }}=\\left\\{S_{1}, S_{2}, \\ldots\\right\\} \\). Because \\( S_{n} \\) is infinite while \\( S_{i} \\cap S_{n} \\) is finite for \\( i \\neq n \\), there exists \\( b_{n} \\in S_{n}-\\bigcup_{i=1}^{n-1} S_{i} \\) for each \\( n \\geq 1 \\). For \\( m<n, b_{m} \\in S_{m} \\) but \\( b_{n} \\notin S_{m} \\), so \\( b_{m} \\neq b_{n} \\), and the set \\( B=\\left\\{b_{1}, b_{2}, \\ldots\\right\\} \\) is countably infinite. Moreover, \\( b_{N} \\notin S_{n} \\) for \\( N>n \\), so \\( B \\cap S_{n} \\) is finite. Hence \\( \\left\\{B, S_{1}, S_{2}, \\ldots\\right\\} \\) is a good collection, contradicting the maximality of \\( \\mathcal{C}_{\\max } \\). Thus \\( \\mathcal{C}_{\\text {max }} \\) is uncountable, as desired.\n\nRemark. This question appears as [New, Problem 49], where the countably infinite set is taken to be \\( \\mathbb{Z} \\).\n\nRemark (Zorn's Lemma). A chain in a partially ordered set ( \\( S, \\leq \\) ) is a subset in which every two elements are comparable. Zorn's Lemma states that if \\( S \\) is a nonempty partially ordered set such that every chain in \\( S \\) has an upper bound in \\( S \\), then \\( S \\) contains a maximal element, i.e., an element \\( m \\) such that the only element \\( s \\in S \\) satisfying \\( s \\geq m \\) is \\( m \\) itself. Zorn's Lemma is equivalent to the Axiom of Choice, which states that the product of a family of nonempty sets indexed by a nonempty set is nonempty. It is also equivalent to the Well Ordering Principle, which states that every set admits a well ordering. (A well ordering of a set \\( S \\) is a total ordering such that every nonempty subset \\( A \\subseteq S \\) has a least element.) See pages 151 and 196 of [En].\n\nRelated question. Show that the following similar question, a restatement of [Hal, Problem 11C], has a negative answer:\n\nCan a countably infinite set have an uncountable collection of nonempty subsets such that the intersection of any two of them has at most 1989 elements?",
  "vars": [
    "A",
    "B",
    "C",
    "N",
    "S",
    "T",
    "a",
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    "i",
    "m",
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    "y"
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  "params": [
    "\\\\alpha",
    "\\\\beta",
    "S_\\\\alpha",
    "S_\\\\beta",
    "S_a",
    "S_i",
    "S_m",
    "S_n",
    "S_1",
    "S_2",
    "a_1",
    "a_2",
    "a_3",
    "a_m",
    "b_1",
    "b_2",
    "b_m",
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  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "mainset",
        "B": "altsetb",
        "C": "classset",
        "N": "indexnum",
        "S": "supersub",
        "T": "othersub",
        "a": "infstring",
        "b": "altstring",
        "i": "iterindex",
        "m": "matchpos",
        "n": "countern",
        "x": "planex",
        "y": "planey",
        "\\\\alpha": "realalpha",
        "\\\\beta": "realbeta",
        "S_\\\\alpha": "subsetalpha",
        "S_\\\\beta": "subsetbeta",
        "S_a": "subseta",
        "S_i": "subseti",
        "S_m": "subsetm",
        "S_n": "subsetn",
        "S_1": "subsetone",
        "S_2": "subsettwo",
        "a_1": "stringone",
        "a_2": "stringtwo",
        "a_3": "stringthr",
        "a_m": "stringm",
        "b_1": "altone",
        "b_2": "alttwo",
        "b_m": "altmany",
        "b_n": "altnum",
        "b_N": "altindex"
      },
      "question": "Can a countably infinite set have an uncountable collection of\nnon-empty subsets such that the intersection of any two of them is\nfinite?",
      "solution": "Solution 1. The set \\( \\mathbb{Q} \\) of rational numbers is countably infinite. For each real number \\( realalpha \\), choose a sequence of distinct rational numbers tending to \\( realalpha \\), and let \\( subsetalpha \\) be the set of terms. If \\( realalpha, realbeta \\) are distinct real numbers, then \\( subsetalpha \\cap subsetbeta \\) is finite, since otherwise a sequence obtained by listing its elements would converge to both \\( realalpha \\) and \\( realbeta \\). In particular, \\( subsetalpha \\neq subsetbeta \\). Thus \\( \\left\\{subsetalpha: realalpha \\in \\mathbb{R}\\right\\} \\) is an uncountable collection of nonempty subsets of \\( \\mathbb{Q} \\) with the desired property.\n\nRemark. A minor variant on this solution would be to take the (countable) set of real numbers with terminating decimal expansions, and for each of the (uncountably many) irrational numbers \\( infstring \\), let \\( subseta \\) be the set of decimal approximations to \\( infstring \\) obtained by truncating the decimal expansion at some point.\n\nSolution 2. Let \\( mainset \\) denote the countably infinite set of finite strings of 0 's and 1's. For each infinite string \\( infstring=stringone stringtwo stringthr \\ldots \\) of 0's and 1's, let \\( subseta=\\left\\{stringone stringtwo \\ldots infstring_{countern}: countern \\geq 1\\right\\} \\) denote the set of finite initial substrings. There are uncountably many \\( infstring \\), and if \\( infstring=stringone stringtwo \\ldots \\) and \\( altstring=altone alttwo \\ldots \\) are distinct infinite strings, say with \\( stringm \\neq altmany \\), then all strings in \\( subseta \\cap supersub_{altstring} \\) have length less than \\( matchpos \\), so \\( subseta \\cap supersub_{altstring} \\) is finite.\n\nSolution 3. The set \\( \\mathbb{Z}^{2} \\) of lattice points in the plane is countably infinite. For each real \\( realalpha \\), let \\( subsetalpha \\) denote the set of points in \\( \\mathbb{Z}^{2} \\) whose distance to the line \\( planey=realalpha\\,planex \\) is at most 1. If \\( realalpha, realbeta \\) are distinct real numbers, then \\( subsetalpha \\cap subsetbeta \\) is a set of lattice points in a bounded region, so it is finite.\n\nSolution 4. Let \\( mainset \\) be a disjoint union of a countably infinite number of countably infinite sets, so \\( mainset \\) is countably infinite. Call a collection of subsets \\( classset \\) of \\( mainset \\) good if it consists of an infinite number of countably infinite subsets of \\( mainset \\), and \\( supersub \\cap othersub \\) is finite for any distinct \\( supersub, othersub \\in classset \\). By construction of \\( mainset \\), there exists a good collection (of disjoint subsets). Order the good collections by inclusion. For any chain of good collections, the union is also a good collection. Hence by Zorn's Lemma, there exists a maximal good collection \\( classset_{\\text {max }} \\).\n\nSuppose \\( classset_{\\text {max }} \\) were countable, say \\( classset_{\\text {max }}=\\left\\{subsetone, subsettwo, \\ldots\\right\\} \\). Because \\( subsetn \\) is infinite while \\( subseti \\cap subsetn \\) is finite for \\( iterindex \\neq countern \\), there exists \\( altnum \\in subsetn-\\bigcup_{iterindex=1}^{countern-1} subseti \\) for each \\( countern \\geq 1 \\). For \\( matchpos<countern, altmany \\in subsetm \\) but \\( altnum \\notin subsetm \\), so \\( altmany \\neq altnum \\), and the set \\( altsetb=\\left\\{altone, alttwo, \\ldots\\right\\} \\) is countably infinite. Moreover, \\( altindex \\notin subsetn \\) for \\( indexnum>countern \\), so \\( altsetb \\cap subsetn \\) is finite. Hence \\( \\left\\{altsetb, subsetone, subsettwo, \\ldots\\right\\} \\) is a good collection, contradicting the maximality of \\( classset_{\\max } \\). Thus \\( classset_{\\text {max }} \\) is uncountable, as desired.\n\nRemark. This question appears as [New, Problem 49], where the countably infinite set is taken to be \\( \\mathbb{Z} \\).\n\nRemark (Zorn's Lemma). A chain in a partially ordered set ( \\( supersub, \\leq \\) ) is a subset in which every two elements are comparable. Zorn's Lemma states that if \\( supersub \\) is a nonempty partially ordered set such that every chain in \\( supersub \\) has an upper bound in \\( supersub \\), then \\( supersub \\) contains a maximal element, i.e., an element \\( matchpos \\) such that the only element \\( s \\in supersub \\) satisfying \\( s \\geq matchpos \\) is \\( matchpos \\) itself. Zorn's Lemma is equivalent to the Axiom of Choice, which states that the product of a family of nonempty sets indexed by a nonempty set is nonempty. It is also equivalent to the Well Ordering Principle, which states that every set admits a well ordering. (A well ordering of a set \\( supersub \\) is a total ordering such that every nonempty subset \\( altsetb \\subseteq supersub \\) has a least element.) See pages 151 and 196 of [En].\n\nRelated question. Show that the following similar question, a restatement of [Hal, Problem 11classset], has a negative answer:\n\nCan a countably infinite set have an uncountable collection of nonempty subsets such that the intersection of any two of them has at most 1989 elements?"
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "mackerels",
        "B": "pineapples",
        "C": "submarines",
        "N": "cardboard",
        "S": "chocolate",
        "T": "smoothies",
        "a": "jellyfish",
        "b": "dragonfly",
        "i": "teaspoon",
        "m": "nachosdip",
        "n": "porcupine",
        "x": "butternut",
        "y": "marshmallow",
        "\\alpha": "sunflower",
        "\\beta": "toothbrush",
        "S_\\alpha": "crystalline",
        "S_\\beta": "blueberries",
        "S_a": "bricklayer",
        "S_i": "woodcutter",
        "S_m": "rattlesnake",
        "S_n": "tangerines",
        "S_1": "salamander",
        "S_2": "lighthouse",
        "a_1": "grandfather",
        "a_2": "basketball",
        "a_3": "peppercorn",
        "a_m": "harmonicas",
        "b_1": "honeybees",
        "b_2": "storytales",
        "b_m": "shorebirds",
        "b_n": "mountaineer",
        "b_N": "blacksmith"
      },
      "question": "Can a countably infinite set have an uncountable collection of\nnon-empty subsets such that the intersection of any two of them is\nfinite?",
      "solution": "Solution 1. The set \\( \\mathbb{Q} \\) of rational numbers is countably infinite. For each real number \\( sunflower \\), choose a sequence of distinct rational numbers tending to \\( sunflower \\), and let \\( chocolate_{sunflower} \\) be the set of terms. If \\( sunflower, toothbrush \\) are distinct real numbers, then \\( chocolate_{sunflower} \\cap chocolate_{toothbrush} \\) is finite, since otherwise a sequence obtained by listing its elements would converge to both \\( sunflower \\) and \\( toothbrush \\). In particular, \\( chocolate_{sunflower} \\neq chocolate_{toothbrush} \\). Thus \\( \\left\\{chocolate_{sunflower}: sunflower \\in \\mathbb{R}\\right\\} \\) is an uncountable collection of nonempty subsets of \\( \\mathbb{Q} \\) with the desired property.\n\nRemark. A minor variant on this solution would be to take the (countable) set of real numbers with terminating decimal expansions, and for each of the (uncountably many) irrational numbers \\( jellyfish \\), let \\( chocolate_{jellyfish} \\) be the set of decimal approximations to \\( jellyfish \\) obtained by truncating the decimal expansion at some point.\n\nSolution 2. Let \\( mackerels \\) denote the countably infinite set of finite strings of 0 's and 1's. For each infinite string \\( jellyfish=jellyfish_{1} jellyfish_{2} jellyfish_{3} \\ldots \\) of 0's and 1's, let \\( chocolate_{jellyfish}=\\left\\{jellyfish_{1} jellyfish_{2} \\ldots jellyfish_{porcupine}: porcupine \\geq 1\\right\\} \\) denote the set of finite initial substrings. There are uncountably many \\( jellyfish \\), and if \\( jellyfish=jellyfish_{1} jellyfish_{2} \\ldots \\) and \\( dragonfly=dragonfly_{1} dragonfly_{2} \\ldots \\) are distinct infinite strings, say with \\( jellyfish_{nachosdip} \\neq dragonfly_{nachosdip} \\), then all strings in \\( chocolate_{jellyfish} \\cap chocolate_{dragonfly} \\) have length less than \\( nachosdip \\), so \\( chocolate_{jellyfish} \\cap chocolate_{dragonfly} \\) is finite.\n\nSolution 3. The set \\( \\mathbb{Z}^{2} \\) of lattice points in the plane is countably infinite. For each real \\( sunflower \\), let \\( chocolate_{sunflower} \\) denote the set of points in \\( \\mathbb{Z}^{2} \\) whose distance to the line \\( marshmallow=sunflower butternut \\) is at most 1. If \\( sunflower, toothbrush \\) are distinct real numbers, then \\( chocolate_{sunflower} \\cap chocolate_{toothbrush} \\) is a set of lattice points in a bounded region, so it is finite.\n\nSolution 4. Let \\( mackerels \\) be a disjoint union of a countably infinite number of countably infinite sets, so \\( mackerels \\) is countably infinite. Call a collection of subsets \\( \\mathcal{submarines} \\) of \\( mackerels \\) good if it consists of an infinite number of countably infinite subsets of \\( mackerels \\), and \\( chocolate \\cap smoothies \\) is finite for any distinct \\( chocolate, smoothies \\in \\mathcal{submarines} \\). By construction of \\( mackerels \\), there exists a good collection (of disjoint subsets). Order the good collections by inclusion. For any chain of good collections, the union is also a good collection. Hence by Zorn's Lemma, there exists a maximal good collection \\( \\mathcal{submarines}_{\\text {max }} \\).\n\nSuppose \\( \\mathcal{submarines}_{\\text {max }} \\) were countable, say \\( \\mathcal{submarines}_{\\text {max }}=\\left\\{chocolate_{1}, chocolate_{2}, \\ldots\\right\\} \\). Because \\( chocolate_{porcupine} \\) is infinite while \\( chocolate_{teaspoon} \\cap chocolate_{porcupine} \\) is finite for \\( teaspoon \\neq porcupine \\), there exists \\( dragonfly_{porcupine} \\in chocolate_{porcupine}-\\bigcup_{teaspoon=1}^{porcupine-1} chocolate_{teaspoon} \\) for each \\( porcupine \\geq 1 \\). For \\( nachosdip<porcupine, dragonfly_{nachosdip} \\in chocolate_{nachosdip} \\) but \\( dragonfly_{porcupine} \\notin chocolate_{nachosdip} \\), so \\( dragonfly_{nachosdip} \\neq dragonfly_{porcupine} \\), and the set \\( pineapples=\\left\\{dragonfly_{1}, dragonfly_{2}, \\ldots\\right\\} \\) is countably infinite. Moreover, \\( dragonfly_{cardboard} \\notin chocolate_{porcupine} \\) for \\( cardboard>porcupine \\), so \\( pineapples \\cap chocolate_{porcupine} \\) is finite. Hence \\( \\left\\{pineapples, chocolate_{1}, chocolate_{2}, \\ldots\\right\\} \\) is a good collection, contradicting the maximality of \\( \\mathcal{submarines}_{\\max } \\). Thus \\( \\mathcal{submarines}_{\\text {max }} \\) is uncountable, as desired.\n\nRemark. This question appears as [New, Problem 49], where the countably infinite set is taken to be \\( \\mathbb{Z} \\).\n\nRemark (Zorn's Lemma). A chain in a partially ordered set ( \\( chocolate, \\leq \\) ) is a subset in which every two elements are comparable. Zorn's Lemma states that if \\( chocolate \\) is a nonempty partially ordered set such that every chain in \\( chocolate \\) has an upper bound in \\( chocolate \\), then \\( chocolate \\) contains a maximal element, i.e., an element \\( nachosdip \\) such that the only element \\( s \\in chocolate \\) satisfying \\( s \\geq nachosdip \\) is \\( nachosdip \\) itself. Zorn's Lemma is equivalent to the Axiom of Choice, which states that the product of a family of nonempty sets indexed by a nonempty set is nonempty. It is also equivalent to the Well Ordering Principle, which states that every set admits a well ordering. (A well ordering of a set \\( chocolate \\) is a total ordering such that every nonempty subset \\( mackerels \\subseteq chocolate \\) has a least element.) See pages 151 and 196 of [En].\n\nRelated question. Show that the following similar question, a restatement of [Hal, Problem 11C], has a negative answer:\n\nCan a countably infinite set have an uncountable collection of nonempty subsets such that the intersection of any two of them has at most 1989 elements?"
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "microset",
        "B": "emptiness",
        "C": "voidclass",
        "N": "finitecount",
        "S": "singleton",
        "T": "unitaryset",
        "a": "terminus",
        "b": "beginner",
        "i": "collective",
        "m": "immensity",
        "n": "nilpotent",
        "x": "knownvalue",
        "y": "stagnant",
        "\\alpha": "ultimate",
        "\\beta": "initials",
        "S_\\alpha": "singularity",
        "S_\\beta": "plurality",
        "S_a": "solitude",
        "S_i": "unionism",
        "S_m": "monolith",
        "S_n": "aggregate",
        "S_1": "unityfirst",
        "S_2": "unitysecond",
        "a_1": "terminalone",
        "a_2": "terminaltwo",
        "a_3": "terminalthree",
        "a_m": "terminalmass",
        "b_1": "originone",
        "b_2": "origintwo",
        "b_m": "originmass",
        "b_n": "originnull",
        "b_N": "originwhole"
      },
      "question": "Can a countably infinite set have an uncountable collection of\nnon-empty subsets such that the intersection of any two of them is\nfinite?",
      "solution": "Solution 1. The set \\( \\mathbb{Q} \\) of rational numbers is countably infinite. For each real number \\( ultimate \\), choose a sequence of distinct rational numbers tending to \\( ultimate \\), and let \\( singularity \\) be the set of terms. If \\( ultimate, initials \\) are distinct real numbers, then \\( singularity \\cap plurality \\) is finite, since otherwise a sequence obtained by listing its elements would converge to both \\( ultimate \\) and \\( initials \\). In particular, \\( singularity \\neq plurality \\). Thus \\( \\left\\{singularity: ultimate \\in \\mathbb{R}\\right\\} \\) is an uncountable collection of nonempty subsets of \\( \\mathbb{Q} \\) with the desired property.\n\nRemark. A minor variant on this solution would be to take the (countable) set of real numbers with terminating decimal expansions, and for each of the (uncountably many) irrational numbers \\( terminus \\), let \\( solitude \\) be the set of decimal approximations to \\( terminus \\) obtained by truncating the decimal expansion at some point.\n\nSolution 2. Let \\( microset \\) denote the countably infinite set of finite strings of 0 's and 1's. For each infinite string \\( terminus=terminalone terminaltwo terminalthree \\ldots \\) of 0's and 1's, let \\( solitude=\\left\\{terminalone terminaltwo \\ldots terminus_{nilpotent}: nilpotent \\geq 1\\right\\} \\) denote the set of finite initial substrings. There are uncountably many \\( terminus \\), and if \\( terminus=terminalone terminaltwo \\ldots \\) and \\( beginner=originone origintwo \\ldots \\) are distinct infinite strings, say with \\( terminus_{immensity} \\neq beginner_{immensity} \\), then all strings in \\( solitude \\cap singleton_{beginner} \\) have length less than \\( immensity \\), so \\( solitude \\cap singleton_{beginner} \\) is finite.\n\nSolution 3. The set \\( \\mathbb{Z}^{2} \\) of lattice points in the plane is countably infinite. For each real \\( ultimate \\), let \\( singularity \\) denote the set of points in \\( \\mathbb{Z}^{2} \\) whose distance to the line \\( stagnant = ultimate\\, knownvalue \\) is at most 1. If \\( ultimate, initials \\) are distinct real numbers, then \\( singularity \\cap plurality \\) is a set of lattice points in a bounded region, so it is finite.\n\nSolution 4. Let \\( microset \\) be a disjoint union of a countably infinite number of countably infinite sets, so \\( microset \\) is countably infinite. Call a collection of subsets \\( \\mathcal{voidclass} \\) of \\( microset \\) good if it consists of an infinite number of countably infinite subsets of \\( microset \\), and \\( singleton \\cap unitaryset \\) is finite for any distinct \\( singleton, unitaryset \\in \\mathcal{voidclass} \\). By construction of \\( microset \\), there exists a good collection (of disjoint subsets). Order the good collections by inclusion. For any chain of good collections, the union is also a good collection. Hence by Zorn's Lemma, there exists a maximal good collection \\( \\mathcal{voidclass}_{\\text {max }} \\).\n\nSuppose \\( \\mathcal{voidclass}_{\\text {max }} \\) were countable, say \\( \\mathcal{voidclass}_{\\text {max }}=\\left\\{unityfirst, unitysecond, \\ldots\\right\\} \\). Because \\( aggregate \\) is infinite while \\( unionism \\cap aggregate \\) is finite for \\( collective \\neq nilpotent \\), there exists \\( originnull \\in aggregate-\\bigcup_{collective=1}^{nilpotent-1} unionism \\) for each \\( nilpotent \\geq 1 \\). For \\( immensity<nilpotent, originmass \\in monolith \\) but \\( originnull \\notin monolith \\), so \\( originmass \\neq originnull \\), and the set \\( emptiness=\\left\\{originone, origintwo, \\ldots\\right\\} \\) is countably infinite. Moreover, \\( originwhole \\notin aggregate \\) for \\( finitecount>nilpotent \\), so \\( emptiness \\cap aggregate \\) is finite. Hence \\( \\left\\{emptiness, unityfirst, unitysecond, \\ldots\\right\\} \\) is a good collection, contradicting the maximality of \\( \\mathcal{voidclass}_{\\max } \\). Thus \\( \\mathcal{voidclass}_{\\text {max }} \\) is uncountable, as desired.\n\nRemark. This question appears as [New, Problem 49], where the countably infinite set is taken to be \\( \\mathbb{Z} \\).\n\nRemark (Zorn's Lemma). A chain in a partially ordered set ( \\( singleton, \\leq \\) ) is a subset in which every two elements are comparable. Zorn's Lemma states that if \\( singleton \\) is a nonempty partially ordered set such that every chain in \\( singleton \\) has an upper bound in \\( singleton \\), then \\( singleton \\) contains a maximal element, i.e., an element \\( immensity \\) such that the only element \\( s \\in singleton \\) satisfying \\( s \\geq immensity \\) is \\( immensity \\) itself. Zorn's Lemma is equivalent to the Axiom of Choice, which states that the product of a family of nonempty sets indexed by a nonempty set is nonempty. It is also equivalent to the Well Ordering Principle, which states that every set admits a well ordering. (A well ordering of a set \\( singleton \\) is a total ordering such that every nonempty subset \\( microset \\subseteq singleton \\) has a least element.) See pages 151 and 196 of [En].\n\nRelated question. Show that the following similar question, a restatement of [Hal, Problem 11C], has a negative answer:\n\nCan a countably infinite set have an uncountable collection of nonempty subsets such that the intersection of any two of them has at most 1989 elements?"
    },
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        "a": "zorvukypt",
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        "S_a": "laspoirmq",
        "S_i": "qafykzuvx",
        "S_m": "gosrafkpt",
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        "a_m": "rbcqwlyst",
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      },
      "question": "Can a countably infinite set have an uncountable collection of\nnon-empty subsets such that the intersection of any two of them is\nfinite?",
      "solution": "Solution 1. The set \\( \\mathbb{Q} \\) of rational numbers is countably infinite. For each real number \\( qzxwvtnp \\), choose a sequence of distinct rational numbers tending to \\( qzxwvtnp \\), and let \\( porvqtmnd \\) be the set of terms. If \\( qzxwvtnp, hjgrksla \\) are distinct real numbers, then \\( porvqtmnd \\cap yapwfrsle \\) is finite, since otherwise a sequence obtained by listing its elements would converge to both \\( qzxwvtnp \\) and \\( hjgrksla \\). In particular, \\( porvqtmnd \\neq yapwfrsle \\). Thus \\( \\left\\{porvqtmnd: qzxwvtnp \\in \\mathbb{R}\\right\\} \\) is an uncountable collection of nonempty subsets of \\( \\mathbb{Q} \\) with the desired property.\n\nRemark. A minor variant on this solution would be to take the (countable) set of real numbers with terminating decimal expansions, and for each of the (uncountably many) irrational numbers \\( zorvukypt \\), let \\( laspoirmq \\) be the set of decimal approximations to \\( zorvukypt \\) obtained by truncating the decimal expansion at some point.\n\nSolution 2. Let \\( kfjwqzops \\) denote the countably infinite set of finite strings of 0's and 1's. For each infinite string \\( zorvukypt=takqrplsw dzmqucyra vrxtpqnas \\ldots \\) of 0's and 1's, let\n\\( laspoirmq=\\left\\{takqrplsw dzmqucyra \\ldots zorvukypt_{vfazkyreo}: vfazkyreo \\ge 1\\right\\} \\)\ndenote the set of finite initial substrings. There are uncountably many \\( zorvukypt \\), and if \\( zorvukypt=takqrplsw dzmqucyra \\ldots \\) and \\( lyhcxsmad=zgylwspmq kytnrofaq \\ldots \\) are distinct infinite strings, say with \\( rbcqwlyst \\neq vbczrlyuk \\), then all strings in \\( laspoirmq \\cap ghqlxmtru_{lyhcxsmad} \\) have length less than \\( girpsotkw \\), so \\( laspoirmq \\cap ghqlxmtru_{lyhcxsmad} \\) is finite.\n\nSolution 3. The set \\( \\mathbb{Z}^{2} \\) of lattice points in the plane is countably infinite. For each real \\( qzxwvtnp \\), let \\( porvqtmnd \\) denote the set of points in \\( \\mathbb{Z}^{2} \\) whose distance to the line \\( lzetpomga=qzxwvtnp mjkqudica \\) is at most 1. If \\( qzxwvtnp, hjgrksla \\) are distinct real numbers, then \\( porvqtmnd \\cap yapwfrsle \\) is a set of lattice points in a bounded region, so it is finite.\n\nSolution 4. Let \\( kfjwqzops \\) be a disjoint union of a countably infinite number of countably infinite sets, so \\( kfjwqzops \\) is countably infinite. Call a collection of subsets \\( \\mathcal{sdfjnalwe} \\) of \\( kfjwqzops \\) good if it consists of an infinite number of countably infinite subsets of \\( kfjwqzops \\), and \\( ghqlxmtru \\cap jwpralnxc \\) is finite for any distinct \\( ghqlxmtru, jwpralnxc \\in \\mathcal{sdfjnalwe} \\). By construction of \\( kfjwqzops \\), there exists a good collection (of disjoint subsets). Order the good collections by inclusion. For any chain of good collections, the union is also a good collection. Hence by Zorn's Lemma, there exists a maximal good collection \\( \\mathcal{sdfjnalwe}_{\\text {max }} \\).\n\nSuppose \\( \\mathcal{sdfjnalwe}_{\\text {max }} \\) were countable, say \\( \\mathcal{sdfjnalwe}_{\\text {max }}=\\left\\{xjrnalkwp, gvyarktqs, \\ldots\\right\\} \\). Because \\( woqzlbind \\) is infinite while \\( qafykzuvx \\cap woqzlbind \\) is finite for \\( qevbnrtsa \\neq vfazkyreo \\), there exists \\( qsomnhyle \\in woqzlbind-\\bigcup_{qevbnrtsa=1}^{vfazkyreo-1} qafykzuvx \\) for each \\( vfazkyreo \\ge 1 \\). For \\( girpsotkw<vfazkyreo, vbczrlyuk \\in gosrafkpt \\) but \\( qsomnhyle \\notin gosrafkpt \\), so \\( vbczrlyuk \\neq qsomnhyle \\), and the set \\( hfjkeqrtu=\\left\\{zgylwspmq, kytnrofaq, \\ldots\\right\\} \\) is countably infinite. Moreover, \\( wyrtqkcsa \\notin woqzlbind \\) for \\( pvzchmnqo>vfazkyreo \\), so \\( hfjkeqrtu \\cap woqzlbind \\) is finite. Hence \\( \\left\\{hfjkeqrtu, xjrnalkwp, gvyarktqs, \\ldots\\right\\} \\) is a good collection, contradicting the maximality of \\( \\mathcal{sdfjnalwe}_{\\max } \\). Thus \\( \\mathcal{sdfjnalwe}_{\\text {max }} \\) is uncountable, as desired.\n\nRemark. This question appears as [New, Problem 49], where the countably infinite set is taken to be \\( \\mathbb{Z} \\).\n\nRemark (Zorn's Lemma). A chain in a partially ordered set ( \\( ghqlxmtru, \\le \\) ) is a subset in which every two elements are comparable. Zorn's Lemma states that if \\( ghqlxmtru \\) is a nonempty partially ordered set such that every chain in \\( ghqlxmtru \\) has an upper bound in \\( ghqlxmtru \\), then \\( ghqlxmtru \\) contains a maximal element, i.e., an element \\( girpsotkw \\) such that the only element \\( s \\in ghqlxmtru \\) satisfying \\( s \\ge girpsotkw \\) is \\( girpsotkw \\) itself. Zorn's Lemma is equivalent to the Axiom of Choice, which states that the product of a family of nonempty sets indexed by a nonempty set is nonempty. It is also equivalent to the Well Ordering Principle, which states that every set admits a well ordering. (A well ordering of a set \\( ghqlxmtru \\) is a total ordering such that every nonempty subset \\( kfjwqzops \\subseteq ghqlxmtru \\) has a least element.) See pages 151 and 196 of [En].\n\nRelated question. Show that the following similar question, a restatement of [Hal, Problem 11C], has a negative answer:\n\nCan a countably infinite set have an uncountable collection of nonempty subsets such that the intersection of any two of them has at most 1989 elements?"
    },
    "kernel_variant": {
      "question": "Let\n  D = { k / 3^n : n ∈ ℕ , k ∈ ℤ , 0 ≤ k ≤ 3^n }\nbe the countably-infinite, dense subset of [0,1] that consists of all real numbers whose base–3 expansion terminates.\nLet C ⊂ [0,1] be the classical middle–third Cantor set.\n\n(a)  For every point γ ∈ C construct an explicit infinite subset S_γ ⊂ D such that for any two distinct Cantor points γ ≠ δ the intersection S_γ ∩ S_δ is finite.\n\n(b)  Deduce that a countably infinite set can carry an uncountable family of non-empty subsets whose pairwise intersections are all finite.",
      "solution": "Write every element of the Cantor set in its (unique) ternary expansion that involves only the digits 0 and 2:\n      γ = 0.γ₁γ₂γ₃ …   with γ_j ∈ {0,2}  (j ≥ 1).\n\n1.  Construction of the sets S_γ\n   -------------------------------------------------\n   For n ≥ 1 define\n        σ_n(γ)  :=  0. γ₁ γ₂ … γ_{n−1} (2−γ_n) 0 0 0 …   (base 3).\n   That is, we keep the first n−1 digits of γ, flip the n-th digit (0 ↔ 2) and afterwards write only 0’s.  Because the resulting base-3 expansion terminates after the n-th place, σ_n(γ) lies in D.  Distinct indices give different real numbers, so\n        S_γ := {σ_n(γ) : n = 1,2,3,…}\n   is an infinite subset of D.\n\n2.  Bounding |S_γ ∩ S_δ| when γ ≠ δ\n   -------------------------------------------------\n   Fix distinct γ, δ ∈ C and let\n        m = min{ j ≥ 1 : γ_j ≠ δ_j }\n   be the first position at which their ternary digits differ.  Without loss of generality assume γ_m = 0 and δ_m = 2.\n\n   Claim.  If σ_n(γ) = σ_k(δ) then at least one of n or k does not exceed m.\n\n   Proof of the claim.\n   Look at the m-th ternary digit of σ_n(γ) and σ_k(δ).\n        •  If n > m, the digit in position m of σ_n(γ) equals γ_m = 0 (because the flip occurs strictly later).\n        •  If k > m, the digit in position m of σ_k(δ) equals δ_m = 2.\n   These two digits differ, so equality is impossible when simultaneously n > m and k > m.  Hence for σ_n(γ)=σ_k(δ) we must have n ≤ m or k ≤ m, proving the claim.\n\n   Consequences of the claim.\n     –  Pairs with max{n,k} ≤ m.  There are only finitely many such pairs (at most m²), so they can contribute at most m elements to the intersection (in fact at most m, because for fixed n the terminating ternary representation is unique).\n\n     –  Pairs with n > m, k = m.  Equality σ_n(γ)=σ_m(δ) forces\n           γ_{m+1}= … = γ_{n−1} = 0   and   2−γ_n = 0  ⇔ γ_n = 2.\n       Because the first index after m at which γ has a 2 is unique, there is **at most one** value of n producing such an equality.\n\n     –  Symmetrically, there is at most one pair with n = m, k > m that yields equality.\n\n   Combining the three possibilities we obtain\n        |S_γ ∩ S_δ| ≤ m  + 1 + 1  =  m + 2  < ∞.\n\n3.  Injectivity of γ ↦ S_γ\n   -------------------------------------------------\n   The smallest index among the elements of S_γ Δ S_δ reveals the first place where γ and δ differ, so γ ≠ δ implies S_γ ≠ S_δ.  Hence the map γ ↦ S_γ is injective.\n\n4.  Conclusion\n   -------------------------------------------------\n   The family 𝔽 = {S_γ : γ ∈ C} is uncountable (because C is), each S_γ is a non-empty subset of the countable set D, and any two distinct members of 𝔽 intersect in only finitely many points.  This completes part (a).\n\n   For part (b) let A be any countably infinite set (for example A = D) and fix a bijection ϕ : D → A.  Put\n        T_γ := ϕ(S_γ)   (γ ∈ C).\n   Then {T_γ : γ ∈ C} is an uncountable collection of non-empty subsets of A whose pairwise intersections are finite, proving the assertion.",
      "_meta": {
        "core_steps": [
          "Pick a countable dense set D ⊂ ℝ (e.g. ℚ).",
          "For every real α choose an infinite sequence of distinct points of D converging to α and set S_α to be its range.",
          "If α ≠ β then any element common to both sequences must eventually be arbitrarily close to both α and β; this is impossible in a Hausdorff space, so S_α ∩ S_β is finite.",
          "Thus {S_α : α ∈ ℝ} is an uncountable family of non-empty subsets of D with pairwise finite intersections."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Underlying countably infinite dense set in ℝ",
            "original": "ℚ"
          },
          "slot2": {
            "description": "Uncountable index set of limit points",
            "original": "ℝ"
          },
          "slot3": {
            "description": "Concrete choice of approximating sequences (e.g. 1/n–close truncations, best rational approximants, etc.)",
            "original": "Sequence of distinct rationals converging to α"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}