{
  "index": "1961-A-5",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "5. Let \\( \\Omega \\) be a set of \\( n \\) points, where \\( n>2 \\). Let \\( \\Sigma \\) be a nonempty subcollection of the \\( 2^{n} \\) subsets of \\( \\Omega \\) that is closed with respect to unions, intersections, and complements (that is, if \\( A \\) and \\( B \\) are members of \\( \\Sigma \\), then so are \\( A \\cup B \\), \\( A \\cap B, \\Omega-A \\) and \\( \\Omega-B \\), where \\( \\Omega-B \\) denotes all points in \\( \\Omega \\) but not in \\( B) \\). If \\( k \\) is the number of members of \\( \\Sigma \\), what are the possible values of \\( k \\) ? Give a proof.",
  "solution": "First Solution. Since \\( \\Sigma \\) is not empty, say it contains \\( A \\). Then \\( \\Sigma \\) contains also \\( \\Omega-A \\) and \\( A \\cap(\\Omega-A)=\\emptyset \\). Hence also \\( \\Omega=\\Omega-\\emptyset \\in \\Sigma \\).\n\nAmong the non-empty members of \\( \\Sigma \\) certain are minimal in the sense that they do not contain any other member of \\( \\Sigma \\) except \\( \\emptyset \\). Any non-empty member \\( A \\) of \\( \\Sigma \\) contains a minimal element, for example, a set \\( B \\) of least cardinal satisfying \\( B \\subseteq A, B \\in \\Sigma, B \\neq \\emptyset \\).\n\nLet \\( B_{1}, B_{2}, \\ldots, B_{p} \\) be a proper enumeration of all the minimal elements of \\( \\Sigma \\). These elements are mutually disjoint. Suppose \\( i \\neq j \\), then \\( B_{i} \\cap B_{i} \\in \\Sigma \\) and \\( B_{i} \\cap B_{i} \\subseteq B_{i} \\). Hence by the minimality of \\( B_{i} \\) either \\( B_{i} \\cap B_{j}=\\emptyset \\) or \\( B_{i} \\cap B_{j}=B_{i} \\). The latter is impossible since it implies \\( B_{j} \\supseteq B_{i} \\), hence \\( B_{i} \\) \\( =B_{j} \\) or \\( B_{i}=\\emptyset \\), both contradictions.\n\nNow if \\( \\Omega-\\left(B_{1} \\cup B_{2} \\cup \\cdots \\cup B_{p}\\right) \\neq \\emptyset \\), then it would contain some minimal element of \\( \\Sigma \\), say \\( B_{i} \\), and we would have\n\\[\nB_{i}=B_{i} \\cap\\left[\\Omega-\\left(B_{1} \\cup B_{2} \\cup \\cdots \\cup B_{p}\\right)\\right]=\\emptyset\n\\]\na contradiction. So \\( B_{1} \\cup B_{2} \\cup \\cdots \\cup B_{p}=\\Omega \\).\nSuppose \\( C \\in \\Sigma \\). Then\n\\[\n\\begin{array}{c}\nC=C \\cap \\Omega=C \\cap\\left(B_{1} \\cup \\cdots \\cup B_{p}\\right) \\\\\n=\\left(C \\cap B_{1}\\right) \\cup\\left(C \\cap B_{2}\\right) \\cup \\cdots \\cup\\left(C \\cap B_{p}\\right) .\n\\end{array}\n\\]\n\nNow each of the sets \\( C \\cap B_{i} \\) is either \\( \\emptyset \\) or \\( B_{i} \\) (by the minimality of \\( B_{i} \\) ). Thus we have shown: Every element of \\( \\Sigma \\) is the union of some subcollection of the sets \\( \\left\\{B_{i}\\right\\} \\). Conversely every such union is a member of \\( \\Sigma \\). Moreover, since the sets \\( \\left\\{B_{i}\\right\\} \\) are non-empty and mutually disjoint, distinct subcollections of \\( \\left\\{B_{i}\\right\\} \\) have distinct unions. There are therefore exactly \\( 2^{p} \\) elements of \\( \\Sigma \\), one for each subset of \\( \\left\\{B_{1}, B_{2}, \\ldots, B_{p}\\right\\} \\). Thus \\( k=2^{p} \\).\n\nEvery power of 2 from 2 to \\( 2^{n} \\) is possible.\nIf \\( p \\) is an integer \\( 1 \\leq p \\leq n \\), choose any partition of \\( \\Omega \\) into \\( p \\) disjoint subsets \\( B_{1} \\ldots \\ldots B_{r} \\). Then the \\( 2^{p} \\) unions of the various subsets of \\( \\left\\{B_{1}, \\ldots, B_{p}\\right\\} \\) form a collection \\( \\Sigma \\) with the required properties containing \\( 2^{p} \\) members.\n\nSecond Solution. For the subsets of any set \\( \\Omega \\) define the operation (called the symmetric difference) by\n\\[\n\\begin{aligned}\nA \\oplus B & =(A \\cup B)-(A \\cap B) \\\\\n& =(A \\cup B) \\cap((\\Omega-A) \\cup(\\Omega-B)) .\n\\end{aligned}\n\\]\n\nWith this operation the set \\( P(\\Omega) \\) of all subsets of \\( \\Omega \\) becomes a group in which \\( \\emptyset \\) is the identity and every element is its own inverse. The second way of writing the definition of \\( \\oplus \\) shows that the collection \\( \\Sigma \\) of the problem is closed with respect to \\( \\oplus \\). Since every element is its own inverse, \\( \\Sigma \\) is a subgroup of \\( P(\\Omega) \\). Hence \\( \\Sigma \\) is a finite group with every element of order 2 , so it is a 2 -group and has size \\( 2^{\\prime} \\) for some \\( t \\). Thus \\( k=2^{\\prime} \\) for some \\( t \\), and here \\( t \\geq 1 \\) since \\( \\emptyset . \\Omega \\in \\Sigma \\) and \\( t \\leq n \\) since \\( |P(\\Omega)|=2^{\\prime \\prime} \\).\n\nWe cannot show by purely group theoretic methods that every power of two within this range can occur. because a subgroup of \\( P(\\Omega) \\) need not be closed with respect to unions and intersections. even those subgroups that contain \\( \\Omega \\). It is, of course, true that, if \\( \\left\\{B_{1}, B_{2}, \\ldots, B_{n}\\right\\} \\) is a partition of \\( \\Omega \\) into \\( p \\) disjoint sets. then the subgroup of \\( P(\\Omega) \\) generated by the \\( B \\) 's is closed with respect to unions. intersections. and complements and has size \\( 2^{\\prime \\prime} \\). but this requires calculations similar to those in the first solution.\n\nRemark. A collection of sets closed with respect to unions, intersections. and complements is called a Boolean algebra of sets.",
  "vars": [
    "A",
    "B",
    "C",
    "p",
    "t",
    "\\\\Sigma",
    "\\\\Omega",
    "B_i",
    "B_j",
    "B_1",
    "B_2",
    "B_p",
    "B_r",
    "P"
  ],
  "params": [
    "n",
    "k"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "subseta",
        "B": "subsetb",
        "C": "subsetc",
        "p": "minicount",
        "t": "twopower",
        "\\Sigma": "setfamily",
        "\\Omega": "universe",
        "B_i": "blockvar",
        "B_j": "blockvarj",
        "B_1": "blockone",
        "B_2": "blocktwo",
        "B_p": "blockp",
        "B_r": "blockr",
        "P": "powerset",
        "n": "pointcount",
        "k": "familysize"
      },
      "question": "5. Let \\( universe \\) be a set of \\( pointcount \\) points, where \\( pointcount>2 \\). Let \\( setfamily \\) be a nonempty subcollection of the \\( 2^{pointcount} \\) subsets of \\( universe \\) that is closed with respect to unions, intersections, and complements (that is, if \\( subseta \\) and \\( subsetb \\) are members of \\( setfamily \\), then so are \\( subseta \\cup subsetb \\), \\( subseta \\cap subsetb, universe-subseta \\) and \\( universe-subsetb \\), where \\( universe-subsetb \\) denotes all points in \\( universe \\) but not in \\( subsetb) \\). If \\( familysize \\) is the number of members of \\( setfamily \\), what are the possible values of \\( familysize \\) ? Give a proof.",
      "solution": "First Solution. Since \\( setfamily \\) is not empty, say it contains \\( subseta \\). Then \\( setfamily \\) contains also \\( universe-subseta \\) and \\( subseta \\cap(universe-subseta)=\\emptyset \\). Hence also \\( universe=universe-\\emptyset \\in setfamily \\).\n\nAmong the non-empty members of \\( setfamily \\) certain are minimal in the sense that they do not contain any other member of \\( setfamily \\) except \\( \\emptyset \\). Any non-empty member \\( subseta \\) of \\( setfamily \\) contains a minimal element, for example, a set \\( subsetb \\) of least cardinal satisfying \\( subsetb \\subseteq subseta, subsetb \\in setfamily, subsetb \\neq \\emptyset \\).\n\nLet \\( blockone, blocktwo, \\ldots, blockp \\) be a proper enumeration of all the minimal elements of \\( setfamily \\). These elements are mutually disjoint. Suppose \\( i \\neq j \\); then \\( blockvar \\cap blockvar \\in setfamily \\) and \\( blockvar \\cap blockvar \\subseteq blockvar \\). Hence by the minimality of \\( blockvar \\) either \\( blockvar \\cap blockvarj=\\emptyset \\) or \\( blockvar \\cap blockvarj=blockvar \\). The latter is impossible since it implies \\( blockvarj \\supseteq blockvar \\), hence \\( blockvar =blockvarj \\) or \\( blockvar=\\emptyset \\), both contradictions.\n\nNow if \\( universe-\\left(blockone \\cup blocktwo \\cup \\cdots \\cup blockp\\right) \\neq \\emptyset \\), then it would contain some minimal element of \\( setfamily \\), say \\( blockvar \\), and we would have\n\\[\nblockvar=blockvar \\cap\\left[universe-\\left(blockone \\cup blocktwo \\cup \\cdots \\cup blockp\\right)\\right]=\\emptyset ,\n\\]\na contradiction. So \\( blockone \\cup blocktwo \\cup \\cdots \\cup blockp=universe \\).\n\nSuppose \\( subsetc \\in setfamily \\). Then\n\\[\n\\begin{array}{c}\nsubsetc=subsetc \\cap universe=subsetc \\cap\\left(blockone \\cup \\cdots \\cup blockp\\right) \\\\\n=\\left(subsetc \\cap blockone\\right) \\cup\\left(subsetc \\cap blocktwo\\right) \\cup \\cdots \\cup\\left(subsetc \\cap blockp\\right) .\n\\end{array}\n\\]\n\nNow each of the sets \\( subsetc \\cap blockvar \\) is either \\( \\emptyset \\) or \\( blockvar \\) (by the minimality of \\( blockvar \\) ). Thus we have shown: every element of \\( setfamily \\) is the union of some subcollection of the sets \\( \\{blockvar\\} \\). Conversely every such union is a member of \\( setfamily \\). Moreover, since the sets \\( \\{blockvar\\} \\) are non-empty and mutually disjoint, distinct subcollections of \\( \\{blockvar\\} \\) have distinct unions. There are therefore exactly \\( 2^{minicount} \\) elements of \\( setfamily \\), one for each subset of \\( \\{blockone, blocktwo, \\ldots, blockp\\} \\). Thus \\( familysize=2^{minicount} \\).\n\nEvery power of 2 from 2 to \\( 2^{pointcount} \\) is possible. If \\( minicount \\) is an integer \\( 1 \\leq minicount \\leq pointcount \\), choose any partition of \\( universe \\) into \\( minicount \\) disjoint subsets \\( blockone \\ldots \\ldots blockr \\). Then the \\( 2^{minicount} \\) unions of the various subsets of \\( \\{blockone, \\ldots, blockp\\} \\) form a collection \\( setfamily \\) with the required properties containing \\( 2^{minicount} \\) members.\n\nSecond Solution. For the subsets of any set \\( universe \\) define the operation (called the symmetric difference) by\n\\[\n\\begin{aligned}\nsubseta \\oplus subsetb &= (subseta \\cup subsetb)-(subseta \\cap subsetb) \\\\\n                       &= (subseta \\cup subsetb) \\cap\\bigl((universe-subseta) \\cup (universe-subsetb)\\bigr).\n\\end{aligned}\n\\]\n\nWith this operation the set \\( powerset(universe) \\) of all subsets of \\( universe \\) becomes a group in which \\( \\emptyset \\) is the identity and every element is its own inverse. The second way of writing the definition of \\( \\oplus \\) shows that the collection \\( setfamily \\) of the problem is closed with respect to \\( \\oplus \\). Since every element is its own inverse, \\( setfamily \\) is a subgroup of \\( powerset(universe) \\). Hence \\( setfamily \\) is a finite group with every element of order 2, so it is a 2-group and has size \\( 2^{twopower} \\) for some \\( twopower \\). Thus \\( familysize=2^{twopower} \\) for some \\( twopower \\), and here \\( twopower \\ge 1 \\) since \\( \\emptyset , universe \\in setfamily \\) and \\( twopower \\le pointcount \\) since \\( |powerset(universe)|=2^{pointcount} \\).\n\nWe cannot show by purely group-theoretic methods that every power of two within this range can occur, because a subgroup of \\( powerset(universe) \\) need not be closed with respect to unions and intersections, even those subgroups that contain \\( universe \\). It is, of course, true that, if \\( \\{blockone, blocktwo, \\ldots, blockvar\\} \\) is a partition of \\( universe \\) into \\( minicount \\) disjoint sets, then the subgroup of \\( powerset(universe) \\) generated by the \\( subsetb \\)'s is closed with respect to unions, intersections, and complements and has size \\( 2^{minicount} \\), but this requires calculations similar to those in the first solution.\n\nRemark. A collection of sets closed with respect to unions, intersections, and complements is called a Boolean algebra of sets."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "sunflower",
        "B": "marigold",
        "C": "pinecone",
        "p": "tortoise",
        "t": "gemstone",
        "\\Sigma": "\\rainstorm",
        "\\Omega": "\\landscape",
        "B_i": "marigoldsubi",
        "B_j": "marigoldsubj",
        "B_1": "marigoldsubone",
        "B_2": "marigoldsubtwo",
        "B_p": "marigoldsubtortoise",
        "B_r": "marigoldsubr",
        "P": "notebook",
        "n": "avalanche",
        "k": "labyrinth"
      },
      "question": "5. Let \\( \\landscape \\) be a set of \\( avalanche \\) points, where \\( avalanche>2 \\). Let \\( \\rainstorm \\) be a nonempty subcollection of the \\( 2^{avalanche} \\) subsets of \\( \\landscape \\) that is closed with respect to unions, intersections, and complements (that is, if \\( sunflower \\) and \\( marigold \\) are members of \\( \\rainstorm \\), then so are \\( sunflower \\cup marigold \\), \\( sunflower \\cap marigold, \\landscape-sunflower \\) and \\( \\landscape-marigold \\), where \\( \\landscape-marigold \\) denotes all points in \\( \\landscape \\) but not in \\( marigold) \\). If \\( labyrinth \\) is the number of members of \\( \\rainstorm \\), what are the possible values of \\( labyrinth \\) ? Give a proof.",
      "solution": "First Solution. Since \\( \\rainstorm \\) is not empty, say it contains \\( sunflower \\). Then \\( \\rainstorm \\) contains also \\( \\landscape-sunflower \\) and \\( sunflower \\cap(\\landscape-sunflower)=\\emptyset \\). Hence also \\( \\landscape=\\landscape-\\emptyset \\in \\rainstorm \\).\n\nAmong the non-empty members of \\( \\rainstorm \\) certain are minimal in the sense that they do not contain any other member of \\( \\rainstorm \\) except \\( \\emptyset \\). Any non-empty member \\( sunflower \\) of \\( \\rainstorm \\) contains a minimal element, for example, a set \\( marigold \\) of least cardinal satisfying \\( marigold \\subseteq sunflower, marigold \\in \\rainstorm, marigold \\neq \\emptyset \\).\n\nLet \\( marigoldsubone, marigoldsubtwo, \\ldots, marigoldsubtortoise \\) be a proper enumeration of all the minimal elements of \\( \\rainstorm \\). These elements are mutually disjoint. Suppose \\( i \\neq j \\), then \\( marigoldsubi \\cap marigoldsubi \\in \\rainstorm \\) and \\( marigoldsubi \\cap marigoldsubi \\subseteq marigoldsubi \\). Hence by the minimality of \\( marigoldsubi \\) either \\( marigoldsubi \\cap marigoldsubj=\\emptyset \\) or \\( marigoldsubi \\cap marigoldsubj=marigoldsubi \\). The latter is impossible since it implies \\( marigoldsubj \\supseteq marigoldsubi \\), hence \\( marigoldsubi \\) \\( =marigoldsubj \\) or \\( marigoldsubi=\\emptyset \\), both contradictions.\n\nNow if \\( \\landscape-\\left(marigoldsubone \\cup marigoldsubtwo \\cup \\cdots \\cup marigoldsubtortoise\\right) \\neq \\emptyset \\), then it would contain some minimal element of \\( \\rainstorm \\), say \\( marigoldsubi \\), and we would have\n\\[\nmarigoldsubi=marigoldsubi \\cap\\left[\\landscape-\\left(marigoldsubone \\cup marigoldsubtwo \\cup \\cdots \\cup marigoldsubtortoise\\right)\\right]=\\emptyset\n\\]\na contradiction. So \\( marigoldsubone \\cup marigoldsubtwo \\cup \\cdots \\cup marigoldsubtortoise=\\landscape \\).\nSuppose \\( pinecone \\in \\rainstorm \\). Then\n\\[\n\\begin{array}{c}\npinecone=pinecone \\cap \\landscape=pinecone \\cap\\left(marigoldsubone \\cup \\cdots \\cup marigoldsubtortoise\\right) \\\\\n=\\left(pinecone \\cap marigoldsubone\\right) \\cup\\left(pinecone \\cap marigoldsubtwo\\right) \\cup \\cdots \\cup\\left(pinecone \\cap marigoldsubtortoise\\right) .\n\\end{array}\n\\]\n\nNow each of the sets \\( pinecone \\cap marigoldsubi \\) is either \\( \\emptyset \\) or \\( marigoldsubi \\) (by the minimality of \\( marigoldsubi \\) ). Thus we have shown: Every element of \\( \\rainstorm \\) is the union of some subcollection of the sets \\( \\left\\{marigoldsubi\\right\\} \\). Conversely every such union is a member of \\( \\rainstorm \\). Moreover, since the sets \\( \\left\\{marigoldsubi\\right\\} \\) are non-empty and mutually disjoint, distinct subcollections of \\( \\left\\{marigoldsubi\\right\\} \\) have distinct unions. There are therefore exactly \\( 2^{tortoise} \\) elements of \\( \\rainstorm \\), one for each subset of \\( \\left\\{marigoldsubone, marigoldsubtwo, \\ldots, marigoldsubtortoise\\right\\} \\). Thus \\( labyrinth=2^{tortoise} \\).\n\nEvery power of 2 from 2 to \\( 2^{avalanche} \\) is possible.\nIf \\( tortoise \\) is an integer \\( 1 \\leq tortoise \\leq avalanche \\), choose any partition of \\( \\landscape \\) into \\( tortoise \\) disjoint subsets \\( marigoldsubone \\ldots \\ldots marigoldsubr \\). Then the \\( 2^{tortoise} \\) unions of the various subsets of \\( \\left\\{marigoldsubone, \\ldots, marigoldsubtortoise\\right\\} \\) form a collection \\( \\rainstorm \\) with the required properties containing \\( 2^{tortoise} \\) members.\n\nSecond Solution. For the subsets of any set \\( \\landscape \\) define the operation (called the symmetric difference) by\n\\[\n\\begin{aligned}\nsunflower \\oplus marigold & =(sunflower \\cup marigold)-(sunflower \\cap marigold) \\\\\n& =(sunflower \\cup marigold) \\cap((\\landscape-sunflower) \\cup(\\landscape-marigold)) .\n\\end{aligned}\n\\]\n\nWith this operation the set \\( notebook(\\landscape) \\) of all subsets of \\( \\landscape \\) becomes a group in which \\( \\emptyset \\) is the identity and every element is its own inverse. The second way of writing the definition of \\( \\oplus \\) shows that the collection \\( \\rainstorm \\) of the problem is closed with respect to \\( \\oplus \\). Since every element is its own inverse, \\( \\rainstorm \\) is a subgroup of \\( notebook(\\landscape) \\). Hence \\( \\rainstorm \\) is a finite group with every element of order 2 , so it is a 2 -group and has size \\( 2^{gemstone} \\) for some \\( gemstone \\). Thus \\( labyrinth=2^{gemstone} \\) for some \\( gemstone \\), and here \\( gemstone \\geq 1 \\) since \\( \\emptyset . \\landscape \\in \\rainstorm \\) and \\( gemstone \\leq avalanche \\) since \\( |notebook(\\landscape)|=2^{avalanche} \\).\n\nWe cannot show by purely group theoretic methods that every power of two within this range can occur, because a subgroup of \\( notebook(\\landscape) \\) need not be closed with respect to unions and intersections, even those subgroups that contain \\( \\landscape \\). It is, of course, true that, if \\( \\left\\{marigoldsubone, marigoldsubtwo, \\ldots, marigold_{avalanche}\\right\\} \\) is a partition of \\( \\landscape \\) into \\( tortoise \\) disjoint sets, then the subgroup of \\( notebook(\\landscape) \\) generated by the \\( marigold \\) 's is closed with respect to unions, intersections, and complements and has size \\( 2^{avalanche} \\), but this requires calculations similar to those in the first solution.\n\nRemark. A collection of sets closed with respect to unions, intersections, and complements is called a Boolean algebra of sets."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "wholeset",
        "B": "voidness",
        "C": "exterior",
        "p": "maxcount",
        "t": "smallnum",
        "\\Sigma": "chaossets",
        "\\Omega": "emptiverse",
        "B_i": "voidsubset",
        "B_j": "voidsubjt",
        "B_1": "voidsubone",
        "B_2": "voidsubtwo",
        "B_p": "voidsubtop",
        "B_r": "voidsubrun",
        "P": "singleset",
        "n": "fewcount",
        "k": "manyness"
      },
      "question": "5. Let \\( emptiverse \\) be a set of \\( fewcount \\) points, where \\( fewcount>2 \\). Let \\( chaossets \\) be a nonempty subcollection of the \\( 2^{fewcount} \\) subsets of \\( emptiverse \\) that is closed with respect to unions, intersections, and complements (that is, if \\( wholeset \\) and \\( voidness \\) are members of \\( chaossets \\), then so are \\( wholeset \\cup voidness \\), \\( wholeset \\cap voidness, emptiverse-wholeset \\) and \\( emptiverse-voidness \\), where \\( emptiverse-voidness \\) denotes all points in \\( emptiverse \\) but not in \\( voidness) \\). If \\( manyness \\) is the number of members of \\( chaossets \\), what are the possible values of \\( manyness \\) ? Give a proof.",
      "solution": "First Solution. Since \\( chaossets \\) is not empty, say it contains \\( wholeset \\). Then \\( chaossets \\) contains also \\( emptiverse-wholeset \\) and \\( wholeset \\cap(emptiverse-wholeset)=\\emptyset \\). Hence also \\( emptiverse=emptiverse-\\emptyset \\in chaossets \\).\n\nAmong the non-empty members of \\( chaossets \\) certain are minimal in the sense that they do not contain any other member of \\( chaossets \\) except \\( \\emptyset \\). Any non-empty member \\( wholeset \\) of \\( chaossets \\) contains a minimal element, for example, a set \\( voidness \\) of least cardinal satisfying \\( voidness \\subseteq wholeset, voidness \\in chaossets, voidness \\neq \\emptyset \\).\n\nLet \\( voidsubone, voidsubtwo, \\ldots, voidsubtop \\) be a proper enumeration of all the minimal elements of \\( chaossets \\). These elements are mutually disjoint. Suppose \\( i \\neq j \\), then \\( voidsubset \\cap voidsubset \\in chaossets \\) and \\( voidsubset \\cap voidsubset \\subseteq voidsubset \\). Hence by the minimality of \\( voidsubset \\) either \\( voidsubset \\cap voidsubjt=\\emptyset \\) or \\( voidsubset \\cap voidsubjt=voidsubset \\). The latter is impossible since it implies \\( voidsubjt \\supseteq voidsubset \\), hence \\( voidsubset =voidsubjt \\) or \\( voidsubset=\\emptyset \\), both contradictions.\n\nNow if \\( emptiverse-\\left(voidsubone \\cup voidsubtwo \\cup \\cdots \\cup voidsubtop\\right) \\neq \\emptyset \\), then it would contain some minimal element of \\( chaossets \\), say \\( voidsubset \\), and we would have\n\\[\nvoidsubset=voidsubset \\cap\\left[emptiverse-\\left(voidsubone \\cup voidsubtwo \\cup \\cdots \\cup voidsubtop\\right)\\right]=\\emptyset\n\\]\na contradiction. So \\( voidsubone \\cup voidsubtwo \\cup \\cdots \\cup voidsubtop=emptiverse \\).\nSuppose \\( exterior \\in chaossets \\). Then\n\\[\n\\begin{array}{c}\nexterior=exterior \\cap emptiverse=exterior \\cap\\left(voidsubone \\cup \\cdots \\cup voidsubtop\\right) \\\\\n=\\left(exterior \\cap voidsubone\\right) \\cup\\left(exterior \\cap voidsubtwo\\right) \\cup \\cdots \\cup\\left(exterior \\cap voidsubtop\\right) .\n\\end{array}\n\\]\n\nNow each of the sets \\( exterior \\cap voidsubset \\) is either \\( \\emptyset \\) or \\( voidsubset \\) (by the minimality of \\( voidsubset \\) ). Thus we have shown: Every element of \\( chaossets \\) is the union of some subcollection of the sets \\( \\{voidsubset\\} \\). Conversely every such union is a member of \\( chaossets \\). Moreover, since the sets \\( \\{voidsubset\\} \\) are non-empty and mutually disjoint, distinct subcollections of \\( \\{voidsubset, \\ldots, voidsubset\\} \\) have distinct unions. There are therefore exactly \\( 2^{maxcount} \\) elements of \\( chaossets \\), one for each subset of \\( \\{voidsubone, voidsubtwo, \\ldots, voidsubtop\\} \\). Thus \\( manyness=2^{maxcount} \\).\n\nEvery power of 2 from 2 to \\( 2^{fewcount} \\) is possible.\nIf \\( maxcount \\) is an integer \\( 1 \\leq maxcount \\leq fewcount \\), choose any partition of \\( emptiverse \\) into \\( maxcount \\) disjoint subsets \\( voidsubone \\ldots \\ldots voidsubrun \\). Then the \\( 2^{maxcount} \\) unions of the various subsets of \\( \\{voidsubone, \\ldots, voidsubtop\\} \\) form a collection \\( chaossets \\) with the required properties containing \\( 2^{maxcount} \\) members.\n\nSecond Solution. For the subsets of any set \\( emptiverse \\) define the operation (called the symmetric difference) by\n\\[\n\\begin{aligned}\nwholeset \\oplus voidness & =(wholeset \\cup voidness)-(wholeset \\cap voidness) \\\\\n& =(wholeset \\cup voidness) \\cap((emptiverse-wholeset) \\cup(emptiverse-voidness)) .\n\\end{aligned}\n\\]\n\nWith this operation the set \\( singleset(emptiverse) \\) of all subsets of \\( emptiverse \\) becomes a group in which \\( \\emptyset \\) is the identity and every element is its own inverse. The second way of writing the definition of \\( \\oplus \\) shows that the collection \\( chaossets \\) of the problem is closed with respect to \\( \\oplus \\). Since every element is its own inverse, \\( chaossets \\) is a subgroup of \\( singleset(emptiverse) \\). Hence \\( chaossets \\) is a finite group with every element of order 2 , so it is a 2 -group and has size \\( 2^{\\prime} \\) for some \\( smallnum \\). Thus \\( manyness=2^{\\prime} \\) for some \\( smallnum \\), and here \\( smallnum \\geq 1 \\) since \\( \\emptyset . emptiverse \\in chaossets \\) and \\( smallnum \\leq fewcount \\) since \\( |singleset(emptiverse)|=2^{\\prime \\prime} \\).\n\nWe cannot show by purely group theoretic methods that every power of two within this range can occur. because a subgroup of \\( singleset(emptiverse) \\) need not be closed with respect to unions and intersections. even those subgroups that contain \\( emptiverse \\). It is, of course, true that, if \\( \\{voidsubone, voidsubtwo, \\ldots, voidsubtop\\} \\) is a partition of \\( emptiverse \\) into \\( maxcount \\) disjoint sets. then the subgroup of \\( singleset(emptiverse) \\) generated by the \\( voidness \\) 's is closed with respect to unions. intersections. and complements and has size \\( 2^{\\prime \\prime} \\). but this requires calculations similar to those in the first solution.\n\nRemark. A collection of sets closed with respect to unions, intersections. and complements is called a Boolean algebra of sets."
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "C": "mftcazoi",
        "p": "dlqrenuv",
        "t": "sbnleowc",
        "\\Sigma": "lksruqpo",
        "\\Omega": "vzbctnag",
        "B_i": "kzmpdoni",
        "B_j": "psrlhdmq",
        "B_1": "zhtnaoer",
        "B_2": "vdqslmuc",
        "B_p": "wgrnecaq",
        "B_r": "nyfokusl",
        "P": "xyrpqglm",
        "n": "ajsdoflv",
        "k": "bxneulvo"
      },
      "question": "5. Let \\( vzbctnag \\) be a set of \\( ajsdoflv \\) points, where \\( ajsdoflv>2 \\). Let \\( lksruqpo \\) be a nonempty subcollection of the \\( 2^{ajsdoflv} \\) subsets of \\( vzbctnag \\) that is closed with respect to unions, intersections, and complements (that is, if \\( qzxwvtnp \\) and \\( hjgrksla \\) are members of \\( lksruqpo \\), then so are \\( qzxwvtnp \\cup hjgrksla \\), \\( qzxwvtnp \\cap hjgrksla, vzbctnag-qzxwvtnp \\) and \\( vzbctnag-hjgrksla \\), where \\( vzbctnag-hjgrksla \\) denotes all points in \\( vzbctnag \\) but not in \\( hjgrksla) \\). If \\( bxneulvo \\) is the number of members of \\( lksruqpo \\), what are the possible values of \\( bxneulvo \\) ? Give a proof.",
      "solution": "First Solution. Since \\( lksruqpo \\) is not empty, say it contains \\( qzxwvtnp \\). Then \\( lksruqpo \\) contains also \\( vzbctnag-qzxwvtnp \\) and \\( qzxwvtnp \\cap(vzbctnag-qzxwvtnp)=\\emptyset \\). Hence also \\( vzbctnag=vzbctnag-\\emptyset \\in lksruqpo \\).\n\nAmong the non-empty members of \\( lksruqpo \\) certain are minimal in the sense that they do not contain any other member of \\( lksruqpo \\) except \\( \\emptyset \\). Any non-empty member \\( qzxwvtnp \\) of \\( lksruqpo \\) contains a minimal element, for example, a set \\( hjgrksla \\) of least cardinal satisfying \\( hjgrksla \\subseteq qzxwvtnp, hjgrksla \\in lksruqpo, hjgrksla \\neq \\emptyset \\).\n\nLet \\( zhtnaoer, vdqslmuc, \\ldots, wgrnecaq \\) be a proper enumeration of all the minimal elements of \\( lksruqpo \\). These elements are mutually disjoint. Suppose \\( i \\neq j \\), then \\( kzmpdoni \\cap kzmpdoni \\in lksruqpo \\) and \\( kzmpdoni \\cap kzmpdoni \\subseteq kzmpdoni \\). Hence by the minimality of \\( kzmpdoni \\) either \\( kzmpdoni \\cap psrlhdmq=\\emptyset \\) or \\( kzmpdoni \\cap psrlhdmq=kzmpdoni \\). The latter is impossible since it implies \\( psrlhdmq \\supseteq kzmpdoni \\), hence \\( kzmpdoni =psrlhdmq \\) or \\( kzmpdoni =\\emptyset \\), both contradictions.\n\nNow if \\( vzbctnag-\\left(zhtnaoer \\cup vdqslmuc \\cup \\cdots \\cup wgrnecaq\\right) \\neq \\emptyset \\), then it would contain some minimal element of \\( lksruqpo \\), say \\( kzmpdoni \\), and we would have\n\\[\nkzmpdoni =kzmpdoni \\cap\\left[vzbctnag-\\left(zhtnaoer \\cup vdqslmuc \\cup \\cdots \\cup wgrnecaq\\right)\\right]=\\emptyset\n\\]\na contradiction. So \\( zhtnaoer \\cup vdqslmuc \\cup \\cdots \\cup wgrnecaq=vzbctnag \\).\nSuppose \\( mftcazoi \\in lksruqpo \\). Then\n\\[\n\\begin{array}{c}\nmftcazoi =mftcazoi \\cap vzbctnag=mftcazoi \\cap\\left(zhtnaoer \\cup \\cdots \\cup wgrnecaq\\right) \\\\\n=\\left(mftcazoi \\cap zhtnaoer\\right) \\cup\\left(mftcazoi \\cap vdqslmuc\\right) \\cup \\cdots \\cup\\left(mftcazoi \\cap wgrnecaq\\right) .\n\\end{array}\n\\]\n\nNow each of the sets \\( mftcazoi \\cap kzmpdoni \\) is either \\( \\emptyset \\) or \\( kzmpdoni \\) (by the minimality of \\( kzmpdoni \\) ). Thus we have shown: Every element of \\( lksruqpo \\) is the union of some subcollection of the sets \\( \\{kzmpdoni\\} \\). Conversely every such union is a member of \\( lksruqpo \\). Moreover, since the sets \\( \\{kzmpdoni\\} \\) are non-empty and mutually disjoint, distinct subcollections of \\( \\{kzmpdoni\\} \\) have distinct unions. There are therefore exactly \\( 2^{dlqrenuv} \\) elements of \\( lksruqpo \\), one for each subset of \\( \\{zhtnaoer, vdqslmuc, \\ldots, wgrnecaq\\} \\). Thus \\( bxneulvo=2^{dlqrenuv} \\).\n\nEvery power of 2 from 2 to \\( 2^{ajsdoflv} \\) is possible.\nIf \\( dlqrenuv \\) is an integer \\( 1 \\leq dlqrenuv \\leq ajsdoflv \\), choose any partition of \\( vzbctnag \\) into \\( dlqrenuv \\) disjoint subsets \\( zhtnaoer \\ldots \\ldots nyfokusl \\). Then the \\( 2^{dlqrenuv} \\) unions of the various subsets of \\( \\{zhtnaoer, \\ldots, wgrnecaq\\} \\) form a collection \\( lksruqpo \\) with the required properties containing \\( 2^{dlqrenuv} \\) members.\n\nSecond Solution. For the subsets of any set \\( vzbctnag \\) define the operation (called the symmetric difference) by\n\\[\n\\begin{aligned}\nqzxwvtnp \\oplus hjgrksla & =(qzxwvtnp \\cup hjgrksla)-(qzxwvtnp \\cap hjgrksla) \\\\\n& =(qzxwvtnp \\cup hjgrksla) \\cap((vzbctnag-qzxwvtnp) \\cup(vzbctnag-hjgrksla)) .\n\\end{aligned}\n\\]\n\nWith this operation the set \\( xyrpqglm(vzbctnag) \\) of all subsets of \\( vzbctnag \\) becomes a group in which \\( \\emptyset \\) is the identity and every element is its own inverse. The second way of writing the definition of \\( \\oplus \\) shows that the collection \\( lksruqpo \\) of the problem is closed with respect to \\( \\oplus \\). Since every element is its own inverse, \\( lksruqpo \\) is a subgroup of \\( xyrpqglm(vzbctnag) \\). Hence \\( lksruqpo \\) is a finite group with every element of order 2, so it is a 2-group and has size \\( 2^{sbnleowc} \\) for some \\( sbnleowc \\). Thus \\( bxneulvo=2^{sbnleowc} \\) for some \\( sbnleowc \\), and here \\( sbnleowc \\geq 1 \\) since \\( \\emptyset , vzbctnag \\in lksruqpo \\) and \\( sbnleowc \\leq ajsdoflv \\) since \\( |xyrpqglm(vzbctnag)|=2^{ajsdoflv} \\).\n\nWe cannot show by purely group theoretic methods that every power of two within this range can occur, because a subgroup of \\( xyrpqglm(vzbctnag) \\) need not be closed with respect to unions and intersections, even those subgroups that contain \\( vzbctnag \\). It is, of course, true that, if \\( \\{zhtnaoer, vdqslmuc, \\ldots, wgrnecaq\\} \\) is a partition of \\( vzbctnag \\) into \\( dlqrenuv \\) disjoint sets, then the subgroup of \\( xyrpqglm(vzbctnag) \\) generated by the \\( hjgrksla \\)'s is closed with respect to unions, intersections, and complements and has size \\( 2^{ajsdoflv} \\), but this requires calculations similar to those in the first solution.\n\nRemark. A collection of sets closed with respect to unions, intersections, and complements is called a Boolean algebra of sets."
    },
    "kernel_variant": {
      "question": "Let $\\Omega$ be a finite set with $\\lvert\\Omega\\rvert=n\\ge 3$ and let $G\\le\\operatorname{Sym}(\\Omega)$ act transitively on $\\Omega$.  \nA non–empty family $\\Sigma\\subseteq\\mathcal P(\\Omega)$ is called a \\emph{$G$–Boolean algebra} if  \n\n1. (Boolean algebra) $\\Sigma$ is closed under finite unions, finite intersections and complements;  \n2. ($G$–invariance) $g\\!\\cdot\\!A\\in\\Sigma$ for every $A\\in\\Sigma$ and every $g\\in G$.\n\nFor $\\omega\\in\\Omega$ write  \n\\[\nG_{\\omega}=\\{g\\in G:\\,g\\omega=\\omega\\},\\qquad k:=\\lvert\\Sigma\\rvert .\n\\]\n\n(a)  Show that there exists a subgroup $H$ with $G_{\\omega}\\le H\\le G$ and an atom $A_{0}$ of $\\Sigma$ such that  \n\\[\nH=\\{g\\in G:\\,g\\!\\cdot\\!A_{0}=A_{0}\\},\n\\qquad\n\\mathcal A:=\\{g\\!\\cdot\\!A_{0}:g\\in G\\}\n\\]\nis the set of atoms of $\\Sigma$, so that $gH\\longmapsto g\\!\\cdot\\!A_{0}$ induces a bijection $G/H\\to\\mathcal A$.  Deduce that  \n\\[\nk=\\lvert\\Sigma\\rvert=2^{[G:H]}.\n\\]\n\n(b)  Conversely, fix $\\omega\\in\\Omega$ and let $H$ be any subgroup with $G_{\\omega}\\le H\\le G$.  \nFor $gH\\in G/H$ put  \n\\[\nA_{gH}:=g\\,H\\!\\cdot\\!\\omega=\\{g h\\!\\cdot\\!\\omega:\\,h\\in H\\}\\subseteq\\Omega.\\tag{$\\star$}\n\\]\n\n(i)  Prove that the sets $A_{gH}$ are pairwise disjoint and cover $\\Omega$.  \n\n(ii)  Let $\\mathcal A_H:=\\{A_{gH}:gH\\in G/H\\}$.  Show that  \n\\[\n\\Sigma_H:=\\Bigl\\{\\bigcup\\mathcal C:\\,\\mathcal C\\subseteq\\mathcal A_H\\Bigr\\}\n\\]\nis a $G$–Boolean algebra whose atoms are exactly the members of $\\mathcal A_H$.  \nIn particular $\\lvert\\Sigma_H\\rvert=2^{[G:H]}$.\n\n(c)  Specialise to $G=\\operatorname{Sym}(\\Omega)$ with $n\\ge 3$.  \n\n(i) Prove that $G_{\\omega}\\cong\\operatorname{Sym}(n-1)$ is a maximal subgroup of $\\operatorname{Sym}(n)$.  \n\n(ii) Deduce that the only possibilities for $H$ in parts (a),(b) are  \n\\[\nH=G_{\\omega}\\quad\\text{or}\\quad H=G,\n\\]\nso that $[G:H]$ equals $n$ or $1$ and therefore  \n\\[\nk\\in\\{\\,2,\\,2^{\\,n}\\}.\n\\]\nDescribe explicitly $\\Sigma$ in both cases.\n\n(d)  Specialise to the regular cyclic group $G=C_{n}=\\langle\\rho\\rangle$ acting on $\\Omega=\\{0,1,\\dots ,n-1\\}$ by $\\rho\\!\\cdot\\!i=i+1\\pmod n$.  Determine  \n\\[\n\\mathfrak M(C_{n})=\\bigl\\{[G:H]:\\,G_{\\omega}\\le H\\le G\\bigr\\}\n\\]\nand hence all possible values of $k$ when $G=C_{n}$.",
      "solution": "\\textbf{Preliminaries.}  \nBecause $\\Sigma$ is a Boolean algebra, its atoms (minimal non–empty members) are pairwise disjoint and their union equals $\\Omega$.  Property (2) forces $G$ to permute the atoms transitively.\n\n\\medskip\n\\textbf{(a) Cosets versus atoms.}  \nFix $\\omega\\in\\Omega$ and let $\\mathcal A$ be the set of atoms of $\\Sigma$.  Denote by $A_{0}$ the unique atom containing $\\omega$ and define  \n\\[\nH:=\\{g\\in G:\\,g\\!\\cdot\\!A_{0}=A_{0}\\}.\n\\]\nClearly $G_{\\omega}\\le H\\le G$.  For $g\\in G$ the set $g\\!\\cdot\\!A_{0}$ is again an atom, and  \n\\[\ng_{1}\\!\\cdot\\!A_{0}=g_{2}\\!\\cdot\\!A_{0}\n\\Longleftrightarrow\ng_{2}^{-1}g_{1}\\in H\n\\Longleftrightarrow\ng_{1}H=g_{2}H.\n\\]\nHence  \n\\[\n\\varphi:G/H\\longrightarrow\\mathcal A,\\qquad gH\\longmapsto g\\!\\cdot\\!A_{0},\n\\]\nis a bijection, so $\\lvert\\mathcal A\\rvert=[G:H]$.  Every member of $\\Sigma$ is a union of atoms, and different subsets of $\\mathcal A$ give different unions; consequently  \n\\[\nk=\\lvert\\Sigma\\rvert=2^{\\lvert\\mathcal A\\rvert}=2^{[G:H]}.\n\\]\n\n\\medskip\n\\textbf{(b) Converse construction.}\n\n\\emph{Step 1.  The candidate atoms.}  \nGiven $H$ with $G_{\\omega}\\le H\\le G$ set $A_{gH}$ as in $(\\star)$.  \nIf $g_{1}H=g_{2}H$ then $g_{1}H\\!\\cdot\\!\\omega=g_{2}H\\!\\cdot\\!\\omega$, so $A_{gH}$ is well defined.\n\n\\emph{Pairwise disjointness and covering.}  \nSuppose $g_{1}H\\neq g_{2}H$ and $x\\in A_{g_{1}H}\\cap A_{g_{2}H}$.  Then $x=g_{1}h_{1}\\!\\cdot\\!\\omega=g_{2}h_{2}\\!\\cdot\\!\\omega$ for some $h_{1},h_{2}\\in H$, so $g_{2}^{-1}g_{1}h_{1}h_{2}^{-1}\\in G_{\\omega}\\le H$, giving $g_{2}^{-1}g_{1}\\in H$, contradiction.  Hence the $A_{gH}$ are disjoint.  Transitivity of $G$ guarantees $\\bigcup_{gH}A_{gH}=\\Omega$.\n\nThus  \n\\[\n\\mathcal A_H:=\\{A_{gH}\\mid gH\\in G/H\\}\n\\]\nis a partition of $\\Omega$ of size $[G:H]$.\n\n\\emph{Step 2.  The Boolean algebra $\\Sigma_H$.}  \nDefine $\\Sigma_H$ as in (b)(ii).  Because $\\Sigma_H$ is obtained by taking all unions of blocks of the partition $\\mathcal A_H$, it is a Boolean algebra.  Moreover  \n\\[\ng'\\!\\cdot\\!A_{gH}=g' gH\\!\\cdot\\!\\omega=A_{g'gH},\n\\]\nso $\\Sigma_H$ is $G$–invariant.  \nEvery non–empty element of $\\Sigma_H$ is a union of some members of $\\mathcal A_H$; therefore no proper subset of any $A_{gH}$ lies in $\\Sigma_H$, showing that the $A_{gH}$ are \\emph{exactly} the atoms of $\\Sigma_H$.  Consequently  \n\\[\n\\lvert\\Sigma_H\\rvert=2^{\\lvert\\mathcal A_H\\rvert}=2^{[G:H]}.\n\\]\n\n\\medskip\n\\textbf{(c) The symmetric group.}\n\n\\emph{(i)  Maximality of $G_{\\omega}$.}  \nLet $G=\\operatorname{Sym}(\\Omega)$, $n\\ge 3$, and suppose $H$ satisfies $G_{\\omega}\\le H<G$.  Choose $g\\in H\\setminus G_{\\omega}$ with $g\\omega=\\alpha\\neq\\omega$ and pick $\\beta\\in\\Omega\\setminus\\{\\omega,\\alpha\\}$ (possible since $n\\ge 3$).  \n\nBecause $G_{\\omega}\\cong\\operatorname{Sym}(\\Omega\\setminus\\{\\omega\\})$, the transposition $(\\alpha\\ \\beta)$ lies in $G_{\\omega}\\le H$.  Conjugating by $g^{-1}$ gives  \n\\[\ng^{-1}(\\alpha\\ \\beta)g=(g^{-1}\\alpha\\ g^{-1}\\beta)=(\\omega\\ ,\\ g^{-1}\\beta)\\in H,\n\\]\nproducing a transposition that \\emph{moves $\\omega$}.  Now $G_{\\omega}$ contains all permutations of $\\Omega\\setminus\\{\\omega\\}$, so by conjugating the above transposition inside $H$ we obtain \\emph{every} transposition of the form $(\\omega\\ x)$ with $x\\neq\\omega$.  Since $G_{\\omega}$ already contains all transpositions not involving $\\omega$, we conclude  \n\\[\nH=\\langle\\,G_{\\omega},(\\omega\\ x)\\,\\rangle=\\operatorname{Sym}(\\Omega)=G,\n\\]\ncontradicting $H<G$.  Therefore $G_{\\omega}$ is maximal.\n\n\\medskip\n\\emph{(ii)  Consequences for $\\Sigma$.}  \nBy maximality,\n\\[\nH=G_{\\omega}\\quad\\text{or}\\quad H=G.\n\\]\n\n\\underline{Case $H=G$}.  \nThen $[G:H]=1$ and $k=2$, so $\\Sigma=\\{\\varnothing,\\Omega\\}$.\n\n\\underline{Case $H=G_{\\omega}$}.  \nNow $[G:H]=n$ and $k=2^{\\,n}$.  \nBecause $G_{\\omega}$ acts transitively on $\\Omega\\setminus\\{\\omega\\}$, a $G_{\\omega}$–invariant subset either avoids this orbit or contains all of it.  Hence the $G_{\\omega}$–invariant subsets \\emph{containing $\\omega$} are exactly $\\{\\omega\\}$ and $\\Omega$.  Since the atom $A_{0}$ contains $\\omega$ but is proper, we must have $A_{0}=\\{\\omega\\}$.  The conjugates of $A_{0}$ are the singletons, so the atoms of $\\Sigma$ are precisely the singletons and therefore $\\Sigma=\\mathcal P(\\Omega)$.\n\nThus for $G=\\operatorname{Sym}(\\Omega)$\n\\[\nk\\in\\{2,2^{\\,n}\\}\\quad\\text{and no other value occurs}.\n\\]\n\n\\medskip\n\\textbf{(d)  The regular cyclic group.}  \n\nLet $G=C_{n}=\\langle\\rho\\rangle$ act regularly on $\\Omega=\\{0,1,\\dots ,n-1\\}$ by $\\rho\\!\\cdot\\!i=i+1\\pmod n$.  Here $G_{\\omega}=\\{1\\}$, so admissible subgroups are precisely the subgroups of $C_{n}$.  Writing $n=p_{1}^{\\,\\alpha_{1}}\\cdots p_{r}^{\\,\\alpha_{r}}$, the group $C_{n}$ has exactly one subgroup of each order $d\\mid n$.  For such an $H$\n\\[\n[G:H]=\\frac{\\lvert C_{n}\\rvert}{\\lvert H\\rvert}=\\frac{n}{d},\n\\qquad d\\mid n.\n\\]\nHence  \n\\[\n\\mathfrak M(C_{n})=\\Bigl\\{\\tfrac{n}{d}:d\\mid n\\Bigr\\}=\\{m:\\,m\\mid n\\}.\n\\]\nBy part (b) the corresponding Boolean algebras have sizes  \n\\[\nk=2^{m}\\quad(m\\mid n),\n\\]\nand no other values occur.\n\n\\bigskip\n\\textbf{Summary of results.}  \nFor any transitive permutation group $G$ and point $\\omega\\in\\Omega$ the sizes of $G$–Boolean algebras are  \n\\[\nk=2^{[G:H]}\\quad(G_{\\omega}\\le H\\le G).\n\\]\nSpecial cases:\n\\[\n\\operatorname{Sym}(\\Omega)\\,(n\\ge 3):\\; k\\in\\{2,2^{\\,n}\\},\\qquad\nC_{n}:\\; k=2^{m}\\;(m\\mid n).\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.529811",
        "was_fixed": false,
        "difficulty_analysis": "• Extra layer of structure.  The original problem dealt only with Boolean algebras of subsets; the enhanced version insists on invariance under an external group action, injecting group theory into the set-theoretic framework.  \n• Classification now requires the orbit–stabiliser theorem, subgroup lattices, systems of imprimitivity and explicit passage between coset partitions and Boolean algebras—concepts absent from the original exercise.  \n• The answer is no longer a simple “all powers of two up to 2^n.”  One must determine which exponents 2^{[G:H]} actually occur, and this in turn depends intricately on the subgroup lattice of a transitive permutation group.  \n• Parts (c) and (d) force the solver to analyse two very different groups (the full symmetric group and a cyclic group), showing that the same abstract formula yields distinct concrete spectra of k.  \n• The construction in Step 3 (inducing a field of sets from a coset partition via a bijection) and the verification of G-invariance are subtle and require careful handling of both set algebra and group action.  \n• Overall, the problem intertwines Boolean algebras, group actions, coset geometry and partition theory, demanding a deeper and broader toolkit than the original Olympiad-level question."
      }
    },
    "original_kernel_variant": {
      "question": "Let $\\Omega$ be a finite set with $|\\Omega|=n\\ge 3$ and let $G\\le \\operatorname{Sym}(\\Omega)$ act transitively on $\\Omega$.  \nA non–empty family $\\Sigma\\subseteq\\mathcal P(\\Omega)$ is called a \\emph{$G$–Boolean algebra} if  \n\n1. (Boolean algebra) $\\Sigma$ is closed under finite unions, finite intersections and complements;  \n2. ($G$–invariance) $g\\!\\cdot\\!A\\in\\Sigma$ for every $A\\in\\Sigma$ and every $g\\in G$.\n\nFor $\\omega\\in\\Omega$ write  \n\\[\nG_{\\omega}=\\{g\\in G : g\\omega=\\omega\\}, \\qquad k:=|\\Sigma|.\n\\]\n\n(a)  Show that there exists a subgroup $H$ with $G_{\\omega}\\le H\\le G$ and an atom $A_{0}$ of $\\Sigma$ such that  \n\\[\nH=\\{g\\in G : g\\!\\cdot\\!A_{0}=A_{0}\\},\n\\qquad \n\\mathcal A:=\\{g\\!\\cdot\\!A_{0}: g\\in G\\}\n\\]\nis the set of atoms of $\\Sigma$, so that $gH\\longmapsto g\\!\\cdot\\!A_{0}$ induces a bijection $G/H\\to\\mathcal A$.  Deduce that  \n\\[\nk=|\\Sigma|=2^{[G:H]}.\n\\]\n\n(b)  Conversely, fix $\\omega\\in\\Omega$ and let $H$ be any subgroup with $G_{\\omega}\\le H\\le G$.  \nFor $gH\\in G/H$ put  \n\\[\nA_{gH}:=g\\,H\\!\\cdot\\!\\omega=\\{g h\\!\\cdot\\!\\omega : h\\in H\\}\\subseteq\\Omega.\\tag{$\\star$}\n\\]\n\n(i)  Prove that the sets $A_{gH}$ are pairwise disjoint and cover $\\Omega$.  \n(ii)  Let $\\mathcal A_H:=\\{A_{gH}:gH\\in G/H\\}$.  Show that  \n\\[\n\\Sigma_H:=\\Bigl\\{\\bigcup\\mathcal C : \\mathcal C\\subseteq\\mathcal A_H\\Bigr\\}\n\\]\nis a $G$–Boolean algebra whose atoms are exactly the members of $\\mathcal A_H$.  \nIn particular $|\\Sigma_H|=2^{[G:H]}$.\n\n(c)  Specialise to $G=\\operatorname{Sym}(\\Omega)$ with $n\\ge 3$.  \n\n(i) Prove that $G_{\\omega}\\cong\\operatorname{Sym}(n-1)$ is a maximal subgroup of $\\operatorname{Sym}(n)$.  \n\n(ii) Deduce that the only possibilities for $H$ in parts (a),(b) are  \n\\[\nH=G_{\\omega}\\quad\\text{or}\\quad H=G,\n\\]\nso that $[G:H]$ equals $n$ or $1$ and therefore  \n\\[\nk\\in\\{\\,2,\\,2^{\\,n}\\}.\n\\]\nDescribe explicitly $\\Sigma$ in both cases.\n\n(d)  Specialise to the regular cyclic group $G=C_{n}=\\langle\\rho\\rangle$ acting on $\\Omega=\\{0,1,\\dots ,n-1\\}$ by $\\rho\\!\\cdot\\!i=i+1\\pmod n$.  Determine  \n\\[\n\\mathfrak M(C_{n})=\\bigl\\{[G:H]:G_{\\omega}\\le H\\le G\\bigr\\}\n\\]\nand hence all possible values of $k$ when $G=C_{n}$.",
      "solution": "\\textbf{Preliminaries.}  \nBecause $\\Sigma$ is a Boolean algebra, its atoms (minimal non–empty members) are pairwise disjoint and their union equals $\\Omega$.  Condition (2) forces $G$ to permute the atoms transitively.\n\n\\medskip\n\\textbf{(a) Atom–coset correspondence.}  \nFix $\\omega\\in\\Omega$ and let $\\mathcal A$ be the set of atoms of $\\Sigma$.  Denote by $A_{0}$ the unique atom containing $\\omega$ and define  \n\\[\nH:=\\{g\\in G : g\\!\\cdot\\!A_{0}=A_{0}\\}.\n\\]\nClearly $G_{\\omega}\\le H\\le G$.  For $g\\in G$ the set $g\\!\\cdot\\!A_{0}$ is an atom, and  \n\\[\ng_{1}\\!\\cdot\\!A_{0}=g_{2}\\!\\cdot\\!A_{0}\n\\Longleftrightarrow\ng_{2}^{-1}g_{1}\\in H\n\\Longleftrightarrow\ng_{1}H=g_{2}H.\n\\]\nHence  \n\\[\n\\varphi:G/H\\longrightarrow\\mathcal A,\\qquad gH\\longmapsto g\\!\\cdot\\!A_{0}\n\\]\nis a bijection, so $|\\mathcal A|=[G:H]$.  Every member of $\\Sigma$ is a union of atoms, and different subsets of $\\mathcal A$ give different unions, whence  \n\\[\nk=|\\Sigma|=2^{|\\mathcal A|}=2^{[G:H]}.\n\\]\n\n\\medskip\n\\textbf{(b) Converse construction.}\n\n\\emph{Step 1.  The candidate atoms.}  \nGiven $H$ with $G_{\\omega}\\le H\\le G$ set $A_{gH}$ as in $(\\star)$.  \nIf $g_{1}H=g_{2}H$ then $g_{1}H\\!\\cdot\\!\\omega=g_{2}H\\!\\cdot\\!\\omega$, so $A_{gH}$ is well defined.\n\n\\emph{Pairwise disjointness and covering.}  \nSuppose $g_{1}H\\neq g_{2}H$ and $x\\in A_{g_{1}H}\\cap A_{g_{2}H}$.  Then $x=g_{1}h_{1}\\!\\cdot\\!\\omega=g_{2}h_{2}\\!\\cdot\\!\\omega$ for some $h_{1},h_{2}\\in H$, so $g_{2}^{-1}g_{1}h_{1}h_{2}^{-1}\\in G_{\\omega}\\le H$, giving $g_{2}^{-1}g_{1}\\in H$, contradiction.  Hence the $A_{gH}$ are disjoint.  Transitivity of $G$ guarantees $\\bigcup_{gH}A_{gH}=\\Omega$.\n\nThus  \n\\[\n\\mathcal A_H:=\\{A_{gH}\\mid gH\\in G/H\\}\n\\]\nis a partition of $\\Omega$ into $[G:H]$ parts.\n\n\\emph{Step 2.  The Boolean algebra $\\Sigma_H$.}  \nDefine $\\Sigma_H$ as in (b)(ii).  Because $\\Sigma_H$ is the full power set of $\\mathcal A_H$ transported back to $\\Omega$, it is a Boolean algebra.  Moreover  \n\\[\ng'\\!\\cdot\\!A_{gH}=g' gH\\!\\cdot\\!\\omega=A_{g'gH},\n\\]\nso $\\Sigma_H$ is $G$–invariant.  Its atoms are exactly the $A_{gH}$, hence  \n\\[\n|\\Sigma_H|=2^{|\\mathcal A_H|}=2^{[G:H]}.\n\\]\n\n\\medskip\n\\textbf{(c) The symmetric group.}\n\n\\emph{(i)  Maximality of $G_{\\omega}$.}  \nLet $G=\\operatorname{Sym}(\\Omega)$, $n\\ge 3$, and suppose $H$ satisfies $G_{\\omega}\\le H<G$.  Choose $g\\in H$ with $g\\omega\\neq\\omega$ and set $\\alpha:=g\\omega$.  Pick $\\beta\\in\\Omega\\setminus\\{\\omega,\\alpha\\}$ (possible because $n\\ge 3$).  \n\nThe stabiliser $G_{\\omega}$, being isomorphic to $\\operatorname{Sym}(\\Omega\\setminus\\{\\omega\\})$, contains the transposition $(\\alpha\\ \\beta)$.  Conjugating by $g$ gives  \n\\[\ng(\\alpha\\ \\beta)g^{-1}=(\\omega\\ g\\beta)\\in H.\n\\]\nThis is a transposition that \\emph{moves $\\omega$}.  Since every transposition $(\\omega\\ x)$ with $x\\neq\\omega$ is conjugate to $(\\omega\\ g\\beta)$ by an element of $G_{\\omega}$ (hence also lies in $H$), we have $\\langle\\,G_{\\omega},(\\omega\\ x)\\,\\rangle=\\operatorname{Sym}(\\Omega)\\le H$.  Therefore $H=G$.  \n\nConsequently no subgroup strictly between $G_{\\omega}$ and $G$ exists; $G_{\\omega}$ is maximal.\n\n\\emph{(ii)  Consequences for $\\Sigma$.}  \nThus\n\\[\nH=G_{\\omega}\\quad\\text{or}\\quad H=G.\n\\]\nIf $H=G$ then $[G:H]=1$ and $k=2$, giving $\\Sigma=\\{\\varnothing,\\Omega\\}$.  \nIf $H=G_{\\omega}$ then $[G:H]=n$ and $k=2^{\\,n}$, giving $\\Sigma=\\mathcal P(\\Omega)$.  \nNo other values of $k$ occur for $\\operatorname{Sym}(\\Omega)$.\n\n\\medskip\n\\textbf{(d)  The regular cyclic group.}  \n\nLet $G=C_{n}=\\langle\\rho\\rangle$ act regularly on $\\Omega=\\{0,1,\\dots ,n-1\\}$ by $\\rho\\!\\cdot\\!i=i+1\\pmod n$.  The stabiliser $G_{\\omega}$ is trivial, so admissible subgroups are precisely the subgroups of $C_{n}$.  Writing $n=p_{1}^{\\alpha_{1}}\\cdots p_{r}^{\\alpha_{r}}$, the group $C_{n}$ has exactly one subgroup of each order $d\\mid n$.  For such an $H$\n\\[\n[G:H]=\\frac{|C_{n}|}{|H|}=\\frac{n}{d},\n\\qquad d\\mid n.\n\\]\nHence  \n\\[\n\\mathfrak M(C_{n})=\\bigl\\{\\,n/d\\,:\\,d\\mid n\\bigr\\}=\\bigl\\{m\\mid m\\mid n\\bigr\\}.\n\\]\nBy part (b) the corresponding Boolean algebras have sizes  \n\\[\nk=2^{m}\\quad\\text{for every divisor }m\\text{ of }n,\n\\]\nand no other values occur.\n\n\\medskip\n\\textbf{Summary.}  \nFor any transitive permutation group $G$ and point $\\omega\\in\\Omega$ the sizes of $G$–Boolean algebras are  \n\\[\nk=2^{[G:H]}\\qquad(G_{\\omega}\\le H\\le G).\n\\]\nSpecial cases:\n\\[\n\\operatorname{Sym}(\\Omega)\\ (n\\ge 3):\\; k\\in\\{2,2^{\\,n}\\},\\qquad\nC_{n}:\\; k=2^{m}\\ (m\\mid n).\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.442276",
        "was_fixed": false,
        "difficulty_analysis": "• Extra layer of structure.  The original problem dealt only with Boolean algebras of subsets; the enhanced version insists on invariance under an external group action, injecting group theory into the set-theoretic framework.  \n• Classification now requires the orbit–stabiliser theorem, subgroup lattices, systems of imprimitivity and explicit passage between coset partitions and Boolean algebras—concepts absent from the original exercise.  \n• The answer is no longer a simple “all powers of two up to 2^n.”  One must determine which exponents 2^{[G:H]} actually occur, and this in turn depends intricately on the subgroup lattice of a transitive permutation group.  \n• Parts (c) and (d) force the solver to analyse two very different groups (the full symmetric group and a cyclic group), showing that the same abstract formula yields distinct concrete spectra of k.  \n• The construction in Step 3 (inducing a field of sets from a coset partition via a bijection) and the verification of G-invariance are subtle and require careful handling of both set algebra and group action.  \n• Overall, the problem intertwines Boolean algebras, group actions, coset geometry and partition theory, demanding a deeper and broader toolkit than the original Olympiad-level question."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}