{
  "index": "1975-B-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "B-4. Does there exist a subset \\( B \\) of the unit circle \\( x^{2}+y^{2}=1 \\) such that (i) \\( B \\) is topologically closed, and (ii) \\( B \\) contains exactly one point from each pair of diametrically opposite points on the circle?\n[A set B is topologically closed if it contains the limit of every convergent sequence of points in B.]",
  "solution": "B-4.\nNo. Since the mapping with \\( (x, y) \\rightarrow(-x,-y) \\) is a homeomorphism of the unit circle on itself, the complement - \\( B \\) of such a subset \\( B \\) would also be closed. Thus the existence of such a \\( B \\) would make \\( C \\) the union \\( -B \\cup B \\) of disjoint nonempty closed subsets; this would contradict the fact that \\( C \\) is connected.",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "B",
    "C"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "horizonx",
        "y": "verticaly",
        "B": "subsetbee",
        "C": "circlecee"
      },
      "question": "subsetbee-4. Does there exist a subset \\( subsetbee \\) of the unit circle \\( horizonx^{2}+verticaly^{2}=1 \\) such that (i) \\( subsetbee \\) is topologically closed, and (ii) \\( subsetbee \\) contains exactly one point from each pair of diametrically opposite points on the circle?\n[A set subsetbee is topologically closed if it contains the limit of every convergent sequence of points in subsetbee.]",
      "solution": "subsetbee-4.\nNo. Since the mapping with \\( (horizonx, verticaly) \\rightarrow(-horizonx,-verticaly) \\) is a homeomorphism of the unit circle on itself, the complement - \\( subsetbee \\) of such a subset \\( subsetbee \\) would also be closed. Thus the existence of such a \\( subsetbee \\) would make \\( circlecee \\) the union \\( -subsetbee \\cup subsetbee \\) of disjoint nonempty closed subsets; this would contradict the fact that \\( circlecee \\) is connected."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "gumption",
        "y": "hairbrush",
        "B": "cucumber",
        "C": "sledgehammer"
      },
      "question": "B-4. Does there exist a subset \\( cucumber \\) of the unit circle \\( gumption^{2}+hairbrush^{2}=1 \\) such that (i) \\( cucumber \\) is topologically closed, and (ii) \\( cucumber \\) contains exactly one point from each pair of diametrically opposite points on the circle?\n[A set cucumber is topologically closed if it contains the limit of every convergent sequence of points in cucumber.]",
      "solution": "B-4.\nNo. Since the mapping with \\( (gumption, hairbrush) \\rightarrow(-gumption,-hairbrush) \\) is a homeomorphism of the unit circle on itself, the complement - \\( cucumber \\) of such a subset \\( cucumber \\) would also be closed. Thus the existence of such a \\( cucumber \\) would make \\( sledgehammer \\) the union \\( -cucumber \\cup cucumber \\) of disjoint nonempty closed subsets; this would contradict the fact that \\( sledgehammer \\) is connected."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "B": "superset",
        "C": "straightline"
      },
      "question": "B-4. Does there exist a subset \\( superset \\) of the unit circle \\( verticalaxis^{2}+horizontalaxis^{2}=1 \\) such that (i) \\( superset \\) is topologically closed, and (ii) \\( superset \\) contains exactly one point from each pair of diametrically opposite points on the circle?\n[A set superset is topologically closed if it contains the limit of every convergent sequence of points in superset.]",
      "solution": "B-4.\nNo. Since the mapping with \\( (verticalaxis, horizontalaxis) \\rightarrow(-verticalaxis,-horizontalaxis) \\) is a homeomorphism of the unit circle on itself, the complement - \\( superset \\) of such a subset \\( superset \\) would also be closed. Thus the existence of such a \\( superset \\) would make \\( straightline \\) the union \\( -superset \\cup superset \\) of disjoint nonempty closed subsets; this would contradict the fact that \\( straightline \\) is connected."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "B": "rmpqlskj",
        "C": "dfkmreot"
      },
      "question": "Does there exist a subset \\( rmpqlskj \\) of the unit circle \\( qzxwvtnp^{2}+hjgrksla^{2}=1 \\) such that (i) \\( rmpqlskj \\) is topologically closed, and (ii) \\( rmpqlskj \\) contains exactly one point from each pair of diametrically opposite points on the circle?\n[A set rmpqlskj is topologically closed if it contains the limit of every convergent sequence of points in rmpqlskj.]",
      "solution": "No. Since the mapping with \\( (qzxwvtnp, hjgrksla) \\rightarrow(-qzxwvtnp,-hjgrksla) \\) is a homeomorphism of the unit circle on itself, the complement - \\( rmpqlskj \\) of such a subset \\( rmpqlskj \\) would also be closed. Thus the existence of such a \\( rmpqlskj \\) would make \\( dfkmreot \\) the union \\( -rmpqlskj \\cup rmpqlskj \\) of disjoint nonempty closed subsets; this would contradict the fact that \\( dfkmreot \\) is connected."
    },
    "kernel_variant": {
      "question": "Let  \n S³ = {(z₁ ,z₂) ∈ ℂ² : |z₁|² + |z₂|² = 1}  \nbe the unit 3–sphere in ℂ² and let  \n\n π : S³ → S² , π(z₁ ,z₂) = (2 Re(z₁ \\bar z₂), 2 Im(z₁ \\bar z₂), |z₁|² − |z₂|²)\n\nbe the Hopf fibration, whose fibres are circles.  \nDoes there exist a subset B ⊂ S³ such that  \n\n(i) B is a closed 2-dimensional topological sub-manifold of S³;  \n\n(ii) for every Hopf fibre C = π⁻¹(q) (q ∈ S²) the intersection B ∩ C consists of exactly one point; and  \n\n(iii) B is connected?\n\nProve your answer.\n\n───────────────────────────────────────────────",
      "solution": "Step 1.  A section would give a homeomorphism B ≅ S².  \nBecause π|_B is continuous, bijective, and maps the compact space B onto the Hausdorff space S², it is a homeomorphism. Hence B must be homeomorphic to the 2-sphere S².\n\nStep 2.  A section would trivialise the Hopf bundle.  \nFor every b ∈ B and every angle θ ∈ [0,2π) let  \n f : B × S¹ → S³, f(b,e^{iθ}) = e^{iθ}·b  \n(“rotation by θ along the fibre through b”).  \nBecause each fibre is a circle parametrised once by θ, f is a continuous bijection.\n\nSince B is compact and S³ Hausdorff, f is a homeomorphism; thus\n\n S³ ≅ B × S¹ ≅ S² × S¹.                (∗)\n\nStep 3.  The fundamental‐group contradiction.  \nWe compute\n\n π₁(S³) = 0,  π₁(S² × S¹) = π₁(S²) ⊕ π₁(S¹) = 0 ⊕ ℤ = ℤ,\n\nso (∗) is impossible. Therefore no such manifold B can exist.\n\nStep 4.  Conceptual rephrasing.  \nThe Hopf fibration is a non-trivial principal S¹–bundle over S²; a global section would make it trivial. Non-triviality can be detected, for example, by its non-zero first Chern class, or simply by the fundamental-group argument above. Either way, the assumed B would furnish an impossible global section, completing the proof.\n\nHence a connected 2-manifold B satisfying (i)–(iii) cannot exist.\n\n───────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.621712",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional setting: the problem moves from a 1-sphere to a 3-sphere equipped with the Hopf fibration, adding two extra dimensions and a non-trivial bundle structure.\n\n2. Additional structures and constraints:  \n   • B must be a 2-dimensional closed manifold, not merely a closed subset.  \n   • The selection is taken along Hopf fibres, not simple antipodal pairs, so the equivalence classes are circles rather than two-point sets.  \n   • Connectedness of B is required.\n\n3. Deeper theory demanded: solving the problem requires knowledge of principal bundles, the Hopf fibration, compactness–Hausdorff arguments for homeomorphisms, and fundamental-group computations (or Chern-class obstructions). None of these appear in the original problem, whose solution uses only elementary connectedness.\n\n4. More steps:  \n   • Show π|_B is a homeomorphism.  \n   • Build an explicit product homeomorphism S³ ≅ B × S¹.  \n   • Compute π₁ to obtain a contradiction (or invoke bundle non-triviality).  \n   • Conclude non-existence.\n\nThese layers of topology and algebraic topology make the enhanced variant substantially harder than both the original and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n S³ = {(z₁ ,z₂) ∈ ℂ² : |z₁|² + |z₂|² = 1}  \nbe the unit 3–sphere in ℂ² and let  \n\n π : S³ → S² , π(z₁ ,z₂) = (2 Re(z₁ \\bar z₂), 2 Im(z₁ \\bar z₂), |z₁|² − |z₂|²)\n\nbe the Hopf fibration, whose fibres are circles.  \nDoes there exist a subset B ⊂ S³ such that  \n\n(i) B is a closed 2-dimensional topological sub-manifold of S³;  \n\n(ii) for every Hopf fibre C = π⁻¹(q) (q ∈ S²) the intersection B ∩ C consists of exactly one point; and  \n\n(iii) B is connected?\n\nProve your answer.\n\n───────────────────────────────────────────────",
      "solution": "Step 1.  A section would give a homeomorphism B ≅ S².  \nBecause π|_B is continuous, bijective, and maps the compact space B onto the Hausdorff space S², it is a homeomorphism. Hence B must be homeomorphic to the 2-sphere S².\n\nStep 2.  A section would trivialise the Hopf bundle.  \nFor every b ∈ B and every angle θ ∈ [0,2π) let  \n f : B × S¹ → S³, f(b,e^{iθ}) = e^{iθ}·b  \n(“rotation by θ along the fibre through b”).  \nBecause each fibre is a circle parametrised once by θ, f is a continuous bijection.\n\nSince B is compact and S³ Hausdorff, f is a homeomorphism; thus\n\n S³ ≅ B × S¹ ≅ S² × S¹.                (∗)\n\nStep 3.  The fundamental‐group contradiction.  \nWe compute\n\n π₁(S³) = 0,  π₁(S² × S¹) = π₁(S²) ⊕ π₁(S¹) = 0 ⊕ ℤ = ℤ,\n\nso (∗) is impossible. Therefore no such manifold B can exist.\n\nStep 4.  Conceptual rephrasing.  \nThe Hopf fibration is a non-trivial principal S¹–bundle over S²; a global section would make it trivial. Non-triviality can be detected, for example, by its non-zero first Chern class, or simply by the fundamental-group argument above. Either way, the assumed B would furnish an impossible global section, completing the proof.\n\nHence a connected 2-manifold B satisfying (i)–(iii) cannot exist.\n\n───────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.496888",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional setting: the problem moves from a 1-sphere to a 3-sphere equipped with the Hopf fibration, adding two extra dimensions and a non-trivial bundle structure.\n\n2. Additional structures and constraints:  \n   • B must be a 2-dimensional closed manifold, not merely a closed subset.  \n   • The selection is taken along Hopf fibres, not simple antipodal pairs, so the equivalence classes are circles rather than two-point sets.  \n   • Connectedness of B is required.\n\n3. Deeper theory demanded: solving the problem requires knowledge of principal bundles, the Hopf fibration, compactness–Hausdorff arguments for homeomorphisms, and fundamental-group computations (or Chern-class obstructions). None of these appear in the original problem, whose solution uses only elementary connectedness.\n\n4. More steps:  \n   • Show π|_B is a homeomorphism.  \n   • Build an explicit product homeomorphism S³ ≅ B × S¹.  \n   • Compute π₁ to obtain a contradiction (or invoke bundle non-triviality).  \n   • Conclude non-existence.\n\nThese layers of topology and algebraic topology make the enhanced variant substantially harder than both the original and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}