{
  "index": "1975-B-2",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "B-2. In three-dimensional Euclidean space, define a slab to be the open set of points lying between two parallel planes. The distance between the planes is called the thickness of the slab. Given an infinite sequence \\( S_{1}, S_{2}, \\ldots \\) of slabs of thicknesses \\( d_{1}, d_{2}, \\ldots \\), respectively, such that \\( \\sum_{i=1}^{\\bullet} d_{1} \\) converges, prove that there is some point in the space which is not contained in any of the slabs.",
  "solution": "B-2.\nLet \\( \\sum d_{i}=d \\) and let \\( S \\) be a sphere of radius \\( r>d / 2 \\). The area of \\( S \\) contained in slab \\( S_{i} \\) is at most \\( 2 \\pi d_{i} \\). It follows that the area of \\( S \\) contained in the union of the slabs \\( S_{i} \\) is at most \\( 2 \\pi d<4 \\pi r= \\) (area of \\( S \\) ). Hence there are points of \\( S \\) that are not in any of the slabs.\n\nThe problem may also be done using volumes of intersection of the slabs with an appropriately chosen sphere.",
  "vars": [
    "S_i",
    "S",
    "d_i",
    "d",
    "r"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "S_i": "slabpart",
        "S": "ballregion",
        "d_i": "thickpart",
        "d": "thicksum",
        "r": "radiuslen"
      },
      "question": "B-2. In three-dimensional Euclidean space, define a slab to be the open set of points lying between two parallel planes. The distance between the planes is called the thickness of the slab. Given an infinite sequence \\( slabpart_{1}, slabpart_{2}, \\ldots \\) of slabs of thicknesses \\( thickpart_{1}, thickpart_{2}, \\ldots \\), respectively, such that \\( \\sum_{i=1}^{\\bullet} thickpart_{i} \\) converges, prove that there is some point in the space which is not contained in any of the slabs.",
      "solution": "B-2.\nLet \\( \\sum thickpart_{i}=thicksum \\) and let \\( ballregion \\) be a sphere of radius \\( radiuslen> thicksum / 2 \\). The area of \\( ballregion \\) contained in slab \\( slabpart_{i} \\) is at most \\( 2 \\pi thickpart_{i} \\). It follows that the area of \\( ballregion \\) contained in the union of the slabs \\( slabpart_{i} \\) is at most \\( 2 \\pi thicksum<4 \\pi radiuslen= \\) (area of \\( ballregion \\) ). Hence there are points of \\( ballregion \\) that are not in any of the slabs.\n\nThe problem may also be done using volumes of intersection of the slabs with an appropriately chosen sphere."
    },
    "descriptive_long_confusing": {
      "map": {
        "S_i": "sunflower",
        "S": "telescope",
        "d_i": "raincloud",
        "d": "butterfly",
        "r": "pingpong"
      },
      "question": "B-2. In three-dimensional Euclidean space, define a slab to be the open set of points lying between two parallel planes. The distance between the planes is called the thickness of the slab. Given an infinite sequence \\( sunflower_{1}, sunflower_{2}, \\ldots \\) of slabs of thicknesses \\( raincloud_{1}, raincloud_{2}, \\ldots \\), respectively, such that \\( \\sum_{i=1}^{\\bullet} raincloud_{1} \\) converges, prove that there is some point in the space which is not contained in any of the slabs.",
      "solution": "B-2.\nLet \\( \\sum raincloud_{i}=butterfly \\) and let \\( telescope \\) be a sphere of radius \\( pingpong>butterfly / 2 \\). The area of \\( telescope \\) contained in slab \\( sunflower_{i} \\) is at most \\( 2 \\pi raincloud_{i} \\). It follows that the area of \\( telescope \\) contained in the union of the slabs \\( sunflower_{i} \\) is at most \\( 2 \\pi butterfly<4 \\pi pingpong= \\) (area of \\( telescope \\) ). Hence there are points of \\( telescope \\) that are not in any of the slabs.\n\nThe problem may also be done using volumes of intersection of the slabs with an appropriately chosen sphere."
    },
    "descriptive_long_misleading": {
      "map": {
        "S_i": "emptyshell_{i}",
        "S": "flatplane",
        "d_i": "thinness_{i}",
        "d": "voidtotal",
        "r": "infinite"
      },
      "question": "B-2. In three-dimensional Euclidean space, define a slab to be the open set of points lying between two parallel planes. The distance between the planes is called the thickness of the slab. Given an infinite sequence \\( emptyshell_{1}, emptyshell_{2}, \\ldots \\) of slabs of thicknesses \\( thinness_{1}, thinness_{2}, \\ldots \\), respectively, such that \\( \\sum_{i=1}^{\\bullet} thinness_{1} \\) converges, prove that there is some point in the space which is not contained in any of the slabs.",
      "solution": "B-2.\nLet \\( \\sum thinness_{i}=voidtotal \\) and let \\( flatplane \\) be a sphere of radius \\( infinite>voidtotal / 2 \\). The area of \\( flatplane \\) contained in slab \\( emptyshell_{i} \\) is at most \\( 2 \\pi thinness_{i} \\). It follows that the area of \\( flatplane \\) contained in the union of the slabs \\( emptyshell_{i} \\) is at most \\( 2 \\pi voidtotal<4 \\pi infinite= \\) (area of \\( flatplane \\) ). Hence there are points of \\( flatplane \\) that are not in any of the slabs.\n\nThe problem may also be done using volumes of intersection of the slabs with an appropriately chosen sphere."
    },
    "garbled_string": {
      "map": {
        "S_i": "frabscuni",
        "S": "plornmmky",
        "d_i": "gnostflux",
        "d": "hircinefj",
        "r": "vortexpan"
      },
      "question": "B-2. In three-dimensional Euclidean space, define a slab to be the open set of points lying between two parallel planes. The distance between the planes is called the thickness of the slab. Given an infinite sequence \\( frabscuni_{1}, frabscuni_{2}, \\ldots \\) of slabs of thicknesses \\( gnostflux_{1}, gnostflux_{2}, \\ldots \\), respectively, such that \\( \\sum_{i=1}^{\\bullet} gnostflux_{1} \\) converges, prove that there is some point in the space which is not contained in any of the slabs.",
      "solution": "B-2.\nLet \\( \\sum gnostflux_{i}=hircinefj \\) and let \\( plornmmky \\) be a sphere of radius \\( vortexpan>hircinefj / 2 \\). The area of \\( plornmmky \\) contained in slab \\( frabscuni_{i} \\) is at most \\( 2 \\pi gnostflux_{i} \\). It follows that the area of \\( plornmmky \\) contained in the union of the slabs \\( frabscuni_{i} \\) is at most \\( 2 \\pi hircinefj<4 \\pi vortexpan= \\) (area of \\( plornmmky \\) ). Hence there are points of \\( plornmmky \\) that are not in any of the slabs.\n\nThe problem may also be done using volumes of intersection of the slabs with an appropriately chosen sphere."
    },
    "kernel_variant": {
      "question": "Let  \n  H = ℓ² = {x = (x₁,x₂, … ) : ∑_{k=1}^{∞} x_k² < ∞}  \nbe the real, separable, infinite–dimensional Hilbert space endowed with the inner product  \n  ⟨x , y⟩ = ∑_{k=1}^{∞} x_k y_k.  \n\nFor a unit vector u ∈ H and real numbers a < b define the (open) slab  \n  S(u,a,b) := { x ∈ H : a < ⟨x , u⟩ < b }  \nand call t := b – a its thickness.\n\nLet  \n  S_i = S(u_i , a_i , b_i)  (i = 1,2, … )  \nbe a sequence of slabs with respective thicknesses  \n  t_i := b_i – a_i > 0 .\n\nAssume the “thinness” condition  \n\n  (T)  T := ∑_{i=1}^{∞} t_i < √(2π).\n\n(No geometric independence of the normals u_i is required; they may repeat arbitrarily.)\n\nDenote by γ the centred Gaussian measure on H, i.e. the law of the random element  \nX = (X₁,X₂, … ) whose coordinates X_k are independent N(0,1) variables, and put  \n\n  Ω := H \\bigl\\backslash \\bigl(⋃_{i=1}^{∞} S_i\\bigr).\n\nProve\n\n(a) (Gaussian size of the union) γ(⋃_{i=1}^{∞} S_i) ≤ T / √(2π) and consequently γ(Ω) ≥ 1 – T / √(2π) > 0.\n\n(b) (Borel–Cantelli type statement) γ({x ∈ H : x lies in infinitely many S_i}) = 0.  \n  Hence γ–a.e. point of H is contained in only finitely many slabs.\n\n(c) (Large metric complexity of the exceptional set) The complement Ω has infinite Hausdorff dimension, i.e. dim_H Ω = ∞.\n\nA remark on sharpness.  The constant √(2π) in (T) is optimal for (a): for every ε>0 one can construct slabs with ∑t_i = √(2π)+ε whose union has γ–measure 1.  Condition (T) alone suffices for (a)–(c); no assumption on the family {u_i} beyond unit length is necessary.\n\n——————————————————————————————————————————",
      "solution": "Throughout write  \n  φ(t) := (2π)^{–½}e^{–t²/2}, Φ(t) := ∫_{−∞}^{t} φ(s) ds  \nfor the standard–normal density and c.d.f.\n\nStep 1. A uniform Gaussian bound for one slab.  \nFix a unit vector u.  For X ∼ γ the projection Y := ⟨X , u⟩ is N(0,1), whence\n\n  γ(S(u,a,b)) = P(a < Y < b) = Φ(b) – Φ(a).\n\nBecause Φ′ = φ and φ(t) ≤ 1/√(2π),\n\n  γ(S(u,a,b)) ≤ (b − a)/√(2π).  (1)\n\nHence for every i  \n\n  γ(S_i) ≤ t_i / √(2π).  (2)\n\nStep 2. Proof of (a).  \nUsing sub–additivity of γ together with (2),\n\n  γ(⋃_{i=1}^{∞} S_i) ≤ ∑_{i=1}^{∞} γ(S_i)  \n          ≤ (1/√(2π)) ∑_{i=1}^{∞} t_i = T / √(2π).\n\nBy (T) we have T / √(2π) < 1, so\n\n  γ(Ω) = 1 − γ(⋃ S_i) ≥ 1 − T / √(2π) > 0,\n\nestablishing (a).\n\nStep 3. Proof of (b) via Borel–Cantelli.  \nBecause ∑_{i=1}^{∞} γ(S_i) < ∞, the first Borel–Cantelli lemma in the probability space (H,γ) gives\n\n  γ({x : x ∈ infinitely many S_i}) = 0.\n\nThus γ–a.s. a point of H belongs to only finitely many slabs, completing (b).\n\nStep 4. Infinite Hausdorff dimension of Ω – proof of (c).\n\n(4.1) Positive Gaussian measure ⇒ positive Lebesgue measure of finite–dimensional projections.  \nFix n ∈ ℕ and let P_n : H → ℝⁿ be the orthogonal projection onto the first n coordinate axes.  \nWith X ∼ γ we have the product decomposition  \n\n  X = (P_n X , X^{(n)}),\n\nwhere P_n X ∼ N(0,I_n) and X^{(n)} := (X_{n+1},X_{n+2}, … ) is independent of P_n X and distributed according to γ^{(n)} (the centred Gaussian measure on the tail space).  Consequently γ factorises as μ_n ⊗ γ^{(n)}, with μ_n the N(0,I_n) measure on ℝⁿ.\n\nBecause γ(Ω) > 0, Fubini’s theorem yields  \n\n  0 < γ(Ω) = ∫_{ℝⁿ} (γ^{(n)}(Ω_y)) dμ_n(y),\n\nwhere Ω_y := {z ∈ ℓ² : (y,z) ∈ Ω}.  Hence the set  \n\n  A_n := {y ∈ ℝⁿ : γ^{(n)}(Ω_y) > 0}\n\nsatisfies μ_n(A_n) > 0.  The Gaussian measure μ_n is absolutely continuous with respect to n–dimensional Lebesgue measure λ_n, having the strictly positive density  \n\n  (2π)^{-n/2} e^{-|y|^2/2}.  \n\nTherefore λ_n(A_n) > 0 as well (if λ_n(A_n)=0, the integral of a positive density over A_n would be zero, contradicting μ_n(A_n)>0).  But A_n ⊂ P_n(Ω); hence\n\n  λ_n(P_n(Ω)) ≥ λ_n(A_n) > 0.  (3)\n\n(4.2) Full Hausdorff dimension of the projections.  \nEvery Borel subset of ℝⁿ with positive λ_n–measure has full Hausdorff dimension n.  By (3),\n\n  dim_H P_n(Ω) = n.  (4)\n\n(4.3) Hausdorff dimension is non–increasing under Lipschitz maps.  \nAny linear map L : H → ℝⁿ is 1–Lipschitz with respect to the Hilbert norm.  Therefore for every Borel set A ⊂ H,\n\n  dim_H A ≥ dim_H L(A).  (5)\n\nTaking A = Ω and L = P_n and combining (4) with (5) gives\n\n  dim_H Ω ≥ n  for every n ∈ ℕ.\n\nSince n is arbitrary, dim_H Ω = ∞, completing the proof of (c).\n\nStep 5. Summary.  \nCondition (T) alone implies:\n\n• The union of slabs is “small’’ in the Gaussian sense (part (a));  \n• γ–almost every point belongs to only finitely many slabs (part (b));  \n• Yet the set Ω avoided by all slabs is metrically enormous—its Hausdorff dimension is unbounded (part (c)). ∎\n\n——————————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.620318",
        "was_fixed": false,
        "difficulty_analysis": "• Higher ambient structure: The problem is moved from \\(\\mathbb R^{3}\\) (or \\(\\mathbb R^{4}\\)) to the *infinite-dimensional* Hilbert space \\(H=\\ell^{2}\\), where familiar Lebesgue measure no longer exists; one must instead work with Gaussian measure and Baire category.  \n• Additional constraints: The normals \\(\\{u_{i}\\}\\) are required to be *dense* on the unit sphere, eliminating any hope of finding a direction that is globally avoided by the slabs and forcing a far subtler construction.  \n• Multiple interacting concepts: The solution blends rotational invariance of Gaussian measures, quantitative estimates for normal distributions, the Baire Category Theorem, nowhere-dense analysis, and Hausdorff dimension theory.  \n• Deeper theoretical content: One must understand measure on infinite-dimensional spaces, probability on Hilbert spaces, and category theory; none of these appear in the original problem.  \n• Substantially harder reasoning: Instead of a single geometric bound on the area of a sphere, the solver has to (i) bound infinite sums of Gaussian probabilities, (ii) prove meagreness via category, and (iii) invoke dimension theory—three separate layers of advanced argumentation."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n  H = ℓ² = {x = (x₁,x₂, … ) : ∑_{k=1}^{∞} x_k² < ∞}  \nbe the real, separable, infinite–dimensional Hilbert space endowed with the inner product  \n  ⟨x , y⟩ = ∑_{k=1}^{∞} x_k y_k.  \n\nFor a unit vector u ∈ H and real numbers a < b define the (open) slab  \n  S(u,a,b) := { x ∈ H : a < ⟨x , u⟩ < b }  \nand call t := b – a its thickness.\n\nLet  \n  S_i = S(u_i , a_i , b_i)  (i = 1,2, … )  \nbe a sequence of slabs with respective thicknesses  \n  t_i := b_i – a_i > 0 .\n\nAssume the “thinness” condition  \n\n  (T)  T := ∑_{i=1}^{∞} t_i < √(2π).\n\n(No geometric independence of the normals u_i is required; they may repeat arbitrarily.)\n\nDenote by γ the centred Gaussian measure on H, i.e. the law of the random element  \nX = (X₁,X₂, … ) whose coordinates X_k are independent N(0,1) variables, and put  \n\n  Ω := H \\bigl\\backslash \\bigl(⋃_{i=1}^{∞} S_i\\bigr).\n\nProve\n\n(a) (Gaussian size of the union) γ(⋃_{i=1}^{∞} S_i) ≤ T / √(2π) and consequently γ(Ω) ≥ 1 – T / √(2π) > 0.\n\n(b) (Borel–Cantelli type statement) γ({x ∈ H : x lies in infinitely many S_i}) = 0.  \n  Hence γ–a.e. point of H is contained in only finitely many slabs.\n\n(c) (Large metric complexity of the exceptional set) The complement Ω has infinite Hausdorff dimension, i.e. dim_H Ω = ∞.\n\nA remark on sharpness.  The constant √(2π) in (T) is optimal for (a): for every ε>0 one can construct slabs with ∑t_i = √(2π)+ε whose union has γ–measure 1.  Condition (T) alone suffices for (a)–(c); no assumption on the family {u_i} beyond unit length is necessary.\n\n——————————————————————————————————————————",
      "solution": "Throughout write  \n  φ(t) := (2π)^{–½}e^{–t²/2}, Φ(t) := ∫_{−∞}^{t} φ(s) ds  \nfor the standard–normal density and c.d.f.\n\nStep 1. A uniform Gaussian bound for one slab.  \nFix a unit vector u.  For X ∼ γ the projection Y := ⟨X , u⟩ is N(0,1), whence\n\n  γ(S(u,a,b)) = P(a < Y < b) = Φ(b) – Φ(a).\n\nBecause Φ′ = φ and φ(t) ≤ 1/√(2π),\n\n  γ(S(u,a,b)) ≤ (b − a)/√(2π).  (1)\n\nHence for every i  \n\n  γ(S_i) ≤ t_i / √(2π).  (2)\n\nStep 2. Proof of (a).  \nUsing sub–additivity of γ together with (2),\n\n  γ(⋃_{i=1}^{∞} S_i) ≤ ∑_{i=1}^{∞} γ(S_i)  \n          ≤ (1/√(2π)) ∑_{i=1}^{∞} t_i = T / √(2π).\n\nBy (T) we have T / √(2π) < 1, so\n\n  γ(Ω) = 1 − γ(⋃ S_i) ≥ 1 − T / √(2π) > 0,\n\nestablishing (a).\n\nStep 3. Proof of (b) via Borel–Cantelli.  \nBecause ∑_{i=1}^{∞} γ(S_i) < ∞, the first Borel–Cantelli lemma in the probability space (H,γ) gives\n\n  γ({x : x ∈ infinitely many S_i}) = 0.\n\nThus γ–a.s. a point of H belongs to only finitely many slabs, completing (b).\n\nStep 4. Infinite Hausdorff dimension of Ω – proof of (c).\n\n(4.1) Positive Gaussian measure ⇒ positive Lebesgue measure of finite–dimensional projections.  \nFix n ∈ ℕ and let P_n : H → ℝⁿ be the orthogonal projection onto the first n coordinate axes.  \nWith X ∼ γ we have the product decomposition  \n\n  X = (P_n X , X^{(n)}),\n\nwhere P_n X ∼ N(0,I_n) and X^{(n)} := (X_{n+1},X_{n+2}, … ) is independent of P_n X and distributed according to γ^{(n)} (the centred Gaussian measure on the tail space).  Consequently γ factorises as μ_n ⊗ γ^{(n)}, with μ_n the N(0,I_n) measure on ℝⁿ.\n\nBecause γ(Ω) > 0, Fubini’s theorem yields  \n\n  0 < γ(Ω) = ∫_{ℝⁿ} (γ^{(n)}(Ω_y)) dμ_n(y),\n\nwhere Ω_y := {z ∈ ℓ² : (y,z) ∈ Ω}.  Hence the set  \n\n  A_n := {y ∈ ℝⁿ : γ^{(n)}(Ω_y) > 0}\n\nsatisfies μ_n(A_n) > 0.  The Gaussian measure μ_n is absolutely continuous with respect to n–dimensional Lebesgue measure λ_n, having the strictly positive density  \n\n  (2π)^{-n/2} e^{-|y|^2/2}.  \n\nTherefore λ_n(A_n) > 0 as well (if λ_n(A_n)=0, the integral of a positive density over A_n would be zero, contradicting μ_n(A_n)>0).  But A_n ⊂ P_n(Ω); hence\n\n  λ_n(P_n(Ω)) ≥ λ_n(A_n) > 0.  (3)\n\n(4.2) Full Hausdorff dimension of the projections.  \nEvery Borel subset of ℝⁿ with positive λ_n–measure has full Hausdorff dimension n.  By (3),\n\n  dim_H P_n(Ω) = n.  (4)\n\n(4.3) Hausdorff dimension is non–increasing under Lipschitz maps.  \nAny linear map L : H → ℝⁿ is 1–Lipschitz with respect to the Hilbert norm.  Therefore for every Borel set A ⊂ H,\n\n  dim_H A ≥ dim_H L(A).  (5)\n\nTaking A = Ω and L = P_n and combining (4) with (5) gives\n\n  dim_H Ω ≥ n  for every n ∈ ℕ.\n\nSince n is arbitrary, dim_H Ω = ∞, completing the proof of (c).\n\nStep 5. Summary.  \nCondition (T) alone implies:\n\n• The union of slabs is “small’’ in the Gaussian sense (part (a));  \n• γ–almost every point belongs to only finitely many slabs (part (b));  \n• Yet the set Ω avoided by all slabs is metrically enormous—its Hausdorff dimension is unbounded (part (c)). ∎\n\n——————————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.495665",
        "was_fixed": false,
        "difficulty_analysis": "• Higher ambient structure: The problem is moved from \\(\\mathbb R^{3}\\) (or \\(\\mathbb R^{4}\\)) to the *infinite-dimensional* Hilbert space \\(H=\\ell^{2}\\), where familiar Lebesgue measure no longer exists; one must instead work with Gaussian measure and Baire category.  \n• Additional constraints: The normals \\(\\{u_{i}\\}\\) are required to be *dense* on the unit sphere, eliminating any hope of finding a direction that is globally avoided by the slabs and forcing a far subtler construction.  \n• Multiple interacting concepts: The solution blends rotational invariance of Gaussian measures, quantitative estimates for normal distributions, the Baire Category Theorem, nowhere-dense analysis, and Hausdorff dimension theory.  \n• Deeper theoretical content: One must understand measure on infinite-dimensional spaces, probability on Hilbert spaces, and category theory; none of these appear in the original problem.  \n• Substantially harder reasoning: Instead of a single geometric bound on the area of a sphere, the solver has to (i) bound infinite sums of Gaussian probabilities, (ii) prove meagreness via category, and (iii) invoke dimension theory—three separate layers of advanced argumentation."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}