{
  "index": "1968-B-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "B-6. A set of real numbers is called compact if it is closed and bounded. Show that there does not exist a sequence \\( \\left\\{K_{n}\\right\\}_{n=0}^{\\infty} \\) of compact sets of rational numbers such that each compact set of rationals is contained in at least one \\( K_{n} \\).",
  "solution": "B-6 Let \\( \\left\\{K_{n}\\right\\} \\) be any sequence of compact sets of rational numbers. For each \\( n \\), there is a rational \\( r_{n} \\nsubseteq K_{n} \\), with \\( 0 \\leqq r_{n}<1 / n \\). Otherwise, it would be that \\( K_{n} \\) contained all rationals in \\( [0,1 / n] \\), and hence some irrationals (since \\( K_{n} \\) is closed). Let \\( S=\\left\\{0, r_{1}, r_{2}, \\cdots\\right\\} \\). Then \\( S \\) is compact and not included in any \\( K_{n} \\).",
  "vars": [
    "K_n",
    "n",
    "r_n",
    "S"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "K_n": "compactset",
        "n": "seqindex",
        "r_n": "excludedrat",
        "S": "specialset"
      },
      "question": "B-6. A set of real numbers is called compact if it is closed and bounded. Show that there does not exist a sequence \\( \\left\\{compactset\\right\\}_{seqindex=0}^{\\infty} \\) of compact sets of rational numbers such that each compact set of rationals is contained in at least one \\( compactset \\).",
      "solution": "B-6 Let \\( \\left\\{compactset\\right\\} \\) be any sequence of compact sets of rational numbers. For each \\( seqindex \\), there is a rational \\( excludedrat \\nsubseteq compactset \\), with \\( 0 \\leqq excludedrat<1 / seqindex \\). Otherwise, it would be that \\( compactset \\) contained all rationals in \\( [0,1 / seqindex] \\), and hence some irrationals (since \\( compactset \\) is closed). Let \\( specialset=\\left\\{0, r_{1}, r_{2}, \\cdots\\right\\} \\). Then \\( specialset \\) is compact and not included in any \\( compactset \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "K_n": "oceanbreeze",
        "n": "caterpillar",
        "r_n": "blueberry",
        "S": "raincloud"
      },
      "question": "B-6. A set of real numbers is called compact if it is closed and bounded. Show that there does not exist a sequence \\( \\left\\{oceanbreeze\\right\\}_{caterpillar=0}^{\\infty} \\) of compact sets of rational numbers such that each compact set of rationals is contained in at least one \\( oceanbreeze \\).",
      "solution": "B-6 Let \\( \\left\\{oceanbreeze\\right\\} \\) be any sequence of compact sets of rational numbers. For each \\( caterpillar \\), there is a rational \\( blueberry \\nsubseteq oceanbreeze \\), with \\( 0 \\leqq blueberry<1 / caterpillar \\). Otherwise, it would be that \\( oceanbreeze \\) contained all rationals in \\( [0,1 / caterpillar] \\), and hence some irrationals (since \\( oceanbreeze \\) is closed). Let \\( raincloud=\\left\\{0, blueberry, blueberry, \\cdots\\right\\} \\). Then \\( raincloud \\) is compact and not included in any \\( oceanbreeze \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "K_n": "sprawlingset",
        "n": "negativeindex",
        "r_n": "irrationalvalue",
        "S": "infinitecloud"
      },
      "question": "A set of real numbers is called compact if it is closed and bounded. Show that there does not exist a sequence \\( \\left\\{sprawlingset_{negativeindex}\\right\\}_{negativeindex=0}^{\\infty} \\) of compact sets of rational numbers such that each compact set of rationals is contained in at least one \\( sprawlingset_{negativeindex} \\).",
      "solution": "B-6 Let \\( \\left\\{sprawlingset_{negativeindex}\\right\\} \\) be any sequence of compact sets of rational numbers. For each \\( negativeindex \\), there is a rational \\( irrationalvalue_{negativeindex} \\nsubseteq sprawlingset_{negativeindex} \\), with \\( 0 \\leqq irrationalvalue_{negativeindex}<1 / negativeindex \\). Otherwise, it would be that \\( sprawlingset_{negativeindex} \\) contained all rationals in \\( [0,1 / negativeindex] \\), and hence some irrationals (since \\( sprawlingset_{negativeindex} \\) is closed). Let \\( infinitecloud=\\left\\{0, irrationalvalue_{1}, irrationalvalue_{2}, \\cdots\\right\\} \\). Then \\( infinitecloud \\) is compact and not included in any \\( sprawlingset_{negativeindex} \\)."
    },
    "garbled_string": {
      "map": {
        "K_n": "zpqlemno",
        "n": "wjgrtlia",
        "r_n": "duafczxp",
        "S": "eufrnqaz"
      },
      "question": "A set of real numbers is called compact if it is closed and bounded. Show that there does not exist a sequence \\( \\left\\{zpqlemno\\right\\}_{wjgrtlia=0}^{\\infty} \\) of compact sets of rational numbers such that each compact set of rationals is contained in at least one \\( zpqlemno \\).",
      "solution": "B-6 Let \\( \\left\\{zpqlemno\\right\\} \\) be any sequence of compact sets of rational numbers. For each \\( wjgrtlia \\), there is a rational \\( duafczxp \\nsubseteq zpqlemno \\), with \\( 0 \\leqq duafczxp<1 / wjgrtlia \\). Otherwise, it would be that \\( zpqlemno \\) contained all rationals in \\( [0,1 / wjgrtlia] \\), and hence some irrationals (since \\( zpqlemno \\) is closed). Let \\( eufrnqaz=\\left\\{0, r_{1}, r_{2}, \\cdots\\right\\} \\). Then \\( eufrnqaz \\) is compact and not included in any \\( zpqlemno \\)."
    },
    "kernel_variant": {
      "question": "A subset of \\(\\mathbb{Q}\\) is called compact if it is closed in \\(\\mathbb{R}\\) and bounded.  Prove that there is no sequence \\(\\{K_{n}\\}_{n\\ge 1}\\) of compact subsets of \\(\\mathbb{Q}\\), all contained in the interval \\([3,4]\\), with the property that every compact subset of \\(\\mathbb{Q}\\) lying in \\([3,4]\\) is contained in at least one of the sets \\(K_{n}.\\)",
      "solution": "Assume, for contradiction, that such a sequence {Kₙ}ₙ₌₁^∞⊂[3,4]∩ℚ of compact sets exists and that every compact subset of ℚ in [3,4] is contained in at least one of the Kₙ.\n\n1.  For each n≥1, let Iₙ=(3,3+2⁻ⁿ).  If Kₙ contained every rational in Iₙ, then since rationals are dense in Iₙ and Kₙ is closed in ℝ, Kₙ would have to contain all of the closure of the rationals in Iₙ, namely the entire interval [3,3+2⁻ⁿ], including its irrational points.  But Kₙ⊂ℚ, so this is impossible.  Therefore we can pick a rational rₙ∈Iₙ∖Kₙ.\n\n2.  Define S={3}∪{r₁,r₂,…}.  Since rₙ−3<2⁻ⁿ→0, the only limit point of S is 3, so S is closed in ℝ.  It is clearly bounded inside [3,4], and every element of S is rational.  Hence S is compact in ℚ (closed in ℝ and bounded).\n\n3.  But for each n, rₙ∈S yet rₙ∉Kₙ, so S is not contained in any Kₙ.  This contradicts the assumption that {Kₙ} covers every compact rational subset of [3,4].\n\nTherefore no countable family of compact subsets of ℚ in [3,4] can cover all such compact sets.",
      "_meta": {
        "core_steps": [
          "Assume a sequence {K_n} of compact rational sets and work by contradiction.",
          "For each n pick a rational r_n ∉ K_n lying in a tiny interval around a common point (so r_n → that point); this uses: if K_n contained all rationals in such an interval, closedness would force irrationals, impossible.",
          "Form S = {common point} ∪ {r_n}; boundedness is evident and r_n → point makes S closed, hence compact.",
          "Because r_n ∉ K_n, the set S is not contained in any K_n, contradicting the assumed cover."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Width of the shrinking interval used to locate r_n (any positive sequence a_n ↓ 0 would suffice).",
            "original": "1/n"
          },
          "slot2": {
            "description": "Choice of the common accumulation point toward which r_n converges (any rational c would work).",
            "original": "0"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}