<p>For non-negative integers <i>n</i>, <i>m</i>, <i>a</i> and <i>b</i>, we write <InlineEquation ID="IEq1"> <EquationSource Format="TEX">\(\left( n,m \right) \rightarrow \left( a,b \right) \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mfenced close=")" open="("> <mi>n</mi> <mo>,</mo> <mi>m</mi> </mfenced> <mo stretchy="false">→</mo> <mfenced close=")" open="("> <mi>a</mi> <mo>,</mo> <mi>b</mi> </mfenced> </mrow> </math></EquationSource> </InlineEquation> if for every family <InlineEquation ID="IEq2"> <EquationSource Format="TEX">\(\mathcal {F}\subseteq 2^{[n]}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi mathvariant="script">F</mi> <mo>⊆</mo> <msup> <mn>2</mn> <mrow> <mo stretchy="false">[</mo> <mi>n</mi> <mo stretchy="false">]</mo> </mrow> </msup> </mrow> </math></EquationSource> </InlineEquation> with <InlineEquation ID="IEq3"> <EquationSource Format="TEX">\(|\mathcal {F}|\geqslant m\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">|</mo> <mi mathvariant="script">F</mi> <mo stretchy="false">|</mo> <mo>⩾</mo> <mi>m</mi> </mrow> </math></EquationSource> </InlineEquation> there is an <i>a</i>-element set <InlineEquation ID="IEq4"> <EquationSource Format="TEX">\(T\subseteq [n]\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>T</mi> <mo>⊆</mo> <mo stretchy="false">[</mo> <mi>n</mi> <mo stretchy="false">]</mo> </mrow> </math></EquationSource> </InlineEquation> such that <InlineEquation ID="IEq5"> <EquationSource Format="TEX">\(\left| \mathcal {F}_{\mid T} \right| \geqslant b\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mfenced close="|" open="|"> <msub> <mi mathvariant="script">F</mi> <mrow> <mo>∣</mo> <mi>T</mi> </mrow> </msub> </mfenced> <mo>⩾</mo> <mi>b</mi> </mrow> </math></EquationSource> </InlineEquation>, where <InlineEquation ID="IEq6"> <EquationSource Format="TEX">\(\mathcal {F}_{\mid T}=\{ F \cap T: F \in \mathcal {F} \}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi mathvariant="script">F</mi> <mrow> <mo>∣</mo> <mi>T</mi> </mrow> </msub> <mo>=</mo> <mrow> <mo stretchy="false">{</mo> <mi>F</mi> <mo>∩</mo> <mi>T</mi> <mo>:</mo> <mi>F</mi> <mo>∈</mo> <mi mathvariant="script">F</mi> <mo stretchy="false">}</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation>. A longstanding problem in extremal set theory asks to determine <InlineEquation ID="IEq7"> <EquationSource Format="TEX">\(m(s)=\lim _{n\rightarrow +\infty }\frac{m(n,s)}{n}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>m</mi> <mrow> <mo stretchy="false">(</mo> <mi>s</mi> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <msub> <mo movablelimits="true">lim</mo> <mrow> <mi>n</mi> <mo stretchy="false">→</mo> <mo>+</mo> <mi>∞</mi> </mrow> </msub> <mfrac> <mrow> <mi>m</mi> <mo stretchy="false">(</mo> <mi>n</mi> <mo>,</mo> <mi>s</mi> <mo stretchy="false">)</mo> </mrow> <mi>n</mi> </mfrac> </mrow> </math></EquationSource> </InlineEquation>, where <i>m</i>(<i>n</i>,&#xa0;<i>s</i>) denotes the maximum integer <i>m</i> such that <InlineEquation ID="IEq8"> <EquationSource Format="TEX">\(\left( n,m \right) \rightarrow \left( n-1,m-s \right) \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mfenced close=")" open="("> <mi>n</mi> <mo>,</mo> <mi>m</mi> </mfenced> <mo stretchy="false">→</mo> <mfenced close=")" open="("> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>,</mo> <mi>m</mi> <mo>-</mo> <mi>s</mi> </mfenced> </mrow> </math></EquationSource> </InlineEquation> holds for non-negatives <i>n</i> and <i>s</i>. In this paper, we establish the exact value of <InlineEquation ID="IEq9"> <EquationSource Format="TEX">\(m(2^{d-1}-c)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>m</mi> <mo stretchy="false">(</mo> <msup> <mn>2</mn> <mrow> <mi>d</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo>-</mo> <mi>c</mi> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> for all <InlineEquation ID="IEq10"> <EquationSource Format="TEX">\(1\leqslant c\leqslant d\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>1</mn> <mo>⩽</mo> <mi>c</mi> <mo>⩽</mo> <mi>d</mi> </mrow> </math></EquationSource> </InlineEquation> whenever <InlineEquation ID="IEq11"> <EquationSource Format="TEX">\(d\geqslant 50\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>d</mi> <mo>⩾</mo> <mn>50</mn> </mrow> </math></EquationSource> </InlineEquation>, thereby solving an open problem posed by Piga and Schülke. To be precise, we show that <InlineEquation ID="IEq12"> <EquationSource Format="TEX">\(m(n,2^{d-1}-c)={\left\{ \begin{array}{ll} \frac{2^{d}-c}{d}n &amp; \text{ for } 1\leqslant c\leqslant d-1 \text{ and } d\mid n \\ \frac{2^{d}-d-0.5}{d}n &amp; \text{ for } c=d \text{ and } 2d\mid n \end{array}\right. }\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>m</mi> <mrow> <mo stretchy="false">(</mo> <mi>n</mi> <mo>,</mo> <msup> <mn>2</mn> <mrow> <mi>d</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo>-</mo> <mi>c</mi> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <mfenced open="{"> <mrow> <mtable> <mtr> <mtd columnalign="left"> <mrow> <mfrac> <mrow> <msup> <mn>2</mn> <mi>d</mi> </msup> <mo>-</mo> <mi>c</mi> </mrow> <mi>d</mi> </mfrac> <mi>n</mi> </mrow> </mtd> <mtd columnalign="left"> <mrow> <mspace width="0.333333em" /> <mtext>for</mtext> <mspace width="0.333333em" /> <mn>1</mn> <mo>⩽</mo> <mi>c</mi> <mo>⩽</mo> <mi>d</mi> <mo>-</mo> <mn>1</mn> <mspace width="0.333333em" /> <mtext>and</mtext> <mspace width="0.333333em" /> <mi>d</mi> <mo>∣</mo> <mi>n</mi> </mrow> </mtd> </mtr> <mtr> <mtd columnalign="left"> <mrow> <mrow /> <mfrac> <mrow> <msup> <mn>2</mn> <mi>d</mi> </msup> <mo>-</mo> <mi>d</mi> <mo>-</mo> <mn>0.5</mn> </mrow> <mi>d</mi> </mfrac> <mi>n</mi> </mrow> </mtd> <mtd columnalign="left"> <mrow> <mspace width="0.333333em" /> <mtext>for</mtext> <mspace width="0.333333em" /> <mi>c</mi> <mo>=</mo> <mi>d</mi> <mspace width="0.333333em" /> <mtext>and</mtext> <mspace width="0.333333em" /> <mn>2</mn> <mi>d</mi> <mo>∣</mo> <mi>n</mi> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> </mrow> </math></EquationSource> </InlineEquation> holds for <InlineEquation ID="IEq13"> <EquationSource Format="TEX">\(d\geqslant 50\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>d</mi> <mo>⩾</mo> <mn>50</mn> </mrow> </math></EquationSource> </InlineEquation>. Furthermore, we provide a proof that confirms a conjecture of Frankl and Watanabe from 1994, demonstrating that <InlineEquation ID="IEq14"> <EquationSource Format="TEX">\(m(11)=5.3\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>m</mi> <mo stretchy="false">(</mo> <mn>11</mn> <mo stretchy="false">)</mo> <mo>=</mo> <mn>5.3</mn> </mrow> </math></EquationSource> </InlineEquation>.</p>

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Exact Results on Traces of Sets

  • Mingze Li,
  • Jie Ma,
  • Mingyuan Rong

摘要

For non-negative integers n, m, a and b, we write \(\left( n,m \right) \rightarrow \left( a,b \right) \) n , m a , b if for every family \(\mathcal {F}\subseteq 2^{[n]}\) F 2 [ n ] with \(|\mathcal {F}|\geqslant m\) | F | m there is an a-element set \(T\subseteq [n]\) T [ n ] such that \(\left| \mathcal {F}_{\mid T} \right| \geqslant b\) F T b , where \(\mathcal {F}_{\mid T}=\{ F \cap T: F \in \mathcal {F} \}\) F T = { F T : F F } . A longstanding problem in extremal set theory asks to determine \(m(s)=\lim _{n\rightarrow +\infty }\frac{m(n,s)}{n}\) m ( s ) = lim n + m ( n , s ) n , where m(ns) denotes the maximum integer m such that \(\left( n,m \right) \rightarrow \left( n-1,m-s \right) \) n , m n - 1 , m - s holds for non-negatives n and s. In this paper, we establish the exact value of \(m(2^{d-1}-c)\) m ( 2 d - 1 - c ) for all \(1\leqslant c\leqslant d\) 1 c d whenever \(d\geqslant 50\) d 50 , thereby solving an open problem posed by Piga and Schülke. To be precise, we show that \(m(n,2^{d-1}-c)={\left\{ \begin{array}{ll} \frac{2^{d}-c}{d}n & \text{ for } 1\leqslant c\leqslant d-1 \text{ and } d\mid n \\ \frac{2^{d}-d-0.5}{d}n & \text{ for } c=d \text{ and } 2d\mid n \end{array}\right. }\) m ( n , 2 d - 1 - c ) = 2 d - c d n for 1 c d - 1 and d n 2 d - d - 0.5 d n for c = d and 2 d n holds for \(d\geqslant 50\) d 50 . Furthermore, we provide a proof that confirms a conjecture of Frankl and Watanabe from 1994, demonstrating that \(m(11)=5.3\) m ( 11 ) = 5.3 .