1≤i<j≤nT∈F,write T=(T\{})U{i} ifj∈T,itT,andT年F, otherwise. S(F)={S(T)|T∈F} 1.|Si(T)川=|T|and|Si(F)川=|Fl 2.F intersecting Sii(F)intersecting (2)the only bad case: A,B∈FA∩B={} A=A\{}U{}∈FB=B\{}U{i年Fi庄B > Ai∩B=0 contradiction!
Sij (T) = Tij if j T,i ⇥ T, and Tij ⇥ F, T otherwise. 1 i<j n ⇥T F, write Tij = (T \ {j}) ⌅ {i} Sij (F) = {Sij (T) | T F} (2) the only bad case: A, B F A B = {j} i ⇥ B Aij ⇥ B = Aij = A \ {j} ⇤ {i} F Bij = B \ {j} ⌅ {i} ⇥ F contradiction! |Sij (T)| = |T| F intersecting Sij (F) intersecting 1. and 2. |Sij (F)| = |F|
1≤i<)≤nT∈F,write Tii=(T\{})U{} m-8 ifj∈T,itT,andT年F, otherwise. S(F)={S(T)|T∈F} 1.lS(T)川=T and|Si(F)川=lF到 2.F intersecting Si(F)intersecting repeat applying (i,))-shifting Sis(F)for 1<i<jn eventually,F is unchanged by any Sj(F) called:F is shifted
Sij (T) = Tij if j T,i ⇥ T, and Tij ⇥ F, T otherwise. 1 i<j n ⇥T F, write Tij = (T \ {j}) ⌅ {i} Sij (F) = {Sij (T) | T F} repeat applying (i, j)-shifting Sij (F) for 1 i<j n eventually, F is unchanged by any Sij (F) called: F is shifted |Sij (T)| = |T| F intersecting Sij (F) intersecting 1. and 2. |Sij (F)| = |F|
Let F(),n≥2k. -1 S,T∈F,SnT≠0>IF|≤ k-1 Erdos-Ko-Rado's proof: true for k=1; when n 2k, vS∈() at most one of S and s is in F n! 代) 2·k!(m-k)川 (n-1)！ = (k-1)(m-k)！
Erdős-Ko-Rado’s proof: Let F [n] k ⇥ , n ⇥ 2k. |F| ⇥ n 1 k 1 ⇥ ⇤S, T F, S ⌃ T ⇥= ⌅ when n = 2k, ⇥S [n] k ⇥ at most one of S and S¯ is in F |F| 1 2 n k ⇥ = n! 2 · k!(n k)! = (n 1)! (k 1)!(n k)! = n 1 k 1 ⇥ true for k=1;
Let F(),n≥2k. m-1 VS,T∈F,SnT卡0>IF≤ k-1 arbitrary IF=FI intersecting F shifted keep intersecting ↓ <(i》
Let F [n] k ⇥ , n ⇥ 2k. |F| ⇥ n 1 k 1 ⇥ ⇤S, T F, S ⌃ T ⇥= ⌅ F arbitrary intersecting shifted F |F| = |F | keep intersecting |F| ⇥ n 1 k 1 ⇥ |F | ⇥ n 1 k 1 ⇥
Let FC(),n≥2k. -1 S,T∈F,SnT卡0|F≤ k-1 when n>2k,induction on n WLOG:F is shifted F1={S∈F|n∈S}F1={S\{n}|S∈F} Fis intersecting otherwise, 3A,B∈F A0B={n} |AUB|≤2k-1<n-1> Ji<n,ig AUB C=A\{n}U{}∈F☐ F is shifted C∩B=0 contradiction!
Let F [n] k ⇥ , n ⇥ 2k. |F| ⇥ n 1 k 1 ⇥ ⇤S, T F, S ⌃ T ⇥= ⌅ when n > 2k, induction on n F1 = {S F | n S} WLOG: F is shifted F 1 = {S \ {n} | S F1} is intersecting otherwise, < n 1 F = ⇥A, B F A B = {n} |A ⇤ B| ⇥ 2k 1 C = A \ {n} {i} C B ⇤i < n, i ⇥ A ⌅ B F is shifted contradiction! F 1