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If a subset of is totally ordered, it is called a chain. An antichain is a set of elements that are pairwise incomparable. f^ 373 3734 $Dilworth s theorem The theorem is due to R. Dilworth(1950) The proof is due to H. Tverberg(1967) Let P be a partially ordered finite set. The minimum number m of disjoint chains which together contain all elements of P is equal to the maximum number M of elements in an antichain of P.()U7A: $Dilworth s theoremP /fN Ppartially ordered set, P-Ng'Yvanticain size p a, Ng\S(ub Pdisjoint chainsvoƖOm˄P -N@b gvCQ },GR a = b N! " a aAA  a AAA6Proof of Dilworth s theorema b : trivial a b Use strong induction on Suppose when < n, a(P) b(P) When = n, let C be a maximal chain in P. < n Hence, a(P\C) b(P\C) Z " ZSZ4 " ZA AAC  CCAAAPKQ$6Proof of Dilworth s theoremIf every antichain in P\C contains at most a(P)-1 elements, we are done. a(P)-1 a(P\C) b(P\C) b(P) b(P\C)+1 a(P)-1+1 = a(P)Il&  C  C     CCCCCCCb A+6Proof of Dilworth s theoremAssume that { } is an antichain in P\C. Define C is a maximal chain in P the largest element in C is not in < n b( ) a(P ) Nk P /fT Pchainvmaximal element@l&y;A ACCCP. | 86Proof of Dilworth s theorem Tt _NSNRba(P ) Pdisjoint chains voƖ Nk P /fT Pchainvminimal element S~b0Ra(P )hdisjoint chains Om˄P-N@b gCQ }mCCCCCCCK <A  dual to Dilworth s theoremXA  dual to Dilworth s theorem was given by Mirsky(1971) Let P be a partially ordered set. If P possesses no chain of m+1 elements, then P is the union of m antichains.x\Q=$  L RProof of the  dual to Dilworth s theorem**&Use induction on m m =1, the theorem is trivial. Suppose the theorem is true for m -1 Let P be the poset that has no chain of m +1elements. Let M be the set of maximal elements of P, then M is an antichain of P.$#C<C-CC# ,dm M RProof of the  dual to Dilworth s theorem**&(SIl) Suppose there is a chain { } in P\M and P has no chain of m+1 elements { } is a maximal chain of P , a contradiction P\M has no chain of m elements P\M is the union of m-1antichains P\M M =P is the union of m antichainsbZ6Z$Z#ZZ8$ 4  4S Z "Sperner s theorem  zThe theorem is due to Sperner.(1928) The proof is due to Lubell.(1966) If are subsets of N :={1,2,& ,n} such that is not a subset of if , then n%%j o AA`4Proof of Sperner s theorem  <Let S(N) be the set of all subsets of N Consider the poset (S(N), ), then A := { } is an antichain in S(N) The maximal chain of S(N) can be expressed as the form { { } ,{ },& ,{ } } there are maximal chains (a,jA-|,89 j4Proof of Sperner s theorem  For a given k-subset A of N , there are maximal chains which contain A Use two way counting to count the number of ordered pairs (A,C), such that A A , C is a maximal chain, and A C )3t  ; AAo4Proof of Sperner s theorem   k Pmaximal chaingYS gN Pmember/fOꁼe TND}antichain # {(A,C )} H19 @  r4Proof of Sperner s theorem  NA-Nvk-subsets v Pxep P, \kN PA-N vk-subset , S+Tdkk-subsetvmaximal chains g g P # {(A,C )} = Hence, 0000 "WCCCN    Problem 6AtLet be a permutation of the integers . Show that Dilworth s theorem implies that the sequence has a subsequence of length n +1 that is monotone.2AProof of Problem 6AFNS = Define a partial order of S: Note that a chain in S is an increasing subsequence of vv:~b0RN Pchain, vQsizepn+1 b~b0RN Pantichain, vQsize pn+1 !P'6#      ^  Proof of Problem 6A(If there is no chain in S with size n+1, suppose that the size of the maximal chain in S is n. S can be partitioned into at least disjoint chains. By the Dilworth s theorem, the size of maximal antichain in S is There is a decreasing subsequence in_o"% 24 :ZUsing Dilworth s theorem to prove Theorem 5.1..&Theorem 5.1 A necessary and sufficient condition for there to be a complete matching from X to Y in G is that for every ( ) is trivial.   P    &ZUsing Dilworth s theorem to prove Theorem 5.1..& ( ) Let Define a partial order, in G Suppose is a maximal antichain of poset (V(G),<) , 00 f KU F ZUsing Dilworth s theorem to prove Theorem 5.1..& ZUsing Dilworth s theorem to prove Theorem 5.1..&& We have a complete matching. J#!  Theorem 6.4Let A be a collection of m distinct k-subsets of {1,2,& ,n} ,where with the property that any two of the subsets have a nonempty intersection. Then "A f Theorem 6.5Let A be a collection of m subsets of N:={1,2,& ,n } such that and if and for all Then r #A  yProof of Theorem 6.5If all the subsets have size k, then we are done by Theorem 6.4. Let be the subsets with the smallest cardinality, say . 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S A ??"` @ ^ $ 6^ & 6^ ( 6^ * 6^ , 6^ . 6^ 0 6^ 2 6p  c &A ??"` P b  c &A  ??@ScP  p # c &A ??"` p % c &A ??"`j   p ' c &A ??"`% 8c  p ) c &A ??"`% j c  p + c &A  ??"`; -c  p - c &A ??"` h  p / c &A  ??"`{ =  p 1 c &A  ??"`c 3   H  0޽h ? 33___PPT10i.JP+D=' = @B +   <@(  <x < c $pn `}   x < c $o `  ^ < 6j < S A   ??"`/?  ^ < 6j < S A  #??"` G #^  < 6j < S A -??"`( -^  < 6j  < S A +??"`> +^  < 6j  < S A .??"` 4 .^ < 6j < S A /??"` /H < 0޽h ? 33___PPT10i.pJ+D=' = @B +   @(  @x @ c $r `}   x @ c $Hs `  ^ @ 6H @ 0޽h ? 33___PPT10i.PX+D=' = @B +   D0(  Dx D c $\| `}   x D c $\ `  H D 0޽h ? 33___PPT10i.`/+D=' = @B +      H(  Hx H c $ `}   x H c $ `  ^ H 6j H S A 1??"` 1^ H 6j H S A 2??"`@Q 2^  H 6j H S A ??"`}  ^  H 6j  H S AV ??"`   ^  H 6^ H 6^ H 6j H S A ??"`^J  j H S AW ??"` 9  ^ H 6^ H 6\ H S A E??F   E^ H 6\ H S A/ ?? p H H 0޽h ? 33___PPT10i.e+D=' = @B +   LV(  Lx L c $ `}   x L c $Й `  ^ L 6^ L 6^  L 6^  L 6\  L S A L?? o`  L^ L 6^ L 6^ L 6\ L S A- ??8g  ^ L 6\ L S A1 ??Cg   ^ L 6@@\ L S A0 N?? c  NH L 0޽h ? 33___PPT10i.p4%+D=' = @B +7    N F P (  Px P c $@ `}   x P c $M `  ^ P 6\ P S A R??Mk R^ P 6\ P S A ??3 ^  P 6\  P S A  ??YE ^  P 6\  P S A `??p  `^ P 6\ P S A" c??i   c^ P 6\ P S A! d??S I  d^ P 6\ P S A$ i??   i^ P 6\ P S A& n??R    n^ P 6\ P S AT P?? m  P^ P 6\ P S A ??  ^ P 6\ P S A Q??   QH P 0޽h ? 33___PPT10i.F+D=' = @B +      0(  x  c $೮ `}   x  c $ `  ^  6^  6\  S A ?? ^  6^  6^  6^  68 %  " # Yh2  s *"` @ %  !% h2  s *"` d  S A# }??# c % Y  }d  S A( ??g c  d  S A) ??   ^ & 6^ ( 6^ , 6^ . 6\ / S A% "??' @ "H  0޽h ? 33___PPT10i.t+D=' = @B +}   $(  r  S ® `}   r  S ® `  H  0޽h ? ___f3̙;/f9___PPT10i.P|+D=' = @B +}   $(  r  S 4ɮ `}   r  S ɮ `  H  0޽h ? ___f3̙;/f9___PPT10i.. k+D=' = @B +]    t l $ (  r  S LЮ `}   r  S ѮM `  ^  6^  6\  S A3 ??]`  ^  6\   S A4 ??  ^  6^  6\  S A5 ??%  ^  6\  S A2 ?? ^  6\  S A ??C ^  6\  S A ?? .  ^  6\  S A6 ??1 RG  ^  6^  6\  S A ?? >  ^  6\  S A ?? |  ^ " 6\ ! S A ?? f  ^ $ 6\ # S A7 ??   H  0޽h ? ___f3̙;/f9___PPT10i..0<n+D=' = @B +   0|(  r  S  `}   r  S  `  ^  6\  S A9 ?? ^  6^  6\  S A8 ??=-%  ^  6\   S A: ??2  ^  6\   S A; ??vV  ^  6\  S A= ?? nN }  H  0޽h ? ___f3̙;/f9___PPT10i.8 +D=' = @B +    @@(  r  S  `}   r  S    ^  6\  S A< ?? r ^  6\  S A> ??{m  ^  6\   S AC ??   ^  6\   S AB ??   ^  6\  S A? ?? ?g  ^  6^  6\  S A@ ??l s  ^  6\  S AE ??V d  ^  6\  S AD ?? CI  ^  6H  0޽h ? ___f3̙;/f9___PPT10i.;E+D=' = @B +   P (  r  S  `}   r  S  `  ^  6\  S AF ??e ^  6\  S AG ??6 ^  6\  S AH ??C X  H  0޽h ? ___f3̙;/f9___PPT10i.?+D=' = @B +C   Z R `!) (  r  S  `}   r  S  `  ^  6\  S A2 ??]g ^  6\  S AA ??2 ^  6\  S AJ ??}M  ^  6^  6^  6^ " 6^ & 6^ ( 68 C  )I@ )  VM TB  c $D`) TB B c $D3f- TB  c $DCF h2   s *3"`"h } N ^     / 2  @ ^    ^   ZB  s *D Pa  ZB B s *D^  n2  033"` X d  S AK ??C; d  S AP  ??"L  d  S AQ ??{"T d ! S A ;??Wg  ;d # S AI =?? PK  =d % S AO ??& d ' S AM ??k H  0޽h ? ___f3̙;/f9___PPT10i.AQ+D=' = @B +G   ^Vp(  r  S  `}   r  S \   ^  6\  S AN ??  ^  6\  S AS "??}m  "^  6\  S AA #??p #^  6j   S AR ,??"``B&*  ,^  6\  S AU 0??   0^  6\  S A ??   H  0޽h ? ___f3̙;/f9___PPT10i.I'+D=' = @B +R    R(   r   S  `}   r   S l `  \   S A\ r??R  r^   6\   S A] p??) a pH   0޽h ? ___f3̙;/f980___PPT10.Op    $(  $r $ S  `}   r $ S  `  ^ $ 6\ $ S A[ ??N1 ^  $ 6\ $ S A^ ??q  ^ $ 6\  $ S A_ ?? &K  H $ 0޽h ? ___f3̙;/f980___PPT10.S|    ,(  ,r , S  $ `}   r , S $ `  ^ , 6j , S Ab ??"`2  ^  , 6j , S A` ??"`   ^  , 6\  , S A_ ??  H , 0޽h ? ___f3̙;/f980___PPT10.W    4(  4r 4 S + `}   r 4 S \, `  ^  4 6\ 4 S Aa ??*  ^  4 6\  4 S Ac ??a  ^  4 6\  4 S Ad ??b   H 4 0޽h ? ___f3̙;/f980___PPT10.ZL    z r @ (  @r @ S 6 `}   r @ S 6 `  ^ @ 6\ @ S Ae ??a** ^ @ 6\ @ S A' ??@C e ^  @ 6\ @ S Ah ??  ^  @ 6\  @ S Ai ??  \  @ S AX ??l  ^ @ 6^ @ 6%%^ @ 6\ @ S Ak ??6 $q  ^ @ 6\ @ S Al ?? Z  ^ @ 6\ @ S Ag ??   ^ @ 6^ @ 6\ @ S AY ??  H @ 0޽h ? ___f3̙;/f980___PPT10.h#   ",H(  Hr H S C `}    H S D `  Rl the maximal antichain size in V(G) is s by Dilworth s theorem, V(G) can be partitioned into at least s disjoint chains. This will consist of a matching of size . 1X."%">$ ^ H 6\ H S Ao ??c ^ H 6\ H S Am ??" ^  H 6\ H S Ap ??V ^ )H 6 8 t ,HwtN c  H t h2 H s *"` h2 H s *"`  `B H 0DP P`B H 0D `B H 0D `B H 0D N H S  } N H S  h H s *"`P 3 ZB H s *DmCCZB H s *DmZB H s *D ZB  H s *D  ZB !H s *D   ~ "H s ,As ??"` 8&H ZB #H s *Do; ; h $H s *"`P  ~ %H s ,Au ??"` 6 F  t &H 6HW%"`M   'H JAr ???#" `c    *H 0xY*  P8 0 2C  +H 0\]   P9 0 2C H H 0޽h ?/@HH ___f3̙;/f9___PPT10i.u@9+D=' = @B +     *PL(  Pr P S P `}   r "P S DN `  ^ P 6^ P 6^ $P 6\ #P S AZ ??" * \ %P S Af ??H X \ 'P S Aw ??Q ^ *P 6\ )P S Av ??p H P 0޽h ? ___f3̙;/f9___PPT10i.x+D=' = @B +   XP \(  \r \ S d `}   r \ S xe `  ^ \ 6^ \ 6\ \ S An ??& @ ^  \ 6\ \ S Aq ??@Z ^  \ 6\  \ S Ax ?? 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