Porpostion2: For sets X.Y.X]=n,Y [r].Let XY :={all functions f:Y X).Then XY=n". proof:Let B=fall vectors of length r over X}.define a function 9:Xy→B.f→(f1),,f) gis a injection since f≠f→(f(1),…,f(r)≠(f'(1),…,f'(r). g is a surjection.since∀(b1,·,b,）∈B.define f：Y→X,i→ bi(i.e.f(i)=bi)That is easy to get g(f)=(b1,..,b). →g is a bijection..→lXY1=lBl=n'. 0 Porpostion3: Let Y=[r],2Y :={all subsets of Y).then 2=2 proof:Let B=fall vectors of length r over (0,1)).define function f:2Y→B,A→f(A=(b,…,br) ae={老究 claim:f is a injection. claim:f is a surjection (b,…,b,)∈B define A={b:=1} →f(A)=(d,…,br) →f is a bijection.→Y|=lB=2". If is called as indicator(characteristic function)A is called as support of f(A)=(b1,·,b小 Porpostion2: For sets X,Y.|X| = n, Y = [r]. Let XY := {all functions f : Y → X}.Then |XY | = n r . proof: Let B={all vectors of length r over X}. define a function g : X Y → B. f 7→ (f(1), · · · , f(r)) g is a injection since f 6= f 0 ⇒ (f(1), · · · , f(r)) 6= (f 0 (1), · · · , f0 (r)). g is a surjection.since ∀(b1, · · · , br) ∈ B.define f : Y → X, i 7→ bi(i.e.f(i) = bi) That is easy to get g(f) = (b1, · · · , br). ⇒ g is a bijection.⇒ |XY | = |B| = n r . Porpostion3: Let Y = [r], 2 Y :={all subsets of Y }.then |2 Y | = 2r proof: Let B ={all vectors of length r over {0,1}}. define function f : 2Y → B, A → f(A) = (b1, · · · , br) where bi =  0, if i /∈ A; 1, if i ∈ A. claim:f is a injection. ∀A 6= A0 ,⇒ ∃i ∈ [r] s.t.i ∈ A, i /∈ A0 or i /∈ A, i ∈ A0 . ⇒ f(A) 6= f(A0 ). claim:f is a surjection. ∀(b1, · · · , br) ∈ B define A = {i|bi = 1} ⇒ f(A) = (b1, · · · , br) ⇒ f is a bijection.⇒ |Y | = |B| = 2r . [f is called as indicator(characteristic function) A is called as support of f(A) = (b1, · · · , br)]. 3