Characteristic Function

ABOUT THE CHARACTERISTIC FUNCTION OF A SET Prof. Mihály Bencze, Department of Mathematics, University of Braşov, Romania Prof. Florentin Smarandache, Chair of Department of Math & Sciences, University of New Mexico, 200 College Road, Gallup, NM 87301, USA, E-mail: [email protected]

Abstract: In this paper we give a method, based on the characteristic function of a set, to solve some difficult problems of set theory found in undergraduate studies. Definition: Let’s consider A ⊂ E ≠ ∅ (a universal set), then f A : E → {0, 1} , ⎧1, if x ∈ A where the function f A ( x) = ⎨ is called the characteristic function of the set ⎩0, if x ∉ A A. Theorem 1: Let’s consider A, B ⊂ E . In this case f A = fB if and only if A = B . Proof.

⎧1, if x ∈ A = B f A ( x) = ⎨ ⎩0, if x ∉ A = B

= f B ( x)

Reciprocally: For any x ∈ A , fA (x) = 1 , but f A = fB , therefore fB (x) = 1 , namely x ∈ B from where A ⊂ B . The same way we prove that B ⊂ A , namely A = B . Theorem 2: f A% = 1 − f A , A% = CE A . Prof. ⎧⎪1, if x ∈ A% f A% ( x) = ⎨ ⎪⎩0, if x ∉ A%

⎧1, if x ∉ A ⎧1 − 0, if x ∉ A ⎧0, =⎨ =⎨ = 1− ⎨ ⎩0, if x ∈ A ⎩1 − 1, if x ∈ A ⎩1,

Theorem 3: fA ∩ B = fA ∗ fB .

1

if x ∉ A = 1 − f A ( x) if x ∈ A

Proof.

⎧1, if x ∈ A ∩ B f A∩ B ( x ) = ⎨ ⎩0, if x ∉ A ∩ B

⎧1, ⎪ ⎧1, if x ∈ A and x ∈ B ⎪0, =⎨ =⎨ ⎩0, if x ∉ A or x ∉ B ⎪0, ⎪⎩0,

⎛ ⎧1 if x ∈ A ⎞ = ⎜⎨ ⎟ ⎝ ⎩0 if x ∉ A ⎠

if if if if

x ∈ A, x ∈ A, x ∉ A, x ∉ A,

x∈B x∉ B = x∈ B x∉B

⎛ ⎧1 if x ∈ B ⎞ ⎜⎨ ⎟ = f A ( x) f B ( x) . ∉ x B 0 if ⎩ ⎝ ⎠

The theorem can be generalized by induction: n

Theorem 4: f n

I Ak

= ∏ f Ak k =1

k =1

Consequence. For any n ∈ N ∗ , f Mn = f M . Proof. In the previous theorem we chose A1 = A2 = ... = An = M . Theorem 5: f A∪ B = f A + f B − f A f B . Proof. f A∪ B = f A∪ B = f A∩ B = 1 − f A∩ B = 1 − f A f B = 1 − (1 − f A )(1 − f B ) = f A + f B − f A f B

It can be generalized by induction: n

Theorem 6: f n

U Ak

= ∑ (−1) k −1

k =1

k =1

n

∑

1≤i1 <...
(−1) k −1 f Ai f Ai ... f Ai 1

2

k

Theorem 7: fA− B = fA (1 − f B )

Proof. f A− B = f AI B = f A f B = f A (1 − f B ) .

It can be generalized by induction: n

Theorem 8: fA1 − A2 −...− An = ∑ (−1)k −1 fAi fAi ... fAi . 1

k =1

2

k

Theorem 9: fAΔB = f A + fB − 2 fA fB Proof. f AΔB = f AU B − AI B = f AU B (1 − f AI B ) = ( f A + f B − f A f B )(1 − f A f B ) = f A + f B − 2 f A f B .

It can be generalized by induction: n

Theorem 10: FΔn

k =1 Ak

= ∑ (−2) k −1 k =1

∑

1≤ i1 <...< ik ≤ n

Theorem 11: f A× B ( x, y ) = f A ( x) f B ( y ) .

2

f Ai Ai .... Ai . 1

2

k

Proof. If ( x, y ) ∈ A × B , then f A× B ( x, y ) = 1 and x ∈A , namely fA (x) = 1 and y ∈B , namely fB (y) = 1 , therefore f A (x) fB (y) = 1 . If ( x, y ) ∉ A × B , then f A× B ( x, y ) = 0 and x ∉ A , namely fA (x) = 0 or y ∉ B , namely f B ( y ) = 0 , therefore fA (x) f B (y) = 0 . This theorem can be generalized by induction. n

Theorem 12: f×n

k =1 Ak

( x1 , x2 ,..., xn ) = ∏ f Ak ( xk ) . k =1

n

n

k =1

k =1

Theorem 13: (De Morgan) U Ak = I Ak Proof. n

= 1 − ∑ (−1) k −1

= 1− f n

fn

U Ak

U Ak

k =1

k =1

k =1

n

∑

1≤i1 <...
n

n

k =1

k =1

f Ai f Ai ... f Ai = ∏ (1 − f Ak ) = ∏ f Ak = f n 1

2

k

.

I Ak

k =1

We prove in the same way the following theorem: n

n

k =1

k =1

Theorem 14: (De Morgan) I Ak = U Ak . n ⎛ n ⎞ Theorem 15: ⎜ U Ak ⎟ I M = U ( Ak I M ) . k =1 ⎝ k =1 ⎠ Proof.

f⎛ n

⎞ ⎜ U Ak ⎟ I M ⎝ k =1 ⎠

n

= fn

= ∑ (−1) k −1 k =1

n

U Ak

f M = ∑ (−1)

k =1

n

∑

1≤ i1 <...< ik ≤ n

k −1

k =1

n

n

∑

1≤ i1 <...< ik ≤ n

f Ai f Ai ... f Ai f M = ∑ (−1) 1

2

k

f Ai I M f Ai I M ... f Ai I M = f n 1

2

U ( Ak I M )

k

k =1

In the same way we prove that: n ⎛ n ⎞ Theorem 16: ⎜ I Ak ⎟ U M = I ( Ak U M ) . k =1 ⎝ k =1 ⎠

Theorem 17: ( Δ nk =1 Ak ) I M = Δ nk =1 ( Ak I M ) Application. Δ nk =1 Ak U M = Δ nk =1 (Ak U M ) if and only if M = Φ .

(

)

⎛ n ⎞ n Theorem 18: M × ⎜ U Ak ⎟ = U (M × Ak ) ⎝ k =1 ⎠ k =1 Proof.

3

k =1

k −1

n

∑

1≤i1 <...< ik ≤ n

f Ai f Ai ... f Ai f Mk = 1

2

k

f

(x, y ) = ⎞

⎛ M × ⎜ U Ak ⎟ ⎝ k =1 ⎠ n

n

( x ) = ∑ (−1)k −1

fM ( y) f n

U Ak

k =1

k =1

n

= ∑ (−1)

n

∑

k −1

k =1

1≤ i1 < ...< ik ≤ n

n

∑

= ∑ (−1)k −1 k =1

n

∑

1≤ i1 < ...< ik ≤ n

fAi ( x ) f Ai ( x )... f Ai ( x ) fM ( y) = 1

k

fAi (x) f Ai (x)... f Ai (x) f Mk (y) = 1

2

k

n

1≤ i1 < ...< ik ≤ n

2

fAi × M (x, y)... fAi 1

k

×M

(x, y) = f n

U (M × Ak )

k=1

In the same way we prove that: ⎛ n ⎞ n Theorem 19: M × ⎜ I Ak ⎟ = I ( M × Ak ) . ⎝ k =1 ⎠ k =1 Theorem 20: M × (A1 − A2 − ... − An ) = (M × A1 ) − (M × A2 ) − ... − (M × An ). n

n

k =1

k =1

Theorem 21: ( A1 − A2 ) U ( A2 − A3 ) U ... U ( An −1 − An ) U ( An − A1 ) = U Ak − I Ak Proof 1. n

f( A1 − A2 )U...U( An − A1 ) = ∑ (−1) k −1 k =1

n

= ∑ (−1) k −1

n

∑

k =1

1≤ i1 <...< ik ≤ n

n

n

= ∑ (−1) k −1 k =1

n

∑

1≤ i1 <...< ik ≤ n

f Ai − Ai ... f Ai 1

2

k

− Ai

1

=

( f Ai − f Ai − f Ai f Ai )...( f Ai − f Ai − f Ai f Ai ) = 1

2

1

2

k

n ⎛ ⎞ f Ai ... f Ai ⎜1 − ∏ f Ap ⎟ = f n 1 k U Ak p =1 ⎝ ⎠ k =1

∑

1≤ i1 <...< ik ≤ n

1

k

⎛ ⎜⎜1 − f n I Ak k =1 ⎝

1

⎞ ⎟⎟ = f n n . U Ak − I Ak k =1 k =1 ⎠

Proof 2. Let’s consider x ∈ U (Ai − Ai +1 ), (where An +1 = A1 ), then there exists k n

such

x ∈(Ak − Ak +1 ) ,

that

i =1

x ∉ ( Ak I Ak +1 ) ⊂ A1 I A2 I ... I An ,

namely

n

n

k =1

k =1

namely

x ∉ A1 I A2 I ... I An , and x ∈ U Ak − I Ak . Now we prove the inverse statement: n

n

k =1

k =1

Let’s consider x ∈ U Ak − I Ak , we show that there exists k such that x ∈ Ak and x ∉ Ak +1 . On the contrary, it would result that for any k ∈ {1, 2,..., n} , x ∈ Ak and x ∈ Ak +1 n

namely x ∈ U Ak , it results that there exists p such that x ∈ Ap , but from the previous k =1

reasoning it results that x ∈ Ap +1 , and using this we consequently obtain that x ∈ Ak for k = p, n . But from x ∈ An we obtain that x ∈ A1 , therefore, it results that x ∈ Ak , k = 1, p ,

from where x ∈ Ak , k = 1,n , namely x ∈ A1 I ... I An , that is a contradiction. Thus there

exists r such that x ∈ Ar and x ∉ Ar +1 , namely x ∈(Ar − Ar +1 ) and therefore n

x ∈ U ( Ak − Ak +1 ) . k =1

4

In the same way we prove the following theorem: n

n

k =1

k =1

Theorem 22: ( A1ΔA2 ) U ( A2 ΔA3 ) U ... U ( An −1ΔAn ) = U Ak − I Ak . Theorem 23: k ( A1 × A2 × ... × Ak ) I ( Ak +1 × Ak + 2 × ... × A2 k ) I ( An × A1 × ... × Ak −1 ) = ( A1 I A2 I ... I An ) . Proof. f( A1×...× Ak )I...I( An × A1×...× Ak −1 ) ( x1 ,..., xn ) =

= f A1×...× Ak ( x1 ,..., xn )... f An ×...× Ak −1 ( x1 ,..., xn ) =

(

) (

)

= f A1 ( x1 )... f Ak ( xk ) ... f An ( xn )... f Ak −1 ( xk −1 ) = = f Ak1 ( x1 )... f Akn ( xn ) = f Ak1 I...I An ( x1 ,..., xn ) =

= f

( A1 I...I An )k

( x1 ,..., xn ) .

Theorem 24. (P(E), U) is a commutative monoid. Proof. For any A, B ∈ P(E) ; A U B ∈P(E) , namely the intern operation. Because (A U B ) U C = A U (B U C ) is associative, A U B = B U A commutative, and because A U ∅ = A then ∅ is the neutral element. Theorem 25: ( P( E ), I ) is a commutative monoid. Proof. For any A, B ∈ P(E) ; A I B ∈ P( E ) namely intern operation. ( A I B ) I C = A I ( B I C ) associative, A I B = B I A , commutative A I E = A , E is the

neutral element. Theorem 26: (P(E), Δ ) is an abelian group. Proof. For any A, B ∈ P(E) ; AΔB ∈ P(E) , namely the intern operation. AΔB = BΔA commutative. The proof of associativity is in the XIIth grade manual as a problem. We’ll prove it using the characteristic function of the set. f( AΔB )ΔC = 4 f A f B fC − 2 f A f B + f B f C + fC f A + f A + f B + fC = f AΔ( BΔC ) .

Because AΔ∅ = A , ∅ is the neutral element and because AΔA = ∅ ; the symmetric element of A is A itself. Theorem 27: ( P( E ), Δ, I ) is a commutative Boole ring with a divisor of zero. Proof. Because the previous theorem satisfies the commutative ring axioms, the first part of the theorem is proved. Now we prove that it has a divisor of zero. If A ≠ ∅ and B ≠ ∅ are two disjoint sets, then A I B = ∅ , thus it has divisor of zero. From Theorem 17 we get that it is distributive for n = 2 . Because for any A ∈ P(E) ; A I A = A and AΔA = ∅ it also satisfies the Boole-type axioms.

5

Theorem 28: Let’s consider H = { f | f : E → {0,1}} , then (H , ⊕ ) is an abelian

group, where fA ⊕ f B = fA + f B − 2 f A fB and (P(E), Δ ) ≅ (H , ⊕ ). Proof. Let’s consider F : P(E) → H , where f ( A) = fA , then, from the previous theorem we get that it is bijective and because F ( AΔB ) = f AΔB = F ( A) ⊕ F ( B ) it is compatible. Theorem 29: card(A1ΔAn ) ≤ card(A1ΔA2 ) + card(A2 ΔA3 ) + ... + card(An −1ΔAn ) . Proof. By induction. If n = 2 , then it is true, we show that for n = 3 it is also true. Because ( A1 I A2 ) U ( A2 I A3 ) ⊆ A2 U ( A1 I A3 ) ; card ( ( A1 I A2 ) U ( A2 I A3 ) ) ≤ card ( A2 U ( A1 I A3 ) ) but

card ( M U N ) = cardM + cardN − card ( M I N ) , and thus

cardA2 + card ( A1 I A3 ) − card ( A1 I A2 ) − card ( A2 I A3 ) ≥ 0 ,

written as

can

be

cardA1 + cardA3 − 2card ( A1 I A3 ) ≤

≤ ( cardA1 + cardA2 − 2card ( A1 I A2 ) ) + ( cardA2 + cardA3 − 2card ( A2 I A3 ) ) .

But because of

( M ΔN ) = cardM + cardN − 2card ( M I N )

then card (A1ΔA3 ) ≤ card (A1ΔA2 ) + card (A2 ΔA3 ). The proof of this step of the induction relies on the above method. Theorem 30: P 2 (E), card (AΔB ) is a metric space.

(

)

Proof. Let d ( A, B ) = card ( AΔB ) : P( E ) × P( E ) →

1. d (A, B ) = 0 ⇔ card(AΔB) = 0 ⇔ card ((A − B ) U (B − A)) = 0 but

because

( A − B ) I ( B − A) = ∅

we obtain

(A − B ) = 0

(A − B ) + card(B − A) = 0

and because

and card(B − A) = 0 , then A − B = ∅ , B − A = ∅ , and A = B . 2. d (A, B ) = d(B, A) results from AΔB = BΔA . 3. As a consequence of the previous theorem d (A,C ) ≤ d(A, B) + d(B,C) . As a result of the above three properties it is a metric space. PROBLEMS

Problem 1. A = B UC and f : P(A) → P(A) × P(A) , Let’s consider f (x) = (X U B, X U C ) . Prove that f is injective if and only if B I C = ∅ . Solution 1. If f is injective. Then f (∅) = ( ∅ U B, ∅ U C ) = ( B, C ) = ( ( B I C ) U B, ( B I C ) U C ) = f ( B I C )

where

from

which we obtain B I C = ∅ . Now reciprocally: Let’s consider B I C = ∅ , then f ( X ) = f (Y ) ; it results that X U B = Y U B and X U C = Y U C or

6

X = X U ∅ = X U ( B I C ) = ( X U B) I ( X U C ) = (Y U B) I (Y U C ) = Y U ( B I C ) = Y U ∅ = Y

namely it is injective. Solution 2. Let’s consider B I C = ∅ passing over the set function f ( X ) = f (Y ) if and only if X U B = Y U B and X U C = Y U C , namely f X UB = fY UB and f X UC = fY UC or f X + fB − f X fB = fY + fB − fY fB and f X + fC − f X fC = fY + fC − fY fC from which we obtain ( f X − fY )( fB − fC ) = 0 . Because A = B U C and B I C = ∅ , we have ⎧1, if u ∈ B ≠0 ( f B − fC ) (u ) = ⎨ ⎩−1, if u ∈ C therefore f X − fY = 0 , namely X = Y and thus it is injective. n

Generalization. Let M = U Ak and f : P(A) → P n (A) , where k =1

f (X) = (X U A1 , X U A2 ,..., X U An ) . Prove that f is injective if and only if A1 I A2 I ... I An = ∅ .

Problem 2. Let E ≠ ∅ , A ∈ P ( E ) , and f : P(E) → P(E) × P(E) , where f ( X ) = ( X I A, X U A) . a. Prove that f is injective b. Prove that { f ( x), x ∈ P( E )} = {( M , N ) | M ⊂ A ⊂ N ⊂ E} = K .

c. Let g : P(E) → K , where g(X) = f (X) . Prove that g is bijective and compute its inverse. Solution. f ( X ) = f (Y ) , namely ( X I A, X U A) = (Y I A, Y U A ) and then a. X I A =Y I A, X UA =Y UA, from where X ΔA = Y ΔA or (X ΔA )ΔA = (Y ΔA )ΔA , X Δ( AΔA) = Y Δ( AΔA) , X Δ∅ = Y Δ∅ and thus X = Y , namely f is injective. b. { f ( X ), X ∈ P( E )} = f ( P( E )) . We’ll show that f ( P( E )) ⊂ K . For any

( M , N ) ∈ f ( P( E )), ∃ X ∈ P( E ) : f ( X ) = ( M , N ) ; ( X I A, X U A) = ( M , N ) . From here X I A = M , X U A = N , namely M ⊂ A and A ⊂ N thus M ⊂ A ⊂ N , and, therefore (M , N ) ∈ X . Now, we’ll show that K ⊂ f (P(E)) , for any (M , N ) ∈ K, ∃ X ∈ P(E) such that f (X) = (M , N ) . f (X) = (M , N ) , namely ( X I A, X U A) = ( M , N ) from where X I A = M and X U A = N , namely X ΔA = N − M , (X ΔA )ΔA = (N − M )ΔA , X Δ∅ = (N − M )ΔA , X = (N − M )ΔA , X = ( N I M )ΔA ,

) ( ) ( ) ( ) = ( N I ( M I A) ) U ( A I ( N I M ) ) = ( N I A) U ( ( A I N ) U ( A I M ) ) = (

X = (N I M ) − A U A − (N I M ) = (N I M ) I A U A I (N I M ) =

7

= ( N I A) U (∅ U M ) = ( N − A) U M . From here we get the unique solution: X = (N − A) U M . We test f ( ( N − A) U M ) = ( ( ( N − A) U M ) I A, ( ( N − A) U M ) U A ) but

( ( N − A) U M ) I A = ( ( N I A) U M ) I A = ( ( N I A) I A) U ( M I A) =

(

)

= ( N I ( A I A) U M = ( N I ∅) U M = ∅ U M = M and

Thus

( ( N − A) U M ) U A = ( N − A) U ( M U A) = ( N − A) U A = ( N I A) U A = = ( N U A) I ( A U A) = N I E = N , f ((N − A) U M ) = (M , N ) . f ( P( E ) ) = K . c. From point a. we have that g is injective, from point b. we have that g surjective, thus g is bijective. The inverse function is: g −1 (M , N ) = (N − A) U M . Problem 3. Let E ≠ ∅ , A, B ∈ P(E) and f : P(E) → P(E) × P(E) , where f ( X ) = ( X I A, X I B) . a. Give the necessary and sufficient condition such that f is injective. b. Give the necessary and sufficient condition such that f is surjective. c. Supposing that f is bijective, compute its inverse. Solution. a. Suppose that f is injective. Then: f ( A U B ) = ( ( A U B ) I A, ( A U B ) I B ) = ( A, B) = ( E I A, E I B ) = f ( E ) ,

from where A U B = E . Now we suppose that A U B = E , it results that: X = X I E = X I ( A U B) = ( X I A) U ( X I B) = (Y I A) U (Y I B) = Y I ( A U B) = Y I E = Y namely from f (X) = f (Y ) we obtain that X = Y , namely

f

is injective.

b. Suppose that f is surjective, for any M , N ∈ P(A) × P(B) , there exists X ∈ P( E ), f ( X ) = ( M , N ), ( X I A, X I B ) = ( M , N ), X I A = M , X I B = N . In special cases (M , N ) = (A, ∅) , there exists X ∈ P(E) , from X ⊃ A, ∅ = X I B ⊃ A I B, A I B=∅ . Now we suppose that A I B=∅ and show that it is surjective. Let (M , N ) ∈ P(A) × P(B) , then M ⊂ A, N ⊂ B , M I B ⊂ A I B = ∅ , and N I A ⊂ B I A = ∅ , namely M I B = ∅ , N I A = ∅ and f ( M U N ) = ( ( M U N ) I A, ( M U N ) I B ) = = ( ( M I A ) U ( N I A ) , ( M I B ) U ( N I B ) ) = ( M U ∅, ∅ U N ) = ( M , N ) ,

8

for any (M , N ) there exists X = M U N such that f (X) = (M , N ) , namely f is surjective. c. We’ll show that f −1 ( ( M , N ) ) = M U N . Remark. In the previous two problems we can use the characteristic function of the set as in the first problem. We leave this method for the readers. Application. Let E ≠ ∅, Ak ∈ P( E ) ( k = 1,..., n ) and f : P(E) → P n (E) , where f ( X ) = ( X I A1 , X I A2 ,..., X I An ) . n

Prove that f is injective if and only if

=E.

UA

k

k =1

Application. Let E ≠ ∅, Ak ∈ P( E ), (k = 1,..., n) and f : P(E) → P n (E) , where f ( X ) = ( X I A1 , X I A2 ,..., X I An ) .

n

Prove that f is surjective if and only if

IA

k

=∅.

k =1

Problem 4. We name the set M convex if for any x, y ∈ M tx + (1 − t)y ∈ M , for any t ∈[0,1] .

Prove that if Ak , (k = 1,...,n) are convex sets, then

n

IA

is also convex.

k

k =1

Problem 5. If Ak , ( k = 1,..., n ) are convex sets, then

n

IA

k

is also convex.

k =1

Problem 6. Give the necessary and sufficient condition such that if A, B are convex/concave sets, then A U B is also convex/concave. Generalization for the N set. Problem 7. Give the necessary and sufficient condition such that if A, B are convex/concave sets then AΔB is also convex/concave. Generalization for the N set. Problem 8. Let f , g : P(E) → P(E) , where f ( x) = A - X , and g ( x) = AΔX , A ∈ P( E ) . Prove that f , g are bijective and compute their inverse functions. Problem 9. Let A o B = {( x, y ) ∈ ×

particular case let A =

|∃ z∈

: ( x, z ) ∈ A and ( z , y ) ∈ B} . In a

{( x, {x}) | x ∈ } and B = {({ y} , y ) | y ∈ } .

Represent the A o A , B o A , B o B cases. Problem 10. i. If A U B U C = D, A U B U D = C , A U C U D = B, B U C U D = A , then A=B=C=D

9

ii.

Are there different A, B, C , D sets such that A U B UC = A U B U D = A UC U D = B UC U D ?

Problem 11. Prove that AΔB = A U B if and only if A I B = ∅ . Problem 12. Prove the following identity. n n ⎛ n ⎞ A U A = ⎜ I Aj ⎟ I U k j i , j =1,i < j i =1 ⎝ j =1, j ≠ i ⎠ Problem 13. Prove the following identities. ( A U B ) − ( B I C ) = ( A − ( B I C )) U ( B − C ) = ( A − B ) U ( A − C ) U ( B − C )

and

A − ⎡⎣( A I C ) − ( A I B ) ⎤⎦ = ( A − B ) U ( A − C ) .

Problem 14. Prove that A U ( B I C ) = ( A U B ) I C = ( A U C ) I B if and only if

A ⊂ B and A ⊂ C .

Problem 15. Prove the following identities: (A − B ) − C = (A − B ) − (C − B ) ,

( A U B) − ( A U C ) = B − ( A I C ) , ( A I B) − ( A I C ) = ( A I B) − C . Problem 16. Solve the following system of equations: ⎧⎪ A U X U Y = ( A U X ) I ( A U Y ) . ⎨ ⎪⎩ A I X I Y = ( A I X ) U ( A I Y ) Problem 17. Solve the following system of equations: ⎧ AΔX ΔB = A . ⎨ ⎩ AΔY ΔB = B Problem 18. Let X , Y , Z ⊆ A . Prove that:

Z = ( X I Z ) U (Y I Z ) U ( X I Z I Y ) if and only if X = Y = ∅ .

Problem 19. Prove the following identity: n ⎤ ⎛ n ⎞ ⎡⎛ n ⎞ ⎡ ⎤ A U B − C = ( ) U k ⎣ k ⎦ ⎜⎝ U Ak ⎟⎠ U ⎢⎜⎝ U Ak ⎟⎠ − C ⎥ . k =1 k =1 ⎦ ⎣ k =1 Problem 20. Prove that: A U B = ( A − B ) U ( B − A ) U ( A I B ) . Problem 21. Prove that:

10

( AΔB ) ΔC = ( A I B I C ) U ( A I B I C ) U ( A I B I C ) U ( A I B I C ) . REFERENCES:

[1] [2] [3] [4] [5] [6] [7] [8] [9] [10]

Mihály Bencze, F. Popovici – Permutaciok - Matematikai Lapok, Kolozsvar, pp. 7-8, 1991. Pellegrini Miklós – Egy ujabb kiserlet, a retegezett halmaz. – M.L., Kolozsvar, 6, 1978. Halmazokra vonatkozo egyenletekrol – Matematikai Lapok, Kolozsvar, 6, 1970. Alkalmazasok a halmazokkal kapcsolatban - Matematikai Lapok, Kolozsvar, 3, 1970. Ion Savu – Produsul elementelor într-un grup finit comutativ – Gazeta Matematică Perf., 1, 1989. Nicolae Negoescu – Principiul includerii-excluderii – RMT 2, 1987. F. C. Gheorghe, T. Spiru – Teorema de prelungire a unei probabilităţi, dedusă din teorema de completare metrică – Gazeta Matematică, Seria A, 2, 1974. C. P. Popovici – Funcţii Boolene – Gazeta Matematică, Seria A, 1, 1973. Algebra tankonyv IX oszt., Romania. Năstăsescu stb. – Exerciţii şi probleme de algebră pentru clasele IX-XII – Romania. [Published in Octogon, Vol. 6, No. 2, pp. 86-96, 1998.]

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