Mathematics Sets

Laws and Theorems of sets

`text(1. Idempotent Laws :)`

For any set A,
`(i) A uu A = A`
`(ii) A nn A = A`

`text(Proof)`

(i) Let `x in A uu A ⇔ x in A` or `x in A`
` ⇔x in A`
Hence, `A uu A = A`
(ii) Let `x in A nn A ⇔ x in A` and `x in A`
`⇔ x in A`
Hence, `A nn A.= A`

`2.` `text(Identity Laws)`

For any set A,
`(i) Auu phi= A` `(ii) A nn phi = phi`
`(iii) A uu U = U` ` (iv) A nn U = A`

`text(Proof)`
(i) Let `x in A uu phi ⇔ x in A` or `x in phi`
` ⇔x in A`
Hence, `A uu phi = A`
(ii) Let `x in A nn A ⇔ x in A` and `x in phi`
`⇔ x in phi`
Hence, `A nn phi.= A`
(iii) Let `x in A uu U ⇔ x in A` or `x in U`
`⇔ x in U`
Hence, `A uu U = U`
(iv) Let `x in A nn U ⇔ x in A` and `x in U`
`⇔ x in A`
Hence, `A nn U = A`

`3.` `text(Commutative Laws)`

For any two sets A and B, we have
`(i) A uu B = B uu A` `(ii) A nn B = B nn A`


`text(4. Associative Laws)`

For any three sets A, Band C, we have
`(i) A uu (B iu C) = (A uu B) uu C`
`(ii) A nn (B nn C)= (A nn B) nn C`

`text(5. Distributive Laws)`

For any three sets A, Band C, we have
`(i) A uu (B nn C)= (Auu B) nn (Auu C)`
`(ii) A nn (B uu C)= (A nn B) uu (A nn C)`

`text(6. For any two sets A and B, we have)`

`(i) P (A) nn P(B) = P(A nn B)`
`(ii) P (A) uu P(B) subseteq P(A uu B)`
where, P(A) is the power set of A.

`7.` `text(If A is any set, then)` `(A')'= A`

Proof Let `x in (A')' ⇔ x notin A' ⇔ x in A`
Hence, `(A')' = A`

`8.` `text(De-Morgan's Laws)`

For any three sets A, Band C, we have
`(i) (A uu B)'= A' nn B'`
`(ii) (A nn B)' = A' uu B'`
`(iii) A- (B uu C)= (A- B) nn (A- C)`
`(iv) A- (B nn C)= (A- B) uu (A- C)`

Operations on Sets :

For any two sets A and B, we have

`1. A subseteq A uu B , B subseteq A uu B , A nn B subseteq A , A nn B subseteq B`

`2. A- B=B'- A'`

`3. (A uu B) nn (A uu B') =A`

`4. A uu B = (A - B) uu (B- A) uu (A nn B)`

`5. A- (A- B)= A nn B`

`6. A-B = A nn B'`

`text(Proof)`
Let `x in A - B ⇔ x in A` and `x notin B`
` ⇔ x in A` and `x in B'`
`⇔ x in A nn B'`
Hence , `A - B = A nn B'`

`7. (A- B) uu B =A uu B`
`text(Proof)`

`(A- B) uu B = (A nn B') uu B`
`=(A uu B) nn (B' uu B)` [from distributive law]
`=(A uu B) nn U`
` =A uu B`
Hence, `(A- B) uu B = A uu B`


`8. (A- B) nn B = phi`

`text(Proof :)`

`(A-B)nnB=(AnnB')nnB`
`= Ann(B' nn B)` [from associative law]
`= A nn phi = phi`
Hence , `(A-B) nn B phi`

 
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