Question

Class - 9
subject - advance maths
state - assam
exersice- 2.4
type only if you know the answer​

Answer 1

Answer:

are you an army

me too

my bias is jungkook

Answer 2

Answer:

If the notation P(S)P(S) denotes the power-set of a given set SS, it looks to be true.

It's clear that S⊆T⟹P(S)⊆P(T)S⊆T⟹P(S)⊆P(T) given sets SS and TT.

A∩B⊆AA∩B⊆A

A∩B⊆BA∩B⊆B

Thus

P(A∩B)⊆P(A)P(A∩B)⊆P(A)

P(A∩B)⊆P(B)P(A∩B)⊆P(B)

We infer

P(A∩B)⊆(P(A)∩P(B))P(A∩B)⊆(P(A)∩P(B))

That's enough going one way.

Going the other way, consider (for any set TT which is an element of P(A)∩P(B)P(A)∩P(B))

T∈P(A)∩P(B)T∈P(A)∩P(B)

∀t∈T∀t∈T, (t∈A)∧(t∈B)⟹t∈(A∩B)(t∈A)∧(t∈B)⟹t∈(A∩B)

but this is exactly what we need for

T∈P(A∩B)T∈P(A∩B)

Since this is true for all elements TT of P(A)∩P(B)P(A)∩P(B), P(A)∩P(B)⊆P(A∩B)P(A)∩P(B)⊆P(A∩B). Combined with the earlier result 

Since , i don't have my paper pen to give actual proof but u can find symmetric equivalency with this.

Hope so it helps