# Proof;

**To prove that Int(A∩B)=Int(A)∩Int(B), where A and B are subsets of a topological space (X, T) we have need to show that Int(A∩B)⊆(Int(A)∩Int(B))and (Int(A)∩Int(B))⊆Int(A∩B).**

If A and B are subsets of a topological space (X, T) then Let (X,T) be a topological space then Int(A∩B)=(Int(A)∩Int(B)). |

**As by definition of interior of a set, we have**

Int(A)⊆A and Int(B)⊆B

⇒ (Int(A)∩Int(B))⊆(A∩B)

∵ If A⊆B and C⊆D then (A∩B)⊆(C∩D)

### ⇒ (Int(A)∩Int(B))⊆Int(A∩B) ---> (1)

∵ (Int(A)∩Int(B)) is an open set which is contained in (A∩B) But Int(A∩B) is a largest open set which is contained in (A∩B)

Next, as we know that

(A∩B)⊆A and (A∩B)⊆B

⇒ Int(A∩B)⊆Int(A) and Int(A∩B)⊆Int(B)

∵ If A⊆B then Int(A)⊆Int(B)

### ⇒ Int(A∩B)⊆(Int(A)∩Int(B)) ---> (2)

∵ If A⊆B and A⊆C then A⊆(B∩C)