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Theorem: Let (X,τ) be a topological space ,a sub collection β of τ is base for τ iff every open set can be expressed as a union of members of β. Proof: → let β be a base for τ and let G be open set then for all x G ,there exists B β such that x B G, then G= { B τ: B G}. ← let β τ ,and let every open set G be the union of members of τ, we have to prove that β is a base for τ, Let x X, and let N be a nbd of x, then there exists an open set G such that x G N ,but G is the union of members of β then there exist B β such that x B G then β is a base for τ. Theorem: Let 1 , 2 are two topologies defined on X which have a common base β then 1 2 . Proof: let G is open set and x G, since G is open set then it is nbd of x, since β is a base for τ 1 then there exist B β such that x B G, since β is abase for 2 and B β then B 2 hence G is open set in 2 ,since x is arbitrary then G 2 then 1 2 or similarly 1 2 ,hence 1 2 . Hereditary property( ) اﻟﺨﺎﺻﯿﺔ اﻟﻮراﺛﯿﺔ Definition: A property of a topological space is said to be hereditary property if every sub space of the space has that property. Example: Let X={1,2,3,4,5}, τ={X,Φ,{1},{3,4},{1,3,4},{2,3,4,5} } and let Y={1,4,5} X then find the relative topology for Y. I τ Y ={ Φ,Y,{1},{4},{1,4},{4,5} }. Theorem: Let (Y, τ Y ) be a sub space of a topological space (X, τ) and (Z,W) be a sub space of (Y, τ Y ) then(Z,W) is sub space of (X, τ). Theorem: (exercise) Let (Y, τ Y ) be a sub space of a topological space (X, τ) ,then i) a sub set A of Y is closed in Y iff there exist a closed set F in X such that A=F Y . ii) for every sub set A in Y we have cl Y (A)= cl X (A) Y. iii) for every sub set A in Y we have int X (A) int Y (A). Theorem: Let (Y, τ Y ) be a sub space of a topological space (X, τ) and if A Y is open(closed) in X then it is also open(closed) in Y. Theorem: Let (Y, τ Y ) be a sub space of a topological space (X, τ) and let β be a base for τ then β Y ={ B Y: B β } is abase for τ Y . II