For the metric space (x, d), and ACX, we define the boundary A to be the set B d
ID: 2978727 • Letter: F
Question
For the metric space (x, d), and ACX, we define the boundary A to be the set B dry(A) = Prove that A is a closed if and only if Bdry(A). Note : is the closure of A.Explanation / Answer
1) closed ==> subset of A. Now that A is closed, we know closure (A)=A. As a intersection of closure(A) and another set, Bdry(A) is a subset of closure(A)=A. 2) subset of A ==> A is closed. it suffices to show that A=closure(A). Suppose towards contradiction that there is a point x, which is in closure(A) but not in A, i.e. x is a limit point of A. If we can show that x is also in closure(XA), then x is in Bdry(A), which contradicts the assumption that bdry(A) is a subset of A. However, since x is not in A, but x is in X, we know x is in XA, which is a subset of closure(XA), so x is in closure(XA). So we have what we want! :)