I have a problem about proving the following. Can you please help me?
First of all, boundary of A is the set of points that for every r>0 we can find a ball B(x,r) such that B contains points from both A and outside of A.
Secondly, definition of closure of A is the intersection of all closed sets containing A. I am trying to prove that, Let A is a subset of X and X is a metric space. Closure of A = (A union boundary of A). To prove it, i try to show LHS is a subset of RHS, and then RHS is subset of LHS. It is obvious for me that RHS is subset of LHS. But why is closure of A is subset of (A union boundary A) ? For example, let x be an element in closure of A, and let us say closure of A is B. Then, if x is in B, it can be in the set B\A, i think it does not have to be in (A union boundary of A). So, i dont get it. Can anyone help?
Thanks.
Best Answer
So let $a\in\bar{A}$ the closure. Then every open set $U$ such that $a\in U$ we have the following: $$U\cap A \neq \emptyset.$$ And if $a \not\in A \Rightarrow U \cap (X-A) \neq \emptyset$ for all open sets $U$. So $a \in \bar{A} \cap \overline{X-A} = \partial(A)=bdry \ A.$ So either $a \in A \mbox{ or } a\in \partial(A)$.