# Closure of a metric space.

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.

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What is your definition for the boundary of a set? – Sean Gomes Dec 3 '12 at 23:31
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. – bigO Dec 3 '12 at 23:33
@Magpie: its his 7th question. – T. Eskin Dec 4 '12 at 8:51

So suppose that $x\in X\setminus(A\cup\bdry A)$. Then $x\notin\bdry A$, so there is an $\epsilon>0$ such that either $B(x,\epsilon)\cap A=\varnothing$ or $B(x,\epsilon)\subseteq A$. However, $x\in B(x,\epsilon)\setminus A$, so clearly $B(x,\epsilon)\nsubseteq A$, and therefore $B(x,\epsilon)\cap A=\varnothing$. What happens if you set $F=X\setminus B(x,\epsilon)$?
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 $a \in A \mbox{ or } a\in \partial(A)$.
While it’s true that each open nbhd of $a\in\cl A$ meets $A$, this isn’t quite immediate from the OP’s definition of $\cl A$; at the level of experience implied by the question it requires proof. The assertion that $\cl A\cap\cl(X\setminus A)=\bdry A$ also requires proof. – Brian M. Scott Dec 4 '12 at 1:31