# real analysis problem 1/n neighborhood

Prove that if A is a set on [0,1] and An is its $(1/n)$ neighborhood, i.e. $A_n= \{x \in R: \exists y \in A, |x-y|< 1/n \}$ . then $\cap_{n=1}^{\infty} A_n= \overline{A}$, where $\overline{A}$ is the closure of A.

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Hint: A point $x$ is outside $\overline{A}$ if there is an open ball around $x$ that is disjoint from $A$. For any such $x$, there will be an $n$ such that $x \notin A_n$. – Austin Mohr Aug 5 '12 at 19:11
what have you done so far? Do you see why any point not in the closure is going to be excluded by some sufficiently high $n$? – Robert Mastragostino Aug 5 '12 at 19:11
Don't give orders please, instead ask for help and show us what you've already tried. Start by writing down the required definitions (in this case of $\bar A$). – Dario Aug 5 '12 at 19:12
I am just trying to prove it throught definition. I consider that $A_1 \supset A_2 \supset .... \supset A_n$, so that $A_n$ has the same feature with the definition of closure. But I can not work out a clear proof. – David Aug 5 '12 at 19:29
I will try to do the double direction thing. – David Aug 5 '12 at 19:30

1. $$\bigcap_{n=1}^\infty A_n \subseteq \bar A$$ Let $x\notin \bar A$, then show that there is some $n$ such that $A_n$ that doesn't contain $x$ any more.
2. $$\bar A \subseteq \bigcap_{n=1}^\infty A_n$$ Show that for $x\in \bar A$, $x$ is contained in every $A_n$.
Hint: Show that $$x\in A_n \ \Leftrightarrow\ B(x,1/n)\cap A\ne\emptyset,$$ where $B(x,r)$ denotes the ball around $x$ of radius $r$, i.e. $$B(x,r)=\{y\in\mathbb R; |y-x|<r\}=(x-r,x+r).$$
Then use the fact that $x\in\overline A$ $\Leftrightarrow$ every neighborhood $U$ of $x$ intersects $A$.