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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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    I will try to do the double direction thing.2012-08-05

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Hint For the equality of sets, you have to prove two inclusions

  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 $A_n$.
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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|

Then use the fact that $x\in\overline A$ $\Leftrightarrow$ every neighborhood $U$ of $x$ intersects $A$.