# Prove that every $\varepsilon$-neighborhood of $a$ contains a point in $A$ and a point in $\mathbb{R} \setminus A$

Let $A \subset \mathbb{R}$, $A^\circ$ be the interior of $A$ and $\bar{A}$ be the closure of $A$. Define $\delta A =\bar{A} \setminus A^\circ$. Let $a \in \delta A$.

So far: Let $a \in \delta A$. Then $a \in \bar{A} \setminus A^\circ$. So $a \in \bar{A}$ and $a \notin A^\circ$. Then $a$ is not in the set of interior points of $A$ and $a$ is in the accumulation points of $A$. So by definition, $\exists \epsilon > 0$ such that $(a-\epsilon,a+\epsilon) \subset A$. And also contain infinitely many points $A$. Thus every epsilon neighborhood of $a$ contains a point of $A$. Am I done here? Or is there something that I am missing?

To show that every epsilon neighborhood of $a$ contains a point of $\mathbb{R} \setminus A$. Let $a \in \delta A$. Then $a \in \bar{A} \setminus A^\circ$. So $a \in \bar{A}$ and $a \notin A^\circ$. Then $a$ is not in the set of interior points of $A$ and $a$ is in the accumulation points of $A$. Therefore $\forall \delta > 0$, $(a-\delta,a+\delta)$ is not contained in $A$. Therefore there exists $x \in \mathbb{R}$. Then $x \in \mathbb{R} \setminus A$. Then every epsilon neighborhood of $a$ is contained in $\mathbb{R} \setminus A$

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You've got it backwards, I'm afraid. $$\delta A=\bar{A}\smallsetminus A^\circ.$$ That should help smooth the way for your efforts. – Cameron Buie Oct 25 '12 at 17:12
I do. You are right. I'll fix that. – tk2 Oct 25 '12 at 17:13
Every interior point of $A$ is an accumulation point of $A$. – M. Strochyk Oct 25 '12 at 17:19
You note that $a$ is not in the interior of $A$, but then you say there is some $\epsilon$ for which $(a - \epsilon, a + \epsilon) \subset A$. But this latter observation would have meant that $a$ is in the interior of $A$... – Benjamin Dickman Oct 25 '12 at 17:42
So if $a$ is an accumulation point of $A$ and $a$ is not an interior point of $A$. What can I say? This seems counterintuitive because all I get is that there is an epsilon neighborhood that contains infinitely many points of A and for all positive epsilon, the interval $(a-\epsilon,a+\epsilon)$ is not contained in $A$. – tk2 Oct 25 '12 at 17:54

$\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}\newcommand{\bdry}{\operatorname{bdry}}$Suppose that $a\in\cl A\setminus\int A$, and for $\epsilon>0$ let $I_\epsilon=(a-\epsilon,a+\epsilon)$. $I_\epsilon\cap A\ne\varnothing$, because $a\in\cl A$, and $I_\epsilon\cap(\Bbb R\setminus A)\ne\varnothing$, because $a\notin\int A$ and therefore $I_\epsilon\nsubseteq A$. Thus, every $\epsilon$-nbhd of $a$ meets both $A$ and its complement. Conversely if every $I_\epsilon$ meets $A$, then by definition $a\in\cl A$, and if every $I_\epsilon$ meets $\Bbb R\setminus A$, then by definition $a\notin\int A$. Thus, $a\in\cl A\setminus\int A$ iff each $I_\epsilon$ meets both $A$ and its complement.
(In case the terminology is new to you, if $S$ and $T$ are sets we say that $S$ meets $T$ if $S\cap T\ne\varnothing$.)
By the way, you can now easily prove that $\bdry A=\cl A\cap\cl(\Bbb R\setminus A)$: the boundary of $A$ consists of those points that are limit points both of $A$ and of the complement of $A$.