0
$\begingroup$

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 $\mathbb{R} \setminus A$

  • 0
    You've got it backwards, I'm afraid. $$\delta A=\bar{A}\smallsetminus A^\circ.$$ That should help smooth the way for your efforts.2012-10-25
  • 0
    I do. You are right. I'll fix that.2012-10-25
  • 0
    Every interior point of $A$ is an accumulation point of $A$.2012-10-25
  • 0
    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 $A$...2012-10-25
  • 0
    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 $A$.2012-10-25

1 Answers 1