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Definitions:

$A'$ is the set of all accumulation or limit points.

$\bar{A} = A \cup A'$ - this is known as the closure of $A$.

Let $A$ be a subset of $\mathbb{R}$. A point $p\in\mathbb{R}$ is an accumulation point of $A$ if and only if every open set $G$ containing $p$ contains a point of $A$ different from $p$.

Prove or disprove: $(\overline{A})^{c} = \overline{(A^{c}})$

proof: This is true since we know what the compliment of the closure set is open.

I am not sure if I am right, assume already that I have proven that the compliment of the closure set is open.

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  • $\begingroup$ Where do these definitions come from? Why would you need this definition and $A^\prime \neq \bar A$? $\endgroup$ – Peter Mar 6 '15 at 19:05
  • $\begingroup$ This is not true. Consider $\mathbf{R}$ with the usual topology. Let $A:=[0,1)$. Then $(\bar{A})^c=(-\infty,0)\cup(1,\infty)$, but $\bar{A^c}=(-\infty,0]\cup[1,\infty)$. $\endgroup$ – Quique Ruiz Mar 6 '15 at 19:15
  • $\begingroup$ Actually $\overline{A}^{c}=\operatorname{int}\left(A^{c}\right)$ $\endgroup$ – drhab Mar 6 '15 at 19:19
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This is not generally true. Furthermore, we can make the following claim:

Let $X$ be a connected topological space and $A\subseteq X, A\neq\emptyset, A\neq X$. Then $\left(\overline{A} \right)^c\neq\overline{A^c}$.

Assume the contrary. Then $\left(\overline{A}\right)^c$ is open, but since $\overline{A}=\left(\overline{A}\right)^{c^c}=\left(\overline{A^c}\right)^c$, we also have that $\overline{A}$ is open. Then $X$ is the union of two disjoint nonempty open sets, which contradicts $X$ being connected.

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The statement is not true. If $A$ is the set of rationals, then $(\bar{A})^c = (\Bbb R)^c = \emptyset$, but $\overline{A^c} = \bar{\Bbb I} = \Bbb R$, where $\Bbb I$ is the set of irrationals.

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What happens when $A$ consists of a single point in $\mathbb{R}$?

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For any $X,Y\subset \Bbb R$ it can be proved that $\overline {X}-\overline{Y}\subset \overline{X\setminus Y}$.

So taking $X=\Bbb R$ and $Y=A$, it follows that $$\left(\overline{A}\right)^\mathsf{c}\subset \overline{A^\mathsf{c}}.$$ There are plenty of counter examples to show that the reverse inclusion fails to hold. As for example take $A$ to be any non-empty finite subset of $\Bbb R$, then $\left(\overline{A}\right)^\mathsf{c}=\Bbb R\setminus A \neq \Bbb R=\overline{A^\mathsf{c}}.$

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