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I am not sure if I understand the definitions of interior and closure of a set. So I am asking for help in the form of an example:

$C(\cdot)$ means the complement of, $int(\cdot)$ means the interior of, $cl(\cdot)$ means the closure of.

Theorem: $C(int(A))=cl(C(A))$

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    Let us consider the notion of $\textit{interior point}$ for $B$ as a point having an open neighborhood included in $B$, and the notion of $\textit{accumulation point}$ for $B$ as a point without open neighborhood disjoint from $B$. Now what you write is: $p$ is not an interior point of $A$ if nd only if $p$ is an accumulation of $C(A)$. Bye2012-01-22

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Take the very simple example where $A$ is the unit disc (without the boundary) union the point $(0, 1)$.

$int(A)$ = the unit disc. $C(A) = \{ (x, y) | x^2+y^2 \geq 1\} - \{(0, 1)\}$.

$C(int(A)) = cl(C(A)) = plane - unit\ disc.$

In general, you can think of $cl(.)$ as including all points near the set and $int(.)$ as excluding all the points near the complement of the set.

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    $C(A)$ would be the plane with the unit disc and the point $(0, 1)$ taken out. $cl(.)$ restores the "missing point".2012-01-22
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The interior of $A$ is the union of all open subsets included in $A$. So the complement of the interior of $A$ is the intersection of all the complement of open subsets included in $A$. But the complement of the open subsets included in $A$ are exactly the closed subsets containing the complement of $A$. Since the closure of the complement of $A$ is the intersection of all closed subsets containing the complement of $A$, it is equal to the complement of the interior of $A$.

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Let us consider the notion of $\textit{interior point}$ for $B$ as a point having an open neighborhood included in B , and the notion of $\textit{accumulation point}$ for $B$ as a point without open neighborhoods disjoint from $B$.

Now what you write is:

$p$ is not an interior point of $A$ if and only if $p$ is an accumulation of $C(A)$