# What is the intuition behind the proof of this “one-step” topology theorem?

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

-
You say you don't understand what "interior" and "closure" mean, right? Try en.wikipedia.org/wiki/Interior_%28topology%29 and en.wikipedia.org/wiki/Closure_%28topology%29 If you don't understand what is written there, could you be more specific in what exactly you are having trouble with in these definitions? – user23211 Jan 22 '12 at 16:19
To gain intuition, it is always best to draw pictures. Draw a set $A$ (not too crazy-looking) with dashed boundary (which would signify that the boundary of $A$ is not necessarily in $A$ (i.e. $A$ could be not closed, perhaps even open altogether)). In your picture, what is $cl(A)$? What is $C(A)$? What is $int(A)$? What is $C(int(A))$? What is $cl(C(A))$? Once you do so, the equality in this theorem will become absolutely clear to you. – Rick Jan 22 '12 at 16:26
@ymar - i understans interior mean now but i don't understand what closure mean in a graph – Victor Jan 22 '12 at 16:28
@Rick - I don't understand the definition of closure in a graph so i can't really draw a graph... – Victor Jan 22 '12 at 16:30
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 $C(A)$. Bye – Giuseppe Jan 22 '12 at 16:34

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.

-
i still not understand why cl(.) is necessary here... – Victor Jan 22 '12 at 16:39
$C(A)$ would be the plane with the unit disc and the point $(0, 1)$ taken out. $cl(.)$ restores the "missing point". – aelguindy Jan 22 '12 at 16:45
Thanks a lot for help – Victor Jan 22 '12 at 16:51

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 $A$.

-

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 $C(A)$

-