# Thread: The Closure of a Set is Closed ... Lemma 1.2.10

1. I am reading "Multidimensional Real Analysis I: Differentiation by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

I need help with an aspect of the proof of Lemma 1.2.10 ...

Duistermaat and Kolk"s proof of Lemma 1.2.10 (including D&K's definition of a cluster point and the closure of a set) reads as follows:

In the above proof of Lemma 1.2.10 we read the following:

"... ... Thus ( $\displaystyle \overline{A} )^c = \text{int(}A^c)$, or $\displaystyle \overline{A} = \text{(int(} A^c))^c$, which implies that $\displaystyle \overline{A}$ is closed in $\displaystyle \mathbb{R}^n$. ... ...

Can someone please explain (preferably in detail) how/why

$\displaystyle \overline{A} = \text{(int(} A^c))^c$

implies that

$\displaystyle \overline{A}$ is closed in $\displaystyle \mathbb{R}^n$. ... ...

Help will be much appreciated ... ...

Peter

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It may be helpful for MHB members reading the above post to have access to D&K's definition of an open set ... so I am providing the same ... as follows ... :

... and a closed set is simply a set whose complement is open ... ...

Hope that helps ...

2.

3. Originally Posted by Peter
Can someone please explain (preferably in detail) how/why

$\displaystyle \overline{A} = \text{(int(} A^c))^c$

implies that

$\displaystyle \overline{A}$ is closed in $\displaystyle \mathbb{R}^n$.
The interior of a set is open. So $\operatorname{int}(A^c)$ is an open set and therefore its complement $(\operatorname{int}(A^c))^c$ is a closed set.

4. Originally Posted by Opalg
The interior of a set is open. So $\operatorname{int}(A^c)$ is an open set and therefore its complement $(\operatorname{int}(A^c))^c$ is a closed set.
The whole point is to prove that the interior of a set is open. This does not follow directly from definition 1.2.2 above; it may be (and should be) proved somewhere else in the book.