The Closure of a Set is Closed .... Lemma 1.2.10

[Math Amateur]

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:View attachment 7674In the above proof of Lemma 1.2.10 we read the following:

"... ... Thus ( $$\overline{A} )^c = \text{int(}A^c)$$, or $$\overline{A} = \text{(int(} A^c))^c$$, which implies that $$\overline{A}$$ is closed in $$\mathbb{R}^n$$. ... ...Can someone please explain (preferably in detail) how/why

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

implies that

$$\overline{A}$$ is closed in $$\mathbb{R}^n$$. ... ...
Help will be much appreciated ... ...

Peter============================================================================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 ... :https://www.physicsforums.com/attachments/7675... and a closed set is simply a set whose complement is open ... ...

Hope that helps ...

 

[Opalg]

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

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

implies that

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

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.

 

[castor28]

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.