
MHB Master
#1
January 12th, 2018,
02:21
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 ...

January 12th, 2018 02:21
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MHB Oldtimer
#2
January 12th, 2018,
04:46
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.

MHB Craftsman
#3
January 12th, 2018,
06:19
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.