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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 ...
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 ...