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

In summary, the conversation is focused on understanding the proof of Lemma 1.2.10 from the book "Multidimensional Real Analysis I: Differentiation" by Duistermaat and Kolk. The conversation includes a question about the implication of $\overline{A} = \text{(int(} A^c))^c$ on the closure of a set in $\mathbb{R}^n$, as well as the request for the definition of an open set. The response explains that since the interior of a set is open, its complement must be closed. The conversation also mentions the importance of proving the openness of a set, which is not directly stated in the book.
  • #1
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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 ( \(\displaystyle \overline{A} )^c = \text{int(}A^c)\), 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============================================================================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 ...
 
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  • #2
Peter said:
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.
 
  • #3
Opalg said:
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.
 

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

What is the Closure of a Set?

The closure of a set is the smallest closed set that contains all elements of the original set. It includes all limit points of the set as well as the set itself.

Why is the Closure of a Set important?

The closure of a set is important because it helps to define the boundary of the set, indicating which points are included in the set and which are not. It also plays a crucial role in many mathematical proofs and theorems.

How is the Closure of a Set different from the Interior of a Set?

The closure of a set includes all boundary points and limit points, while the interior of a set only includes points that are completely surrounded by the set. The closure of a set is also always a closed set, while the interior of a set can be either open or closed.

Can the Closure of a Set be empty?

Yes, the closure of a set can be empty if the set has no limit points. This means that the set is already a closed set and does not need to be closed further.

How is the Closure of a Set related to the Complement of a Set?

The closure of a set and the complement of a set are related in that the complement of the closure of a set is equal to the interior of the complement of the original set. In other words, the closure of a set is the complement of the interior of the complement of the set.

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