Open and Closed Sets in R^n .... Duistermaat and Kolk, Lemma 1.2.11 ....

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In summary, the conversation discusses a proof in "Multidimensional Real Analysis I" regarding Lemma 1.2.11 and its relation to Lemma 1.2.4. The conversation also includes a request for a formal and rigorous explanation of how Lemma 1.2.11 can be derived from Lemma 1.2.4 through taking complements. After providing the definition of open and closed sets, the conversation concludes with a discussion on the closure of a set and its role in the proof.
  • #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 the proof of Lemma 1.2.11 ...

Duistermaat and Kolk"s Lemma 1.2.11 reads as follows:
View attachment 9020
View attachment 9021

Can someone please demonstrate formally and rigorously how we can, by taking complements, obtain Lemma 1.2.11 from Lemma 1.2.4 ...

Help will be appreciated ...

Peter
=====================================================================================The above post mentions Lemma 1.2.4 so I am providing text of the same ... as follows:
View attachment 9022

It may help MHB readers of the above post to have access to Duistermaat and Kolk's definition of open sets so I am providing text of the same ... as follows:View attachment 9023
View attachment 9024... and a closed set is simply a set whose complement is open ... ... Hope that helps ... Peter
 

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  • D&K - Lemma 1.2.4 ... .png
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  • #2
Unfortunately, you did not include the book's definition of "closure" of a set. I would think, since they have defined the "interior of set A" as the "largest open set contained in A", "closure of A" would be defined as "the smallest closed set containing A". But that is the theorem you want to prove. So what is the book's definition of "closure"?
 
  • #3
HallsofIvy said:
Unfortunately, you did not include the book's definition of "closure" of a set. I would think, since they have defined the "interior of set A" as the "largest open set contained in A", "closure of A" would be defined as "the smallest closed set containing A". But that is the theorem you want to prove. So what is the book's definition of "closure"?

Thanks for your reply, HallsofIvy ...

Indeed, I should have included Duistermaat and Kolk's definition of the closure of a set ...

Belatedly ... here is the definition ...
View attachment 9025
Hope that helps ... ...

Peter
 

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  • #4
First show that $\overline{(\overline A)}=\overline A$.

From the definition, it is clear that $A\subseteq\overline A$. Thus (replacing $A$ by $\overline A$) we have $\overline A\subseteq\overline{(\overline A)}$. If $y\in\overline{(\overline A)}$, then $\forall\delta>0$, $B\left(y;\dfrac{\delta}2\right)\cap\overline A\ne\emptyset$. Let $x\in B\left(y;\dfrac{\delta}2\right)\cap\overline A$. Then $x\in\overline A$ $\implies$ $B\left(x;\dfrac{\delta}2\right)\cap A\ne\emptyset$. Let $a\in B\left(x;\dfrac{\delta}2\right)\cap A$; then $\|a-y\|\leq\|a-x\|+\|x-y\|<\dfrac{\delta}2+\dfrac{\delta}2=\delta$. Thus, $\forall\delta>0$, $B(y;\delta)\cap A\ne\emptyset$ $\implies$ $y\in\overline A$. Thus $\overline{(\overline A)}\subseteq\overline A$.

Now show that $\overline A$ is closed. Let $y\in X\setminus\overline A$ (where $X=\mathbb R^n$). Then $y\in X\setminus\overline{\overline A}$, i.e $y$ is not a cluster point of $\overline A$ $\implies$ $\exists\delta>0$ such that $B(y;\delta)\cap\overline A=\emptyset$ $\implies$ $B(y;\delta)\subseteq X\setminus\overline A$. Thus $y$ is an interior point of $X\setminus\overline A$. Thus $X\setminus\overline A$ is open, i.e. $\overline A$ is closed.

Now let $C$ be a closed set containing $A$. Then $X\setminus C$ is open, i.e. $\forall c\in X\setminus C$, $\exists\delta>0$ such that $B(c;\delta)\subseteq X\setminus C$, i.e. $B(c;\delta)\cap C=\emptyset$, i.e. $B(c;\delta)\cap A=\emptyset$ as $A\subseteq C$. In other words, no point outside $C$ is a cluster point of $A$, i.e. all the cluster points of $A$ are inside $C$, i.e. $\overline A\subseteq C$. This proves that the closure of $A$ is contained in every closed set containing $A$, i.e. it is the smallest closed set containing $A$.
 
  • #5
Olinguito said:
First show that $\overline{(\overline A)}=\overline A$.

From the definition, it is clear that $A\subseteq\overline A$. Thus (replacing $A$ by $\overline A$) we have $\overline A\subseteq\overline{(\overline A)}$. If $y\in\overline{(\overline A)}$, then $\forall\delta>0$, $B\left(y;\dfrac{\delta}2\right)\cap\overline A\ne\emptyset$. Let $x\in B\left(y;\dfrac{\delta}2\right)\cap\overline A$. Then $x\in\overline A$ $\implies$ $B\left(x;\dfrac{\delta}2\right)\cap A\ne\emptyset$. Let $a\in B\left(x;\dfrac{\delta}2\right)\cap A$; then $\|a-y\|\leq\|a-x\|+\|x-y\|<\dfrac{\delta}2+\dfrac{\delta}2=\delta$. Thus, $\forall\delta>0$, $B(y;\delta)\cap A\ne\emptyset$ $\implies$ $y\in\overline A$. Thus $\overline{(\overline A)}\subseteq\overline A$.

Now show that $\overline A$ is closed. Let $y\in X\setminus\overline A$ (where $X=\mathbb R^n$). Then $y\in X\setminus\overline{\overline A}$, i.e $y$ is not a cluster point of $\overline A$ $\implies$ $\exists\delta>0$ such that $B(y;\delta)\cap\overline A=\emptyset$ $\implies$ $B(y;\delta)\subseteq X\setminus\overline A$. Thus $y$ is an interior point of $X\setminus\overline A$. Thus $X\setminus\overline A$ is open, i.e. $\overline A$ is closed.

Now let $C$ be a closed set containing $A$. Then $X\setminus C$ is open, i.e. $\forall c\in X\setminus C$, $\exists\delta>0$ such that $B(c;\delta)\subseteq X\setminus C$, i.e. $B(c;\delta)\cap C=\emptyset$, i.e. $B(c;\delta)\cap A=\emptyset$ as $A\subseteq C$. In other words, no point outside $C$ is a cluster point of $A$, i.e. all the cluster points of $A$ are inside $C$, i.e. $\overline A\subseteq C$. This proves that the closure of $A$ is contained in every closed set containing $A$, i.e. it is the smallest closed set containing $A$.

Thanks for your generous help Olinguito ... I really appreciate it ...

Peter
 

Related to Open and Closed Sets in R^n .... Duistermaat and Kolk, Lemma 1.2.11 ....

1. What is the definition of an open set in R^n?

An open set in R^n is a set of points that contains all of its interior points. This means that for any point in the set, there exists a small enough radius around that point where all other points within that radius are also contained in the set.

2. How is an open set different from a closed set in R^n?

An open set contains all of its interior points, whereas a closed set contains its boundary points as well. This means that for a closed set, there are no points outside of the set that are arbitrarily close to the boundary points.

3. What is the significance of Duistermaat and Kolk, Lemma 1.2.11 in regards to open and closed sets in R^n?

Duistermaat and Kolk, Lemma 1.2.11 states that every open set in R^n can be written as a countable union of disjoint closed sets. This is significant because it allows for a more precise understanding and characterization of open sets in R^n.

4. Can an open set in R^n also be a closed set?

No, an open set and a closed set in R^n are mutually exclusive. A set cannot contain both its interior points and its boundary points simultaneously.

5. How are open and closed sets in R^n used in mathematical analysis?

Open and closed sets in R^n are fundamental concepts in mathematical analysis and are used to define and study important properties of functions, such as continuity and differentiability. They also play a crucial role in the development of topological spaces and their associated structures.

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