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

[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 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

 

[HallsofIvy]

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"?

 

[Math Amateur]

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

 

[Olinguito]

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 $A$.

 

[Math Amateur]

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