Open Sets in R^n .... Duistermaat and Kolk, Lemma 1.2.5 ....

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In summary, the conversation is about a request for help with a proof of Lemma 1.2.5 from the book "Multidimensional Real Analysis I" by Duistermaat and Kolk. The book's assertion (i) is discussed, which can be proven using Definition 1.2.2 from Chapter 1 on Open and Closed Sets. The user Peter is looking for a fully rigorous proof of assertion (i) and asks for help.
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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 Lemma 1.2.5 ...

Duistermaat and Kolk"s Lemma 1.2.5 reads as follows:View attachment 9014In the above proof by Duistermaat and Kolk we read the following:

" ... ... Assertion (i) follows from Definition 1.2.2 ... ..."I have tried to demonstrate a rigorous proof of Assertion (i) but have not been happy it is fully rigorous ...Can someone please demonstrate a fully rigorous proof of Assertion (i) ...

Help will be appreciated ...

Peter ========================================================================================It may help readers of the above post to have access to the start of Section 1.2: Open and Closed Sets ... which includes Definition 1.2.2 referred to above ... so I am providing aces to that text ... as follows ... View attachment 9015
View attachment 9016
Hope that helps ... ...

Peter

Attachments

• D&K - Lemma 1.2.5 ... .png
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• D&K - 1 - Start of Section 1.2 ... Open and Closed Sets ... .png
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• D&K - 2 - Start of Section 1.2 ... Open and Closed Sets ... ...Part 2.png
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Suppose that $U = \bigcup_{s\in S}U_s$ is a union of open sets $U_s$. If $x\in U$ then by definition $x$ must belong to one of the sets $U_s$. Since $U_s$ is open, there exists $\delta>0$ with $B(x,\delta)\subseteq U_s$. Then $x\in B(x,\delta)\subseteq U_s \subseteq U$. Thus every point $x$ in $U$ is in the interior of $U$. So $U$ is open.

Opalg said:
Suppose that $U = \bigcup_{s\in S}U_s$ is a union of open sets $U_s$. If $x\in U$ then by definition $x$ must belong to one of the sets $U_s$. Since $U_s$ is open, there exists $\delta>0$ with $B(x,\delta)\subseteq U_s$. Then $x\in B(x,\delta)\subseteq U_s \subseteq U$. Thus every point $x$ in $U$ is in the interior of $U$. So $U$ is open.

Thanks for your help, Opalg ...

Peter

1. What are open sets in R^n?

Open sets in R^n are subsets of the n-dimensional Euclidean space, where any point within the set has an open neighborhood contained entirely within the set. In other words, all points within an open set have a certain distance from each other, and there are no boundary points or edges.

2. How are open sets defined in Duistermaat and Kolk, Lemma 1.2.5?

Duistermaat and Kolk, Lemma 1.2.5 defines open sets in R^n as sets that contain all their limit points. In other words, for any sequence of points within the open set, there exists a point in the set that is arbitrarily close to the limit point of the sequence.

3. What is the importance of open sets in R^n?

Open sets in R^n are important in mathematical analysis and topology as they help define the concept of continuity and convergence. They also play a crucial role in the definition of differentiability and integration in higher dimensions.

4. Can open sets in R^n be visualized?

Yes, open sets in R^n can be visualized as they are subsets of the n-dimensional Euclidean space. For example, a circle in the x-y plane is an open set as any point within the circle has an open neighborhood contained within the circle.

5. How do open sets in R^n relate to closed sets?

Open sets and closed sets are complementary concepts in topology. A set in R^n is open if and only if its complement is closed. In other words, a set is closed if it contains all its boundary points and their limit points. This means that a set can be both open and closed, or neither open nor closed.

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