Open and Closed in V .... D&K Proposition 1.2.17 .... ....

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In summary, Duistermaat and Kolk's book "Multidimensional Real Analysis I: Differentiation" discusses continuity in Chapter 1. A specific aspect of the proof of Proposition 1.2.17 is addressed, where the author mentions that if A is closed in V, then A = V \ P with P open in V, but the definition of closed sets in V states that A is closed if V \ A is open in V. This may seem contradictory, but it can be reconciled by taking P = V \ A. This was pointed out by Krylov, who provided helpful clarification on the matter.
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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 Proposition 1.2.17 ... ...

Duistermaat and Kolk's Proposition 1.2.17 and the preceding definition (regarding open and closed sets in a set V) read as follows:https://www.physicsforums.com/attachments/7733
View attachment 7734In the above proof of (ii) we read the following:

" ... ... If A is closed in V then A = V \ P with P open in V ... ... "

... BUT in Definition 1.2.16 we read ...

" ... ... A is said to be closed in V if V \ A is open in V ... ... "But ... these two statements are not the same? How do we reconcile the two statements ... specifically how does the statement in the theorem follow from the definition ...

Hope someone can help ...

Peter
 
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Peter said:
In the above proof of (ii) we read the following:

" ... ... If A is closed in V then A = V \ P with P open in V ... ... "

... BUT in Definition 1.2.16 we read ...

" ... ... A is said to be closed in V if V \ A is open in V ... ... "But ... these two statements are not the same? How do we reconcile the two statements ... specifically how does the statement in the theorem follow from the definition ...

Assume that $A$ is closed in the sense of the definition. Then $V \setminus A$ is open in $V$ and $A = V \setminus (V \setminus A)$ so you can take $P = V \setminus A$.
 
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Krylov said:
Assume that $A$ is closed in the sense of the definition. Then $V \setminus A$ is open in $V$ and $A = V \setminus (V \setminus A)$ so you can take $P = V \setminus A$.
Hmm ... wish I'd seen that ...:( ...

Thanks for the help Krylov ... appreciate it ...

Peter
 

Related to Open and Closed in V .... D&K Proposition 1.2.17 .... ....

1. What is the significance of "Open and Closed" in V&D Proposition 1.2.17?

In V&D Proposition 1.2.17, the term "Open and Closed" refers to the properties of a set in a topological space. A set is considered open if it contains all of its limit points, and closed if it contains all of its boundary points.

2. How is the concept of "Open and Closed" related to topology?

In topology, the properties of "Open and Closed" sets are used to define the basic concepts of open and closed neighborhoods, convergence, continuity, and compactness. These concepts help to describe the structure of a topological space and its subsets.

3. What is D&K Proposition 1.2.17 and why is it important?

D&K Proposition 1.2.17 is a proposition in topology that states that any intersection of closed sets is also a closed set. This proposition is important because it helps to define the properties of closed sets and their relationship to other sets in a topological space.

4. Can you give an example of an "Open and Closed" set?

Yes, the set of all real numbers between 0 and 1, including 0 and 1, is both open and closed. It is open because it contains all of its limit points (0 and 1), and it is closed because it contains all of its boundary points (also 0 and 1).

5. How does the concept of "Open and Closed" relate to the concept of continuity?

In topology, continuity is defined in terms of open sets. A function is considered continuous if the inverse image of any open set is also open. The concept of "Open and Closed" sets helps to define continuity by determining which subsets of a topological space are open and closed.

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