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

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Closed
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
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
 
Physics news on Phys.org
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 $P = V \setminus A$.
 
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
 
Back
Top