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

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Closed
Click For Summary
The discussion centers on Proposition 1.2.17 from "Multidimensional Real Analysis I: Differentiation" by Duistermaat and Kolk, specifically regarding the proof of the proposition related to open and closed sets in a set V. A participant seeks assistance in formulating a rigorous proof for part (i) of the proposition, which is stated to be immediate from the definitions. Another participant provides a clarification that if a set A is open in V, it can be expressed as the intersection of V with an open subset U of R^n, confirming that A is open in R^n. The conversation emphasizes the importance of understanding the definitions and results related to open and closed sets in the context of the proof. Overall, the thread highlights the collaborative effort to clarify and prove concepts in real analysis.
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 another 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:
View attachment 7735
View attachment 7736D&K write that the proof of (i) is immediate from the definitions ... but I have been unable to formulate a rigorous proof of (i) ... could someone please demonstrate a rigorous proof of (i) ... ...Hope that someone can help ...

Peter========================================================================================
D&K's definitions and early results on open and closed sets may be helpful to MHB members reading and following the above post ... so I am providing the same ... as follows:View attachment 7737
View attachment 7738Hope that helps ... ...

Peter
 
Last edited:
Physics news on Phys.org
Hi, Peter.

Suppose $A$ is open in $V$. By Definition 1.2.16 there is an open subset, say U, of $\mathbb{R}^{n}$ such that $A=V\cap U.$ Since both $U$ and $V$ are open in $\mathbb{R}^{n}$ and finite intersections of open sets are open, $A$ is open in $\mathbb{R}^{n}.$

If $A$ is open in $\mathbb{R}^{n}$, then it is open in $V$ because $A=V\cap A$ (since $A\subseteq V$).
 
GJA said:
Hi, Peter.

Suppose $A$ is open in $V$. By Definition 1.2.16 there is an open subset, say U, of $\mathbb{R}^{n}$ such that $A=V\cap U.$ Since both $U$ and $V$ are open in $\mathbb{R}^{n}$ and finite intersections of open sets are open, $A$ is open in $\mathbb{R}^{n}.$

If $A$ is open in $\mathbb{R}^{n}$, then it is open in $V$ because $A=V\cap A$ (since $A\subseteq V$).
THanks GJA ... appreciate the help ...

Peter
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
1K
Replies
4
Views
2K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K