MHB Definition of Connectedness .... What's wrong with my informal thinking ....?

Click For Summary
The discussion centers on the definition of connectedness and disconnectedness as presented in "Multidimensional Real Analysis I: Differentiation" by Duistermaat and Kolk. The original poster seeks clarification on how open sets can form a disconnected set without including their boundaries. A key point made in the responses is that the boundary does not need to be part of the disconnected set, exemplified by the set A = ℝ \ {0}, which is a disjoint union of open sets. This highlights that disconnected sets can exist without their boundaries being included. The conversation emphasizes the importance of understanding the properties of open sets in relation to connectedness.
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 D&K's definition of disconnectedness/connectedness ... ...

Duistermaat and Kolk's definition of disconnectedness/connectedness reads as follows:https://www.physicsforums.com/attachments/7774I tried to imagine the typical case of a disconnected set $$A$$ ... there would be two open sets $$U, V \text{ in } \mathbb{R}^n$$ ... indeed both sets could be contained in $$A$$ and we would require, among other things that

$$( A \cap U) \cup (A \cap V) = A$$

This, I imagine, would give a situation like the figure below:
View attachment 7775
BUT ... since $$U, V$$ are open they cannot contain their boundary ... but if they do not contain their boundary ... I am assuming a common boundary ... how can we ever have $$( A \cap U) \cup (A \cap V) = A$$ ...Can someone explain where my informal thinking is going wrong ... since there obviously must exist disconnected sets ...Help will be appreciated ...

Peter
 
Physics news on Phys.org
Hi Peter,

There is no reason why the boundary would be in $A$; that is the whole point of disconnected sets.

As an example, take $A=\mathbb{R}\setminus\{0\}$. This is the disjoint union of two open sets.
 

Thread 'About the definition of topological manifold using closed sets'

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.
View full post »

Similar threads

MHB Open and Closed in V .... D&K Proposition 1.2.17 .... ....
  • · Replies 2 ·
Replies
2
Views
1K
MHB Open and Closed Sets in R^n .... Duistermaat and Kolk, Lemma 1.2.11 ....
  • · Replies 4 ·
Replies
4
Views
2K
MHB Open Sets in R^n .... Duistermaat and Kolk, Lemma 1.2.5 ....
  • · Replies 2 ·
Replies
2
Views
1K
MHB Another question on Open and Closed in V .... D&K Proposition 1.2.17 .... ....
  • · Replies 2 ·
Replies
2
Views
1K
Undergrad Tough lemma on locally finite refinement
  • · Replies 8 ·
Replies
8
Views
2K
MHB The Heine-Borel Theorem in R^n .... .... D&K Theorem 1.8.17 .... ....
  • · Replies 11 ·
Replies
11
Views
3K
MHB Neccesity and sufficiency .... D&K Lemma 1.3.3 ....
  • · Replies 1 ·
Replies
1
Views
1K
MHB Help Needed with Determining B' in Proposition 6.1.2: A Case for n=2
  • · Replies 4 ·
Replies
4
Views
2K
MHB The Closure of a Set is Closed .... Lemma 1.2.10
  • · Replies 2 ·
Replies
2
Views
2K
MHB Continuous Functions and Open Sets .... D&K Example 1.3.8 ....
  • · Replies 2 ·
Replies
2
Views
2K