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

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SUMMARY

The discussion centers on the concept of connectedness and disconnectedness as defined in "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk. The definition states that a set \( A \) is disconnected if it can be represented as the union of two open sets \( U \) and \( V \) in \( \mathbb{R}^n \) such that \( (A \cap U) \cup (A \cap V) = A \). A key point clarified is that the boundary of the open sets \( U \) and \( V \) does not need to be included in \( A \), exemplified by the set \( A = \mathbb{R} \setminus \{0\} \), which is indeed disconnected.

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  • Basic comprehension of real analysis, particularly in relation to \( \mathbb{R}^n \).
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  • Study the properties of open and closed sets in topology.
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Mathematics students, particularly those studying real analysis and topology, as well as educators seeking to clarify concepts of connectedness and disconnectedness in their teaching.

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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 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
 
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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.
 

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