Proving Disjointness of Open Intervals in E Subset of R

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Discussion Overview

The discussion revolves around proving that an open set E in the real numbers can be expressed as a countable union of disjoint open intervals. Participants explore various approaches to demonstrate this property, referencing concepts from topology and set theory.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant suggests using neighborhoods of rational points to construct a countable union of sets that intersect with E, questioning whether this approach leads to a disjoint union.
  • A later reply acknowledges the initial approach does not yield a disjoint union and proposes filling gaps around points in E with neighborhoods to achieve a countably dense set within E.
  • Another participant reiterates the original problem and introduces the concept of disjoint open sets, referencing Lindelöf's theorem to argue that a countable subset of disjoint open sets can be formed, suggesting that this implies the countability of the original collection.
  • One participant notes that the exercise is a familiar result, indicating that similar discussions or references exist elsewhere.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the initial approaches or the correctness of the proposed methods. Multiple competing views and uncertainties remain regarding the construction of disjoint open intervals from the open set E.

Contextual Notes

Participants express uncertainty about the disjoint nature of the unions formed and the implications of Lindelöf's theorem in this context. The discussion highlights the complexity of constructing the desired representation of E.

mynameisfunk
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Hey guys, doing another rudin-related question. Here Goes:

Show that if E \subseteq \Re is open, then E can be written as an at most countable union of disjoint open intervals, i.e., E=\bigcupn(an,bn). (It's possible that an=-\infty bn=+\infty for some n.)

My attempt:
Take the set of all Neighborhoods of all of the rationals of a rational radius in R to be A. Now all members of E intersect A make up E. Take the union of all of the neighborhoods in this set E intersect A and this is a countable union of disjoint sets.

Is there a problem with this?
 
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I now realize this doesn't create a disjoint union. I had the idea of taking an arbitrarily large neighborhood of a point in E that is still a proper subset of E and then filling up the gaps on either side with more neighborhoods, always still contained in E, until I have a countably dense set of neighborhoods contained in E...

Would this do?
 
mynameisfunk said:
Hey guys, doing another rudin-related question. Here Goes:

Show that if E \subseteq \Re is open, then E can be written as an at most countable union of disjoint open intervals, i.e., E=\bigcupn(an,bn). (It's possible that an=-\infty bn=+\infty for some n.)

My attempt:
Take the set of all Neighborhoods of all of the rationals of a rational radius in R to be A. Now all members of E intersect A make up E. Take the union of all of the neighborhoods in this set E intersect A and this is a countable union of disjoint sets.

Is there a problem with this?

Here is my suggestion:
First let E be a union of disjoint open sets F. Then E is an open set. We know that by Lindelof there is a countable subset F' of F. However, since F is disjoint, F=F'. So F is countable.

Accurately, F is a collection of sets and E is the union of sets taken from F.

I guess that your first suggestion proves the lindelof that countable subets exist. Although they are "not disjoint", you have proven a lindelof and you can apply this property to the open set which is a union "disjoint" open sets.
 
This is a familiar exercise/result. See e.g. here, here, and your own (!) thread here.
 

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