How Is a Closed Set of Rational Numbers Defined?

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Discussion Overview

The discussion revolves around the definition of a closed set in the context of rational numbers, specifically focusing on a closed set A that contains every rational number in the interval [0,1]. Participants explore the implications of this definition and the properties of closed sets in relation to rational and irrational numbers.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • One participant questions how A can be closed if it does not contain all points in R^n-A, citing the example of sqrt(0.5) which has no open sphere disjoint from A.
  • Another participant asserts that [0,1] is a closed set containing all rational numbers between 0 and 1.
  • A different participant argues that while [0,1] is closed, A cannot be closed because it does not include irrational numbers in that interval.
  • One participant points out that the problem does not specify A uniquely, indicating that there are many closed sets containing all rational numbers in [0,1].
  • Another participant clarifies that the title of the thread may reflect a misunderstanding, emphasizing that A must contain irrational numbers as well if it includes [0,1].
  • It is noted that for A to be closed, it must include all accumulation points of sequences in A, and that irrational numbers in the unit interval are accumulation points of sequences of rational numbers.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the set A and its relationship to rational and irrational numbers. There is no consensus on whether A can be considered closed under the given conditions, and the discussion remains unresolved.

Contextual Notes

Participants highlight the importance of understanding the definitions of closed sets and accumulation points, as well as the implications of including rational versus irrational numbers in the set A.

Andy_ToK
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Hi,
here is the question, if A is a closed set that contains every rational number r: [0,1], show that [0,1] is a subset of A.

But, how could A be closed? If A is closed, R^n-A is open, so any point in R^n-A would have a open sphere around it and this open sphere wouldn't intersect A. apparently, this is not true. eg. sqrt(0.5) has no open sphere around it that is disjoint from A.
 
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[0, 1] is closed set containing every rational number between 0 and 1, is it not? Ponder that.
 
[0,1] is but A isn't, i think. because A doesn't contain those irrational numbers between 0 and 1.
 
Why not? The problem you stated doesn't assert otherwise.

Incidentally, A is not uniquely specified -- there are lots of sets that have the property of both being closed and of containing every rational number in [0, 1], and the hypotheses is merely that A is one such set.
 
I think your title shows a misunderstanding. You titled this "closed set of rational numbers", which implies it contains only rational numbers, but the question is about a closed set that contains rational numbers- it doesn't say only rational numbers and clearly that cannot be true. If A is to contain [0,1] then clearly it contains irrational numbers as well. The point of the exercise is to show that any closed set that contains all rational numbers in [0,1] must also contain all irrational numbers in [0,1].
 
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Note that in order for A to be closed (by premise!), A must include all accumulation points for sequences in A.

In particular, it means that A must contain all accumulation points for all sequences whose terms are rational numbers in the unit interval.

What you then need to show is that any irrational number within the unit interval is an accumulation point for at least one such sequence of rational numbers in the unit interval.

Remember that the rationals are dense in the reals..
 
Thank you all.
 

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