Micromass' big counterexample challenge

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Discussion Overview

The discussion revolves around the challenge of finding counterexamples to various false statements in basic analysis. Participants are tasked with providing detailed arguments for their counterexamples, exploring concepts related to open sets, continuity, and properties of functions.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose specific constructions of open sets that include rational numbers but exclude certain irrational numbers, such as ##\pi##.
  • One participant suggests a simpler open set construction, ##G=(-\infty , \pi)\cup(\pi,\infty)##, arguing that it satisfies the conditions of the challenge.
  • There are discussions about the clarity and correctness of the definitions used, particularly regarding the construction of ##\epsilon_n## in relation to the rational numbers.
  • Feedback is provided on the difficulty of the challenge, with some participants expressing both confidence and uncertainty about their proposed solutions.

Areas of Agreement / Disagreement

Participants express differing views on the complexity of the counterexamples, with some finding certain statements easier than others. There is no consensus on the best approach or solution to the challenge, and multiple competing views remain.

Contextual Notes

Some participants note potential ambiguities in the definitions and constructions presented, particularly regarding the use of the minimum function and the implications of the chosen open sets.

Who May Find This Useful

Readers interested in mathematical analysis, particularly in the context of counterexamples and the properties of functions and sets, may find this discussion engaging.

  • #61
mfb said:
That is basically what @Samy_A constructed, just with a much shorter argument.
Didn't he remove whole intervals? I am saying:
Take x irrational, then ##\mathbb R ## -{##x##} is your desired open set .

EDIT: It is open as the complement of the closed set {x}, and clearly ##\mathbb Q \subset G \subset \mathbb R ##.
 
Last edited:
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  • #62
WWGD said:
Didn't he remove whole intervals?
If you take the union of all his sets (to get the final set) you should get the same with both approaches.
 
  • #63
micromass said:
I hope some of these statements were surprising to some of you
o yes, especially for those who did not read Bernard R. Gelbaum, John M. H. Olmsted: Counterexamples in Analysis. :)
 

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