Is (0,1) Uncountable If and Only If \Re is Uncountable?

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Discussion Overview

The discussion centers around the relationship between the uncountability of the interval (0,1) and the uncountability of the real numbers \Re. Participants explore implications and methods to demonstrate these properties, engaging with concepts of bijections and subsets.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a proof showing that (0,1) is uncountable and seeks to establish implications regarding \Re.
  • Another participant suggests constructing a one-to-one function from (0,1) to the integers, arguing that if (0,1) is uncountable, then \Re must also be uncountable due to the subset relationship.
  • A different participant proposes setting up a bijection between (0,1) and \Re, mentioning the tangent function as a potential tool.
  • One participant humorously comments on the term "proberty," suggesting it could be coined to describe a property of probabilistic outcomes.

Areas of Agreement / Disagreement

Participants express various methods and ideas regarding the uncountability of (0,1) and \Re, but there is no consensus on a definitive approach or conclusion. Multiple competing views and techniques remain present in the discussion.

Contextual Notes

Some arguments depend on the definitions of uncountability and the properties of subsets, which may not be fully resolved in the discussion. The mathematical steps to establish bijections or implications are not detailed.

rbzima
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Show that (0,1) is uncountable if and only if [tex]\Re[/tex] is uncountable.

I have a nice little proof showing (0,1) is uncountable, however I'm wondering how I can make implications that [tex]\Re[/tex] and vice versa.
 
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Well, i think that you might want to see if you can construct a 1-1 function (correspondence) with the naturals( positive integers).
f:(0,1)-->Z (integers) . Well, you might also use the property that if a set A is uncountable, and further if this set A is a subset of B, then also B is uncountable.
So basically if you manage to show that (0,1) is uncountable, then automatically you have shown that R is uncountable, since even if we managed to put all other elements of R in an order and count them, we defenitely could not count the elements of R that are within the interval (0,1).
 
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You can setup a bijection between (0,1) and R - try playing around with tan(x).
 
sutupidmath said:
f:(0,1)-->Z (integers). Well, you might also use the proberty that if a set A is uncountable, and further if this set A is a subset of B, then also B is uncountable.

I basically just wanted to say that I honestly think you should coin the term proberty because it makes this feel like a property of probabilistic outcomes!
 
rbzima said:
I basically just wanted to say that I honestly think you should coin the term proberty because it makes this feel like a property of probabilistic outcomes!

Well, if you really like, you can start using it from now on, i will not suit u for plagiarism!:cool:
 

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