Is an open interval in R really uncountable?

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Discussion Overview

The discussion centers on the cardinality of open intervals in the real numbers, specifically questioning whether the interval (0,1) can be considered uncountable. Participants explore methods of counting elements within this interval and the implications for understanding its cardinality.

Discussion Character

  • Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant proposes a method to count elements in the interval (0,1) by listing decimal expansions, suggesting this defines a bijection with the positive integers.
  • Another participant points out that this counting method only accounts for rational numbers with finite decimal expansions, implying that irrational numbers are not included.
  • A subsequent reply emphasizes that the proposed counting method fails to capture all real numbers, particularly irrational numbers.
  • Participants acknowledge the difficulty many have in grasping the concept of uncountability and express appreciation for the realization of the error in reasoning.

Areas of Agreement / Disagreement

Participants generally agree that the initial counting method is flawed due to its exclusion of irrational numbers, but the broader question of the uncountability of the interval remains a point of contention.

Contextual Notes

The discussion highlights limitations in the proposed counting method, particularly its reliance on finite decimal expansions and the exclusion of irrational numbers, which are essential to the concept of uncountability.

ak416
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Ok i know it should be because it has the same cardinality as R and R is uncountable. But take for example (0,1). Heres a method I would use to count all its elements:

0.1,...,0.9
0.01,...,0.99
0.001,...,0.999
.
.
.
0.(n-1 zeros)1,...,0.(n 9's)
.
.
.

so count starting from the top left and keep going right till you reach the end of the line then count the in the same direction for the next line and so on.
Doesnt this define a bijection from (0,1) to Z+ ? Take any number in (0,1). If it has n digits after its decimal, go down n lines to find its corresponding integer (it will be 9+99+...+(n-1 9's) + the number * 10^n )

Whats wrong with my reasoning?
 
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ok i think i got it. A real number can have an infinite number of decimals!
 
Correct!
At each step, you have only counted up rationals with finite decimal expansions.

Not a single irrational number occurs anywhere for any n.
 
And only a small fraction of the rational numbers!

ak416, a lot of people just can't grasp that. I'm impressed that you were able to realize the error so quickly.
 

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