Proving Uncountability of (0,1): A Puzzling Challenge

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The discussion centers on the challenge of proving the uncountability of the interval (0,1). Participants explore various approaches, including the construction of sequences of rational numbers that approximate real numbers within the interval. However, they conclude that this method fails as (0,1) is uncountable. The conversation highlights the inadequacy of the problem's presentation and suggests that the statement being proved may be fundamentally flawed.

PREREQUISITES
  • Understanding of real analysis concepts, specifically the properties of intervals.
  • Familiarity with the density of rational numbers in real numbers.
  • Knowledge of sequences and convergence in mathematical analysis.
  • Basic comprehension of countability and uncountability in set theory.
NEXT STEPS
  • Study Cantor's diagonal argument to understand the proof of uncountability.
  • Explore the concept of dense subsets in real analysis.
  • Investigate the implications of fixed epsilon in sequences approaching limits.
  • Review the definitions and properties of countable versus uncountable sets.
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Mathematics students, educators, and anyone interested in set theory and real analysis, particularly those tackling problems related to countability and the properties of real intervals.

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Homework Statement


The problem is attached as a picture.


Homework Equations


...


The Attempt at a Solution


I have been trying a lot to prove this without any really fruitful approach. At first I thought that the statement was false, or that you could at least construct a sequence of rationals such that V=(0,1) the following way:
Let a1 be a real in (0,1). Since the rationals are dense there exists a rational number b1 such that d(a1,b1)<ε. Let this be the first rational number in the sequence. Now let a2 be another real. Because the rationals are dense there exists a rational b2 such that d(a2,b2)<ε/2 etc. etc. and by successive use of this method I could generate the whole (0,1) with my approach. But this fails because (0,1) is not countable. So I'm open for any other approach to this problem.
 

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aaaa202 said:

Homework Statement


The problem is attached as a picture.


Homework Equations


...


The Attempt at a Solution


I have been trying a lot to prove this without any really fruitful approach. At first I thought that the statement was false, or that you could at least construct a sequence of rationals such that V=(0,1) the following way:
Let a1 be a real in (0,1). Since the rationals are dense there exists a rational number b1 such that d(a1,b1)<ε. Let this be the first rational number in the sequence. Now let a2 be another real. Because the rationals are dense there exists a rational b2 such that d(a2,b2)<ε/2 etc. etc. and by successive use of this method I could generate the whole (0,1) with my approach. But this fails because (0,1) is not countable. So I'm open for any other approach to this problem.

Can you always construct a sequence from the enumeration that converges to an arbitrary ##x \in (0,1)##?

Also, showing it's a proper subset isn't too bad. Can you think of an element in ##(0,1)## that isn't in ##U##?
 
The statement that you are trying to prove is false (or, rather, not necessarily true) even after fixing it so that epsilon is fixed and the last line refers to V rather than U.

Edit: The problem, as stated, is very poorly posed. Could you maybe copy it verbatim from the source?
 

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