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TIFR 2013 Problem 34 Solution -Countable or not?

TIFR 2013 Problem 34 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programe leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

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Let \(S\) be the set of all sequences \(\{a_1,a_2,...,a_n,...\} \) where each entry \(a_i \) is either \(0\) or \(1\). Then \(S\) is countable.


What if instead of \(0\) and \(1\) the values were allowed to be any digit from \(0\) to 9? What would that correspond to?


Given a sequence  \(\{a_1,a_2,...,a_n,...\} \) , we can associate it to the binary number \(0.a_1a_2...a_n... \). This association (or function if you like) is one-one, and onto the set of all real numbers having binary expansion in the form of \(0.something\), which is same as the set \((0,1)\), which is uncountable.


One could use Cantor's diagonalization argument as well to argue in this problem. If possible let \(x_1,x_2,...\) is an enumeration of the given set (of sequences) then consider the sequence \(\{b_1,b_2,...\} \) in \(S\) defined by \(b_i \ne x_i  \). We get a contradiction because this is a sequence which is not in the enumeration but is a member of \(S\).


  • What is this topic: Real Analysis
  • What are some of the associated concept: Cantor's diagonalization argument, Sequence, Sequential criterion
  • Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

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