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September 12, 2017

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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Problem:True/False?


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.


Hint:


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


Discussion:


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.


Remark:


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 \(S\).


Helpdesk

  • 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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