INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

August 22, 2017

TIFR 2013 Problem 24 Solution -Non-existence of continuous function


TIFR 2013 Problem 24 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 programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects. In general, TIFR entrance exam hits the floor during the month of December.
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.

Also Visit: College Mathematics Program of Cheenta


Problem:True/False


There exists a continuous surjective function from \(S^1 \) onto \(\mathbb{R}\).

Hint:

Search for topological invariants.


Discussion:


We know that continuous image of a compact set is compact. \(S^1\) is a subset of \(\mathbb{R}^2\), and in \(\mathbb{R}^2\) a set is compact if and only if it is closed and bounded.

By definition, every element of \(S^1\) has unit modulus, so it is bounded.

Let's say \(z_n\to z\) as \(n\to \infty \). Where {\(z_n\)} is a sequence in \(S^1\). Since modulus is a continuous function, \(|z_n| \to |z| \), the sequence {\(|z_n|\)} is simply the constant sequence \(1,1,1,... \) hence \(|z|=1\).

What does above discussion mean? Well it means that if \(z\) is a limit point (or even a point of closure) of \(S^1\) then \(z\in S^1\).  Therefore, \(S^1\) is closed.

The immediate consequence is that the given statement is False. Because, \(\mathbb{R}\) is not compact. \(S^1\) is compact, and continuous image of a compact set has to be compact.


Helpdesk

  • What is this topic: Real Analysis
  • What are some of the associated concept: Compact Set, continuous image of a compact set, continuous function, Limit point,Closed and bounded Sequence
  • Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com