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# 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.

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## 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

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