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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.

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

*Hint:*

Search for topological invariants.

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.

**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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