 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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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