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TIFR 2015 Problem 2 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.

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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Let (f: \mathbb{R} \to \mathbb{R} ) be a continuous function. Which of the following can not be the image of ((0,1]) under (f)?

A. {0}

B. ((0,1))

C. ([0,1))

D. ([0,1])

If f is the constant function constantly mapping to 0, which is continuous, then the image set is {0}.

Suppose that (f((0,1])=(0,1)) . Then (f((0,1))=(0,1)- {f(1)} ). Now since (f(1)\in (0,1) ) the set ( (0,1)- {f(1)} ) is not connected. But ((0,1)) is connected, and we know that continuous image of a connected set is connected. This gives a contradiction. So ((0,1)) can not be the image of ((0,1]) under f.

Define (f(x)=1-x). Then (f((0,1])= [0,1)).

Define (f(x)=0) for (x\in [0,\frac{1}{2}] ) and (f(x)= 2(x-\frac{1}{2}) ) for (x\in [\frac{1}{2} ,1 ] ). (f) is continuous on ((0,\frac{1}{2}] ) and ( [\frac{1}{2} ,1 ] ) and (f) agrees on the common points, by pasting lemma (f) is continuous on ( [0,1] ) . And image of ((0,1] ) is ([0,1]).

TIFR 2015 Problem 2 Solution is concluded.

**What is this topic:**Real Analysis**What are some of the associated concept:**Continuous Function, Metric Space**Book Suggestions:**Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

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