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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## Problem:

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. \([0,1]\)

## Discussion:

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.

## Chatuspathi

**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