 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