 TIFR 2013 Problem 29 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 programe leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
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## Problem:True/False?

Let $$f$$ be a function on the closed interval $$[0,1]$$ defined by

$$f(x)=x$$ if $$x$$ is rational and $$f(x)=x^2$$ if $$x$$ is irrational.

Then $$f$$ is continuous at 0 and 1.

## Hint:

Use the sequence criterion for continuity.

## Discussion:

Let $$x_n$$ be a sequence converging to $$0$$. Then $$x_n^2$$ also converges to $$0$$. Since the value of $$f$$ at $$x_n$$ is either of the two, $$f(x_n)\to 0$$ as $$n\to \infty$$. If this seem confusing, think in terms of $$\epsilon$$. For $$\epsilon >0$$, there exists $$n_1$$ and $$n_2$$ such that $$|x_n|< \epsilon$$ and $$|x_n^2|< \epsilon$$ for $$n>n_1$$ and $$n>n_2$$ respectively. Taking the maximum of $$n_1$$ and $$n_2$$ we get the N for which the condition in “epsilon definition” is satisfied.

The same argument applies for the continuity at 1.

The function is not continuous at any other point. Because if a rational sequence and an irrational sequence converge to $$x_0$$ and the function is continuous at that point then by sequential criterion, $$f(x_0)=x_0$$ due to the rational sequence and also $$f(x_0)=x_0^2$$ due to the irrational sequence. Therefore, $$x_0^2=x_0$$ and hence the only possible points of continuity is $$0$$ or $$1$$.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Continuous function, Closed Interval, Sequential criterion
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert