TIFR 2014 Problem 1 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.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let A,B,C be three subsets of \(\mathbb{R}\) The negation of the following statement: For every \(\epsilon >1\), there exists \(a\in A\) and \(b\in B\) such that for all \(c\in C\), \(|a-c|< \epsilon\) and \(|b-c|>\epsilon \) is:

(A) There exists \( \epsilon \leq 1\), such that for all \(a \in A\) and \(b \in B\) there exists \(c \in C\) such that \(|a − c| \geq \epsilon\) or \(|b − c| \leq \epsilon\)

(B) There exists \( \epsilon \leq 1\), such that for all \(a \in A\) and \(b \in B\) there exists \(c \in C\) such that \(|a − c| \geq \epsilon\) and \(|b − c| \leq \epsilon\)

(C) There exists \( \epsilon > 1\), such that for all \(a \in A\) and \(b \in B\) there exists \(c \in C\) such that \(|a − c| \geq \epsilon\) and \(|b − c| \leq \epsilon\)

(D) There exists \( \epsilon > 1\), such that for all \(a \in A\) and \(b \in B\) there exists \(c \in C\) such that \(|a − c| \geq \epsilon\) and \(|b − c| \leq \epsilon\)

## Discussion:

**Before looking at the options, let us understand what rules we have to negate a statement.
(Please note this topic have discussed in Appendix portion of mentioned book in Helpdesk section.)
**

When we negate a for-all statement we get a there-exists statement. For example, the negation of \(\forall x P(x)\) is true is \(\exists x\) for which \(P(x)\) is not true.

A negation of a there-exists statement is for-all. Example: the negation of \(\exists x\) such that \( P(x)\) is true is \(\forall x\) \(P(x)\) is not true.

To negate a slightly complicated such as that given in question we follow the following simple procedure:

- replace the \(\forall\)s with \(\exists\) and replace \(\exists\) by \(\forall\).
- and in the end, negate the conclusion.

Following this procedure we obtain:

\(\exists \epsilon>1\),such that \(\forall a\in A\),and \(b\in B\) \(\exists c\in C\), \(|a-c|\ge \epsilon\) or \(|b-c|\le \epsilon\).

Note that

\(\exists \epsilon>1\),such that \(\forall a\in A\),and \(b\in B\) \(\exists c\in C\) is the part that comes from by 1. And the part \(|a-c|\ge \epsilon\) or \(|b-c|\le \epsilon\) comes from the part 2.

We look at the question paper now, and indeed, this answer is the statement of option **(d)**.

## Helpdesk

**What is this topic:**Mathematical Logic**What are some of the associated concept:**Negation of a Statement**Book Suggestions:**Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert