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 âˆ’ c| \geq \epsilon) or (|b âˆ’ 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 âˆ’ c| \geq \epsilon) and (|b âˆ’ 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 âˆ’ c| \geq \epsilon) and (|b âˆ’ 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 âˆ’ c| \geq \epsilon) and (|b âˆ’ 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

Google