 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.

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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:

1. replace the $$\forall$$s with $$\exists$$ and replace $$\exists$$ by $$\forall$$.
2. 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