TIFR 2014 Problem 1 Solution - Negation

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