Cheenta
How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

TIFR 2014 Problem 12 Solution - Mapping Properties


TIFR 2014 Problem 12 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:


There exists a map (f:\mathbb{Z} \to \mathbb{Q}) such that (f)

A. is bijective and increasing

B. is onto and decreasing

C. is bijective and satisfies (f(n)\ge 0) if (n\le 0)

D. has uncountable image.


Discussion:


Option D is out of the question right away. Because the co-domain (\mathbb{Q}) is countable, and the image set is a subset of the co-domain set, any subset of countable set is countable, so image set has to be countable.

Let's assume (f) is onto and decreasing. Then (...\ge f(-1) \ge f(0) \ge f(1) \ge f(2) \ge ... ).

At this point, we don't know for sure whether (f(0)=f(1)) or not. But, for the map to be onto, we must have strict inequality somewhere in the above chain of inequalities.

Let's say (f(a)>f(a+1)). Then we ask what is the pre-image of the rational number (\frac{f(a)+f(a+1)}{2})?

Let (f(n)=\frac{f(a)+f(a+1)}{2})

Since, (f(n)=\frac{f(a)+f(a+1)}{2}>f(a)) and (f) is decreasing, (a>n).

Also, since (f(n)=\frac{f(a)+f(a+1)}{2}<f(a+1)) and (f) is decreasing, (n>a+1).

The two inequalities (a>n) and (n>a+1) together give a contradiction.

Therefore, option B is false.

What did we use here, only onto-ness and that f is decreasing. If instead our function (f) was bijective and increasing then we will get a similar kind of contradiction:

Suppose (f) is bijective and increasing. Then (f(0)<f(1)) (One-to-one ness guarantees the strict inequality)

We have (\frac{f(0)+f(1)}{2}\in\mathbb{Q}).

Therefore there exists (m\in\mathbb{Z}) such that ( f(m)= \frac{f(0)+f(1)}{2} ).

We then have (f(0)<f(m)<f(1)) which means (0<m<1), a contradiction because there is no natural number in between 0 and 1.

So option A is false.

We know that (\mathbb{Q}) does have a bijection with (\mathbb{Z}), in other words (\mathbb{Q}) is countable.

Now, both the set (A={n\in \mathbb{Z}| n\le 0 }) and (B={q\in \mathbb{Q}| q\ge 0}) are countably infinite, therefore has a bijection between them. To see this, let (f_1:A\to \mathbb{Z}) and (f_2:B\to \mathbb{Z}) be bijections. Then (f_2^{-1}o f_1:A \to B) is a bijection.

Similarly (C={n\in \mathbb{Z}| n> 0 }) has a bijection with (D={q\in \mathbb{Q}| q< 0}).

Now, we have (h:A\to B) and (g:C\to D) two bijections.

Then define (f(n)=h(n)) for (n\in A) and (f(n)=g(n)) for (n\in C) to get a bijection from (\mathbb{Z}) to (\mathbb{Q}). This is true because (A \cup B= \mathbb{Z}) and (C \cup D= \mathbb{Q}).

And this (f) satisfies the condition of option C: (f) is bijective and satisfies (f(n)\ge 0) if (n\le 0)

Therefore, option C is true.


Chatuspathi

  • What is this topic: Real Analysis
  • What are some of the associated concept: Bijective Mapping
  • Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com