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.

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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

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