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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