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There exists a map $f: \Bbb Z\rightarrow \Bbb Q $ such that $f$ is

A. Bijective and increasing

B. Onto and decreasing

C. Bijective and satisfies $f(n)\ge 0$ if $n\le 0$

D. Has uncountable images

Now option D. is absurd . Option C. is given to be the correct answer.I was thinking since both sets are countable bijection is obvious. Now why cannot be increasing ? I could map $0$ to $0$ and the negative integers to the negative rationals and positive integers to the positive rationals. And if increasing would be possible just interchanging signs would give the decreasing map. So none is possible but why?

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    $\begingroup$ A: If $f(1)<f(2)$, then there are infinitely many rational numbers between them. $\endgroup$ – Sungjin Kim Aug 4 '15 at 21:35
  • $\begingroup$ I don't know . I posted it now for the first time . Why deleted? $\endgroup$ – user118494 Aug 4 '15 at 21:37
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The problem is that, while the cardinality of $\mathbb{Z}$ and $\mathbb{Q}$ is the same, the topology is different. Consider the following: Given two integers $a$ and $b$ with $a < b$, can we say how many integers $c$ satisfy $a < c < b$? Now ask the same question for the rational numbers.

This leads to problems. For example, lets say $a$ and $b$ are two such integers with $k$ other integers between them. And we can even assume the $f(a) = r_{1} < r_{2} = f(b)$. But now we can find an increasing sequence of$k+1$ rational numbers (at least) between $r_{1}$ and $r_{2}$, and only $k$ potential integers to use as their preimages.

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  • $\begingroup$ ok . But why is C. valid? Just like the positive rationals are images of negative rationals and nothing contradicts , that is how? $\endgroup$ – user118494 Aug 5 '15 at 21:30
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    $\begingroup$ You don't need any preservation of order or anything. You can map positive integers to positive rationals, then just flip the sign on everything. $\endgroup$ – Morgan Rodgers Aug 6 '15 at 5:19
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For A, some $f(n)$ lies between $f(0)$ and $f(1)$, which requires the integer $n$ to lie between $0$ and $1$.For B, let $M$ be the least positive $n$ for which $f(n)<f(0)$. Then $f(M)<f(m)=(f(M)+f(0))/2<f(0)$ for some integer $m$, so $M>m>0$. But by the definition of $M$, if any integer $m$ satisfies $M>m>0$ then $f(m)\geq f(0)$.

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