# Existence of a map $f: \mathbb{Z}\rightarrow \mathbb{Q}$

Question is to check which option holds true :

There exist a map $f: \mathbb{Z}\rightarrow \mathbb{Q}$ such that

• is bijective and increasing
• is onto and decreasing
• is bijective and satifies $f(n)\geq 0$ if $n\leq 0$
• has uncountable image.

First of all any subset of $\mathbb{Q}$ is countable so there is no point in looking for last option.

Now, As both $\mathbb{Z}$ and $\mathbb{Q}$ are countable, there could be a possible bijective function..

Now, the first problem is i could not think of a bijection (I am very sure this exist) and second problem is even if i find some function will that old first or third possibilities.

Thank you :)

• Why down vote? please.. – user87543 Dec 9 '13 at 10:10
• The recent questions you have been asking are all very interesting! +1! – Prism Dec 11 '13 at 5:27
• @Prism : Thank you :) – user87543 Dec 11 '13 at 7:45

1) This cannot exist, even if we only suppose that $f$ is onto and increasing. Then there is an $i$ with $f(i)<f(i+1)$. Show that there is no $n\in\mathbb{Z}$ that can map to $\frac{f(i)+f(i+1)}{2}$.
2) $f$ onto and decreasing is the same as $-f$ onto and increasing, so by 1) this can't exist.
3) This exists. Find bijection $g$ from $\mathbb{N}$ to $\mathbb{Q}_{> 0}$ and try to use this in your construction. For negative integers you can define $f(n)=g(-n)$, for positive $f(n)=-g(n)$ and $f(0)=0$.
• for $3$ I guess you mean negative integers to $\mathbb{Q}_{\geq 0}$.. Am i wrong? – user87543 Dec 9 '13 at 10:25
• When I said 'try to use this in your construction' I was going for that. You can easily biject $\mathbb{N}\cup \{0\}$ to $\mathbb{Z}_{\leq 0}$. I edited to make it a bit more clear – MichalisN Dec 9 '13 at 10:35