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.

Please just do not give an answer but please give some hint and give some time to think about.

Thank you :)

  • $\begingroup$ Why down vote? please.. $\endgroup$ – user87543 Dec 9 '13 at 10:10
  • $\begingroup$ The recent questions you have been asking are all very interesting! +1! $\endgroup$ – Prism Dec 11 '13 at 5:27
  • $\begingroup$ @Prism : Thank you :) $\endgroup$ – 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 $f(0)=0$.

  • $\begingroup$ Thank you.. I had some similar idea but not this concrete.. I will work on that... Thank you :) $\endgroup$ – user87543 Dec 9 '13 at 10:24
  • $\begingroup$ for $3$ I guess you mean negative integers to $\mathbb{Q}_{\geq 0}$.. Am i wrong? $\endgroup$ – user87543 Dec 9 '13 at 10:25
  • $\begingroup$ 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 $\endgroup$ – MichalisN Dec 9 '13 at 10:35
  • $\begingroup$ Thank you Michalis for this answer. $\endgroup$ – Dutta Dec 9 '13 at 10:40
  • $\begingroup$ @Michalis : Now this makes sense to me.. :) Thank you :) $\endgroup$ – user87543 Dec 9 '13 at 11:35

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