Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sorry for my poor English:

In my last question, I ask for a the proof of "Are the set of all finite subsets in $\mathbb Z$ countable?" . I had a good answer that show me that it is an $f\colon\mathbb N\to\{\text{finite subsets of }\mathbb N\}$. So knowing that exists a bijection $\mathbb{N\leftrightarrow Z}$, then is proved.

But I have curiosity about an example (if exists) of a function $f\colon\mathbb Z\to\{\text{finite subsets of }\mathbb Z\}$ exists this example?

share|improve this question
Since $\mathbb{N}$ is equipotent to $\mathbb{Z}$ these two questions are the same. –  Alex Youcis Feb 12 '13 at 9:51
I guess you are looking for an explicit function that is nice to write down? –  Michael Greinecker Feb 12 '13 at 9:54
Do you want it to be bijective, injective, surjective? –  Joe Tait Feb 12 '13 at 9:55
Joe Tait: bijective –  Pedro Feb 12 '13 at 11:02

1 Answer 1

There is definitely an abundance of functions from $\mathbb Z$ to its finite subsets, e.g., the map $z \mapsto \{ z \}$. If you are looking for a bijective function, things are more complicated.

Here's an explicit example, though: Let $F(X)$ be the set of finite subets of $X$, and let $b: \mathbb N \to F(\mathbb N)$ be any bijection with $b(0) = \varnothing$. Define $f(x) := b(x)$ for $x \ge 0$ and $f(x) := \{ -y | y \in b(-x) \}$ for $x<0$.

It is easy to see that $f|_{\mathbb N}$ and $f|_{-\mathbb N}$ are injective, $f(\mathbb N) = F(\mathbb N)$ and $f(-\mathbb N) = F(-\mathbb N)$. Thus, to show that $f$ is bijective, it is sufficient to show that there is at most one $x$ such that $f(x) = \varnothing$. But since $b$ is injective and $b(0) = \varnothing$, there is no $x >0$ such that $f(x) = \varnothing$, and by definiton of $f$, $f(x) \neq \varnothing$ for all $x<0$ as well.

share|improve this answer
Thank you, very much. :-) –  Pedro Feb 12 '13 at 11:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.