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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?

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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

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.

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Thank you, very much. :-) – Pedro Feb 12 '13 at 11:04

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