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Let $A=\{ 2+\frac{1}{n+1} ,\quad -2+\frac{1}{n+1} \mid\quad n=0,1,\ldots \}$.

Find a bijection from $\mathbb{Z}$ to $A$.

I try to set $f : \mathbb{Z} \to A $ where $f(k)=2+\frac{1}{k+1}$ if $k\geq 0$ and $f(k)=-2+\frac{1}{k+1} $ if $k <0 $. But this does not work.

Can you give some ideas?

Thank you!

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    @Michalis, I had that in mind when I wrote "usually".2012-01-19

2 Answers 2

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Hints:

  1. Can you find a bijection between $\{m\in\mathbb{Z}\mid m\geq 0\}$ and $\{2 + \frac{1}{n+1}\mid n=0,1,2,\ldots\}$?

  2. Can you find a bijection between $\{m\in\mathbb{Z}\mid m\geq 0\}$ and $\{-2+\frac{1}{n+1}\mid 0,1,2,\ldots\}$?

  3. Can you find a bijection between $\{k\in\mathbb{Z}\mid k\lt 0\}$ and $\{m\in\mathbb{Z}\mid m\geq 0\}$?

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    :No problem! Silly question! I do it. Thank you!2012-01-18
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Your idea almost works. Make no change when $k \ge 0$. When $k<0$, send $k$ to $-2+\frac{1}{|k|}$.

Comment: Before we get to formulas, let's think about what the set $A$ looks like. It consists of

$2+\frac{1}{1}$, $2+\frac{1}{2}$, $2+\frac{1}{3}$, and so on, together with

$-2+\frac{1}{1}$, $-2+\frac{1}{2}$, $-2+\frac{1}{3}$, and so on.

It seems reasonable to send $0$ to $2+\frac{1}{1}$, to send $1$ to $2+\frac{1}{2}$, to send $2$ to $2+\frac{1}{3}$, and so on. The definition of $f(k)$ for $k \ge 0$ is then clear.

Then it seems natural to send $-1$ to $-2+\frac{1}{1}$, to send $-2$ to $-2+\frac{1}{2}$, to send $-3$ to $-2+\frac{1}{3}$, and so on. The definition of $f(k)$ for $k < 0$ is then clear. (Instead of writing $2-\frac{1}{|k|}$ we could write $f(k)=-2-\frac{1}{k}$. But I feel more comfortable with positive numbers.)

The fact that $f$ is bijective is intuitively absolutely obvious, and (almost) does not require proof. However, if proof is asked for, it is easy to write down.

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    @AndreNicolas: It is clear!Thank you for your answer!2012-01-18