Example of a bijection between two sets I am trying to come up with a bijective function $f$ between the set : $\left \{ 2\alpha -1:\alpha \in \mathbb{N} \right \}$ and the set $\left \{ \beta\in \mathbb{N} :\beta\geq 9 \right \}$, but I couldn't figure out how to do it. Can anyone come up with such a bijective function? Thanks
 A: Hint: find bijections of both sets with $\mathbb{N}$.
A: Hint: Can you map both of sets bijectively to $\mathbb{N}$, then compose the maps to give a bijection between the two sets?

Solution: 

Let's call $A = \{2\alpha - 1\ |\ \alpha\in\mathbb{N}\}$ and $B = \{\beta\in\mathbb{N}\ |\ \beta\geq 9\}$.  We can put $A$ in bijection with $\mathbb{N}$ by $f(x) = \frac{1}{2}(x + 1)$.  We can also map $\mathbb{N}$ to $B$ by $g(y) = y + 8$.  Both of these maps are bijections, so their composition is a bijection, $$g\circ f:A\ni x\mapsto \frac{1}{2}(x + 1) + 8\in B.$$

A: I am using $\mathbb{N} = \{0,1, 2, ...\}$. (If your $\mathbb{N}$ starts at $1$, then use subtract $8$ instead of $9$ in the function $f$ below.)
Let $A = \{2k - 1 : k \in \mathbb{N}\}$ and let $B = \{k : k \geq 9\}$. 
Let $f : B \rightarrow A$ be defined by $f(k) = 2(k - 9) - 1$. $f$ is the desired bijection.
A: a map between $\{1,3,5,\cdots\}$ and $\{9,10,11,\cdots\}$ could be done by adding $17$ to everything in the first set, then dividing by two. I.e, $\frac{x+17} 2$ or $2x-17$, depending on which way you're going.
A: Given some element $a$ of $\{ 2\alpha -1 \colon \alpha \in \mathbb{N} \}$, try the function $f(a)=\frac{a+1}{2}+9$.
