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
Example of a bijection between two sets
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0By $\mathbb N$, you mean the set of positive integers? – 2012-09-11
5 Answers
Given some element $a$ of $\{ 2\alpha -1 \colon \alpha \in \mathbb{N} \}$, try the function $f(a)=\frac{a+1}{2}+9$.
Hint: find bijections of both sets with $\mathbb{N}$.
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, $g\circ f:A\ni x\mapsto \frac{1}{2}(x + 1) + 8\in B.$
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0[Now you know, and knowing is half the battle! *singing: G.I. Joeeeee*](http://www.youtube.com/watch?v=pele5vptVgc) – 2012-09-11
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 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.