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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

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By $\mathbb N$, you mean the set of positive integers? – Tunococ Sep 11 '12 at 4:05
up vote 2 down vote accepted

Hint: Can you map both of sets bijectively to $\mathbb{N}$, then compose the maps to give a bijection between the two sets?


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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You can use spoiler tags, by the way, to hide the solution (not that it matters when so many other answers give it right out here). – Asaf Karagila Sep 11 '12 at 4:12
@AsafKaragila Oh cool! I didn't know that. Thanks! – Neal Sep 11 '12 at 4:45

Hint: find bijections of both sets with $\mathbb{N}$.

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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.

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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.

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Given some element $a$ of $\{ 2\alpha -1 \colon \alpha \in \mathbb{N} \}$, try the function $f(a)=\frac{a+1}{2}+9$.

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