I have the sets $\mathbb{N}$ and $\mathbb{N}-\{0\}$.

Clearly, $\mathbb{N}-\{0\}$ is a proper subset of $\mathbb{N}$, yet they have the same cardinality. That means that there exists a bijective map between them.

I have an idea on how each elements should be sent, I would send the $0\in \mathbb{N}$ to the $1\in \mathbb{N}-\{0\}$, and so on... So

$0\rightarrow 1$


$2\rightarrow 3$


But I am not too sure how to present the map in general, I thought about $f(x)=x+1$, but it doesn't look right!

  • $\begingroup$ Looks good to me. $\endgroup$ – André Nicolas Aug 25 '14 at 6:00
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    $\begingroup$ $f(x)=x+1$ is fine :). $\endgroup$ – Kim Jong Un Aug 25 '14 at 6:00
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    $\begingroup$ Sometimes the idea of a map is good enough. For example bijections $\Bbb N\to\Bbb N\times\Bbb N$ are usually hard to describe by formulas, but easy enough by pictures/words. $\endgroup$ – Quang Hoang Aug 25 '14 at 6:10

You need to show that (using Pano axioms or whatever you have available)

  • $f$ defines a map $\mathbb N\to\mathbb N-\{0\}$, that is: $x\in \mathbb N$ implies $x+1\in\mathbb N-\{0\}$

  • $f$ is injective, that is: If $x+1=y+1$, then $x=y$

  • $f$ is onto, that is: If $y\in\mathbb N-\{0\}$, then there exists $x\in\mathbb N$ with $y=x+1$.

Each of these steps should be easy.


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