# Equality of cardinality of $\mathbb{N}$ and $\mathbb{N} - \{0\}$

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

$0\rightarrow 1$

$1\rightarrow2$

$2\rightarrow 3$

...

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!

• Looks good to me. – André Nicolas Aug 25 '14 at 6:00
• $f(x)=x+1$ is fine :). – Kim Jong Un Aug 25 '14 at 6:00
• 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. – Quang Hoang Aug 25 '14 at 6:10

• $f$ 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 $y=x+1$.