# Cardinality of $\omega^2$ [duplicate]

I know $\omega ^ 2$ is countable, but I'm unable to find a bijection from $\omega * \omega \rightarrow \omega$

This should be simple, but I'm very stuck.

## marked as duplicate by Asaf Karagila♦, Seirios, Micah, Chris Eagle, Davide GiraudoMar 1 '13 at 14:29

• Hint: lay out $\omega^2$ in a quarter-infinite grid, as the set of points $\langle x,y\rangle$ with $x,y\in\mathbb{N}$. Can you see how to enumerate all the points in this grid? – Steven Stadnicki Mar 1 '13 at 7:01 If you have to describe this bijection $\varphi:\omega\times\omega\to\omega$ more formally, it’s worth spending some time trying to work out a formula for $\varphi(a,b)$ in terms of $a$ and $b$; all you need is a little ingenuity and the familiar formula for the sum of the first $n$ positive integers, $\sum_{k=1}^nk=\frac12n(n+1)$. If you get stuck, you’ll find much help in this Wikipedia article.
$$\begin{array}? \omega^2 = &\{& 0,1,2,3,\ldots, \\ && \omega,\omega+1,\omega+2,\omega+3,\ldots, \\ && 2\omega,2\omega+1,2\omega+2,2\omega+3,\ldots,\\ && 3\omega,3\omega+1,3\omega+2,3\omega+3,\ldots, \\ && \vdots \\ &\}& \end{array}$$
now can easily see why it is in bijection with $\mathbb N^2$ and hence $\mathbb N$ hence countable.